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Generalization of $sum_i=1^n i = fracn(n+1)2$
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During the course of another math.se question
(What's the probability of completing the illustrated "binomial walk" without ever visiting a node above the baseline?),
the following generalization of $sum_i=1^n i = fracn(n+1)2$ tangentially (not directly related to that discussion) came up. Moreover, it also satisfies several "magic formulas" shown below (but not posted earlier) that suggested to me maybe it's already some well-known "thing" (i.e., "somebody's polynomial" or something). If so, anybody know what it's called, and can point me to a more complete development? Here's the definition, and then those formulas that follow from it...
Define
$$
left.
beginarraycclcl
H^1_n & = & & & 1 mbox for all $n$ \
H^2_n & = & sum_i=1^n H^1_i & = & n \
H^3_n & = & sum_i=1^n H^2_i & = & fracn(n+1)2! mbox (the usual)\
H^4_n & = & sum_i=1^n H^3_i & stackrel?= &
underbracean^3+bn^2+cn_
mboxsolve for $a,b,c$... \
& & & = & frac16n^3 + frac12n^2 + frac13n \
& & & = & fracn(n+1)(n+2)3! \
endarrayright}
beginarrayr
i.e., H^m_n stackrelmboxdef=
sum_i=1^n H^m-1_i, mboxwith H^1_n=1.\
mboxAnd since H^m_n-1=sum_i=1^n-1 H^m-1_i,\
mboxnote that H^m_n=H^m_n-1+H^m-1_n mbox too.
endarray
$$
And by examination we infer the general expression
$beginarrayccl
H^m_n & = & frac1(m-1)! prod_k=1^m-1 (n+k-1) \
& = & frac(n+m-2)!(m-1)!(n-1)! \
& & mboxwhich simplifies to \
& = & binomn+m-2m-1 = binomn+m-2n-1 \
endarray$
And also note that $H^m_n=H^n_m$.
The two "magic formulas" that I numerically tested beyond any reasonable doubt (but haven't been able to derive) are,
$$H^m+2_m - sum_k=1^m-1 kH^m+1-k_m = m mbox, for any m$$
and
$$frac1mH^m+2_m = frac1msum_k=1^m kH^m+1-k_m
= frac1m-1sum_k=1^m-1 k(k+1)H^m-k_m-1$$
But note that the more general progression suggested by the second formula
fails.
  ** Edit **
-------------------
Re @mrtaurho's remark about verification, I numerically generated the table illustrated below, and had the computer compare those directly-calculated values against the binomial coefficient expression. Everything's copacetic (and the computer checked many, many more values than illustrated on the table)...
$beginarrayccccccccccccc
n&H^1_n&H^2_n&H^3_n&H^4_n&H^5_n&H^6_n&H^7_n&H^8_n&
H^9_n&H^10_n&H^11_n&H^12_n \
1&bf 1&1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \
2&1&bf 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10& 11& 12 \
3&1&3 &bf 6& 10& 15& 21& 28& 36& 45& 55& 66& 78 \
4&1&4 & 10&bf 20& 35& 56& 84&120&165&220&286&364 \
5&1&5 & 15& 35&bf 70&126&210&330&495&715&1001&1365 \
6&1&6 & 21& 56&126&bf 252&462&792&1287&2002&3003&4368 \
7&1&7 & 28& 84&210&462&bf 924&1716&3003&5005&8008&12376 \
8&1&8 & 36&120&330&792&1716&bf 3432&6435&11440&19448&31824 \
9&1&9 & 45&165&495&1287&3003&6435&bf 12870&24310&43758&75582 \
10&1&10& 55&220&715&2002&5005&11440&24310&bf 48620&92378&167960 \
11&1&11& 66&286&1001&3003&8008&19448&43758&92378&bf 184756&352716 \
12&1&12& 78&364&1365&4368&12376&31824&75582&167960&352716&bf 705432\
13&1&13& 91&455&1820&6188&18564&50388&125970&293930&646646&1352078\
14&1&14&105&560&2380&8568&27132&77520&203490&497420&1144066&2496144\
15&1&15&120&680&3060&11628&38760&116280&319770&817190&1961256&4457400\
16&1&16&136&816&3876&15504&54264&170544&490314&1307504&3268760&7726160\
endarray$
number-theory
asked Jul 25 at 10:16
John Forkosh
304110
add a comment |Â
up vote
2
down vote
favorite
During the course of another math.se question
(What's the probability of completing the illustrated "binomial walk" without ever visiting a node above the baseline?),
the following generalization of $sum_i=1^n i = fracn(n+1)2$ tangentially (not directly related to that discussion) came up. Moreover, it also satisfies several "magic formulas" shown below (but not posted earlier) that suggested to me maybe it's already some well-known "thing" (i.e., "somebody's polynomial" or something). If so, anybody know what it's called, and can point me to a more complete development? Here's the definition, and then those formulas that follow from it...
Define
$$
left.
beginarraycclcl
H^1_n & = & & & 1 mbox for all $n$ \
H^2_n & = & sum_i=1^n H^1_i & = & n \
H^3_n & = & sum_i=1^n H^2_i & = & fracn(n+1)2! mbox (the usual)\
H^4_n & = & sum_i=1^n H^3_i & stackrel?= &
underbracean^3+bn^2+cn_
mboxsolve for $a,b,c$... \
& & & = & frac16n^3 + frac12n^2 + frac13n \
& & & = & fracn(n+1)(n+2)3! \
endarrayright}
beginarrayr
i.e., H^m_n stackrelmboxdef=
sum_i=1^n H^m-1_i, mboxwith H^1_n=1.\
mboxAnd since H^m_n-1=sum_i=1^n-1 H^m-1_i,\
mboxnote that H^m_n=H^m_n-1+H^m-1_n mbox too.
endarray
$$
And by examination we infer the general expression
$beginarrayccl
H^m_n & = & frac1(m-1)! prod_k=1^m-1 (n+k-1) \
& = & frac(n+m-2)!(m-1)!(n-1)! \
& & mboxwhich simplifies to \
& = & binomn+m-2m-1 = binomn+m-2n-1 \
endarray$
And also note that $H^m_n=H^n_m$.
The two "magic formulas" that I numerically tested beyond any reasonable doubt (but haven't been able to derive) are,
$$H^m+2_m - sum_k=1^m-1 kH^m+1-k_m = m mbox, for any m$$
and
$$frac1mH^m+2_m = frac1msum_k=1^m kH^m+1-k_m
= frac1m-1sum_k=1^m-1 k(k+1)H^m-k_m-1$$
But note that the more general progression suggested by the second formula
fails.
  ** Edit **
-------------------
Re @mrtaurho's remark about verification, I numerically generated the table illustrated below, and had the computer compare those directly-calculated values against the binomial coefficient expression. Everything's copacetic (and the computer checked many, many more values than illustrated on the table)...
$beginarrayccccccccccccc
n&H^1_n&H^2_n&H^3_n&H^4_n&H^5_n&H^6_n&H^7_n&H^8_n&
H^9_n&H^10_n&H^11_n&H^12_n \
1&bf 1&1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \
2&1&bf 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10& 11& 12 \
3&1&3 &bf 6& 10& 15& 21& 28& 36& 45& 55& 66& 78 \
4&1&4 & 10&bf 20& 35& 56& 84&120&165&220&286&364 \
5&1&5 & 15& 35&bf 70&126&210&330&495&715&1001&1365 \
6&1&6 & 21& 56&126&bf 252&462&792&1287&2002&3003&4368 \
7&1&7 & 28& 84&210&462&bf 924&1716&3003&5005&8008&12376 \
8&1&8 & 36&120&330&792&1716&bf 3432&6435&11440&19448&31824 \
9&1&9 & 45&165&495&1287&3003&6435&bf 12870&24310&43758&75582 \
10&1&10& 55&220&715&2002&5005&11440&24310&bf 48620&92378&167960 \
11&1&11& 66&286&1001&3003&8008&19448&43758&92378&bf 184756&352716 \
12&1&12& 78&364&1365&4368&12376&31824&75582&167960&352716&bf 705432\
13&1&13& 91&455&1820&6188&18564&50388&125970&293930&646646&1352078\
14&1&14&105&560&2380&8568&27132&77520&203490&497420&1144066&2496144\
15&1&15&120&680&3060&11628&38760&116280&319770&817190&1961256&4457400\
16&1&16&136&816&3876&15504&54264&170544&490314&1307504&3268760&7726160\
endarray$
number-theory
asked Jul 25 at 10:16
John Forkosh
304110
Did you tried to use Faulhaber's formula (en.wikipedia.org/wiki/Faulhaber%27s_formula) to verify your general expression?
– mrtaurho
Jul 25 at 10:34
@mrtaurho No, I wasn't aware of Faulhaber's formula, but looking at that wikipedia page, I'm not seeing how it applies. As far as I can tell, $sum_k=1^nk^p$ doesn't seem related. How do you think it's applicable?
– John Forkosh
Jul 25 at 11:04
Oh, I have mistaken one formula. I am sorry. After I saw your expressions for $H_n^3$ and $H_n^4$ I did not realized that the one factor in you product for $H_n^4$ is $(n+2)$ and not as I assumed $(2n+1)$.
– mrtaurho
Jul 25 at 11:10
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
During the course of another math.se question
(What's the probability of completing the illustrated "binomial walk" without ever visiting a node above the baseline?),
the following generalization of $sum_i=1^n i = fracn(n+1)2$ tangentially (not directly related to that discussion) came up. Moreover, it also satisfies several "magic formulas" shown below (but not posted earlier) that suggested to me maybe it's already some well-known "thing" (i.e., "somebody's polynomial" or something). If so, anybody know what it's called, and can point me to a more complete development? Here's the definition, and then those formulas that follow from it...
Define
$$
left.
beginarraycclcl
H^1_n & = & & & 1 mbox for all $n$ \
H^2_n & = & sum_i=1^n H^1_i & = & n \
H^3_n & = & sum_i=1^n H^2_i & = & fracn(n+1)2! mbox (the usual)\
H^4_n & = & sum_i=1^n H^3_i & stackrel?= &
underbracean^3+bn^2+cn_
mboxsolve for $a,b,c$... \
& & & = & frac16n^3 + frac12n^2 + frac13n \
& & & = & fracn(n+1)(n+2)3! \
endarrayright}
beginarrayr
i.e., H^m_n stackrelmboxdef=
sum_i=1^n H^m-1_i, mboxwith H^1_n=1.\
mboxAnd since H^m_n-1=sum_i=1^n-1 H^m-1_i,\
mboxnote that H^m_n=H^m_n-1+H^m-1_n mbox too.
endarray
$$
And by examination we infer the general expression
$beginarrayccl
H^m_n & = & frac1(m-1)! prod_k=1^m-1 (n+k-1) \
& = & frac(n+m-2)!(m-1)!(n-1)! \
& & mboxwhich simplifies to \
& = & binomn+m-2m-1 = binomn+m-2n-1 \
endarray$
And also note that $H^m_n=H^n_m$.
The two "magic formulas" that I numerically tested beyond any reasonable doubt (but haven't been able to derive) are,
$$H^m+2_m - sum_k=1^m-1 kH^m+1-k_m = m mbox, for any m$$
and
$$frac1mH^m+2_m = frac1msum_k=1^m kH^m+1-k_m
= frac1m-1sum_k=1^m-1 k(k+1)H^m-k_m-1$$
But note that the more general progression suggested by the second formula
fails.
  ** Edit **
-------------------
Re @mrtaurho's remark about verification, I numerically generated the table illustrated below, and had the computer compare those directly-calculated values against the binomial coefficient expression. Everything's copacetic (and the computer checked many, many more values than illustrated on the table)...
$beginarrayccccccccccccc
n&H^1_n&H^2_n&H^3_n&H^4_n&H^5_n&H^6_n&H^7_n&H^8_n&
H^9_n&H^10_n&H^11_n&H^12_n \
1&bf 1&1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \
2&1&bf 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10& 11& 12 \
3&1&3 &bf 6& 10& 15& 21& 28& 36& 45& 55& 66& 78 \
4&1&4 & 10&bf 20& 35& 56& 84&120&165&220&286&364 \
5&1&5 & 15& 35&bf 70&126&210&330&495&715&1001&1365 \
6&1&6 & 21& 56&126&bf 252&462&792&1287&2002&3003&4368 \
7&1&7 & 28& 84&210&462&bf 924&1716&3003&5005&8008&12376 \
8&1&8 & 36&120&330&792&1716&bf 3432&6435&11440&19448&31824 \
9&1&9 & 45&165&495&1287&3003&6435&bf 12870&24310&43758&75582 \
10&1&10& 55&220&715&2002&5005&11440&24310&bf 48620&92378&167960 \
11&1&11& 66&286&1001&3003&8008&19448&43758&92378&bf 184756&352716 \
12&1&12& 78&364&1365&4368&12376&31824&75582&167960&352716&bf 705432\
13&1&13& 91&455&1820&6188&18564&50388&125970&293930&646646&1352078\
14&1&14&105&560&2380&8568&27132&77520&203490&497420&1144066&2496144\
15&1&15&120&680&3060&11628&38760&116280&319770&817190&1961256&4457400\
16&1&16&136&816&3876&15504&54264&170544&490314&1307504&3268760&7726160\
endarray$
number-theory
asked Jul 25 at 10:16
John Forkosh
304110
During the course of another math.se question
(What's the probability of completing the illustrated "binomial walk" without ever visiting a node above the baseline?),
the following generalization of $sum_i=1^n i = fracn(n+1)2$ tangentially (not directly related to that discussion) came up. Moreover, it also satisfies several "magic formulas" shown below (but not posted earlier) that suggested to me maybe it's already some well-known "thing" (i.e., "somebody's polynomial" or something). If so, anybody know what it's called, and can point me to a more complete development? Here's the definition, and then those formulas that follow from it...
Define
$$
left.
beginarraycclcl
H^1_n & = & & & 1 mbox for all $n$ \
H^2_n & = & sum_i=1^n H^1_i & = & n \
H^3_n & = & sum_i=1^n H^2_i & = & fracn(n+1)2! mbox (the usual)\
H^4_n & = & sum_i=1^n H^3_i & stackrel?= &
underbracean^3+bn^2+cn_
mboxsolve for $a,b,c$... \
& & & = & frac16n^3 + frac12n^2 + frac13n \
& & & = & fracn(n+1)(n+2)3! \
endarrayright}
beginarrayr
i.e., H^m_n stackrelmboxdef=
sum_i=1^n H^m-1_i, mboxwith H^1_n=1.\
mboxAnd since H^m_n-1=sum_i=1^n-1 H^m-1_i,\
mboxnote that H^m_n=H^m_n-1+H^m-1_n mbox too.
endarray
$$
And by examination we infer the general expression
$beginarrayccl
H^m_n & = & frac1(m-1)! prod_k=1^m-1 (n+k-1) \
& = & frac(n+m-2)!(m-1)!(n-1)! \
& & mboxwhich simplifies to \
& = & binomn+m-2m-1 = binomn+m-2n-1 \
endarray$
And also note that $H^m_n=H^n_m$.
The two "magic formulas" that I numerically tested beyond any reasonable doubt (but haven't been able to derive) are,
$$H^m+2_m - sum_k=1^m-1 kH^m+1-k_m = m mbox, for any m$$
and
$$frac1mH^m+2_m = frac1msum_k=1^m kH^m+1-k_m
= frac1m-1sum_k=1^m-1 k(k+1)H^m-k_m-1$$
But note that the more general progression suggested by the second formula
fails.
  ** Edit **
-------------------
Re @mrtaurho's remark about verification, I numerically generated the table illustrated below, and had the computer compare those directly-calculated values against the binomial coefficient expression. Everything's copacetic (and the computer checked many, many more values than illustrated on the table)...
$beginarrayccccccccccccc
n&H^1_n&H^2_n&H^3_n&H^4_n&H^5_n&H^6_n&H^7_n&H^8_n&
H^9_n&H^10_n&H^11_n&H^12_n \
1&bf 1&1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \
2&1&bf 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10& 11& 12 \
3&1&3 &bf 6& 10& 15& 21& 28& 36& 45& 55& 66& 78 \
4&1&4 & 10&bf 20& 35& 56& 84&120&165&220&286&364 \
5&1&5 & 15& 35&bf 70&126&210&330&495&715&1001&1365 \
6&1&6 & 21& 56&126&bf 252&462&792&1287&2002&3003&4368 \
7&1&7 & 28& 84&210&462&bf 924&1716&3003&5005&8008&12376 \
8&1&8 & 36&120&330&792&1716&bf 3432&6435&11440&19448&31824 \
9&1&9 & 45&165&495&1287&3003&6435&bf 12870&24310&43758&75582 \
10&1&10& 55&220&715&2002&5005&11440&24310&bf 48620&92378&167960 \
11&1&11& 66&286&1001&3003&8008&19448&43758&92378&bf 184756&352716 \
12&1&12& 78&364&1365&4368&12376&31824&75582&167960&352716&bf 705432\
13&1&13& 91&455&1820&6188&18564&50388&125970&293930&646646&1352078\
14&1&14&105&560&2380&8568&27132&77520&203490&497420&1144066&2496144\
15&1&15&120&680&3060&11628&38760&116280&319770&817190&1961256&4457400\
16&1&16&136&816&3876&15504&54264&170544&490314&1307504&3268760&7726160\
endarray$
number-theory
asked Jul 25 at 10:16
John Forkosh
304110
edited Jul 25 at 14:27
asked Jul 25 at 10:16
John Forkosh
304110
asked Jul 25 at 10:16
John Forkosh
304110
asked Jul 25 at 10:16
John Forkosh
304110
304110
Did you tried to use Faulhaber's formula (en.wikipedia.org/wiki/Faulhaber%27s_formula) to verify your general expression?
– mrtaurho
Jul 25 at 10:34
@mrtaurho No, I wasn't aware of Faulhaber's formula, but looking at that wikipedia page, I'm not seeing how it applies. As far as I can tell, $sum_k=1^nk^p$ doesn't seem related. How do you think it's applicable?
– John Forkosh
Jul 25 at 11:04
Oh, I have mistaken one formula. I am sorry. After I saw your expressions for $H_n^3$ and $H_n^4$ I did not realized that the one factor in you product for $H_n^4$ is $(n+2)$ and not as I assumed $(2n+1)$.
– mrtaurho
Jul 25 at 11:10
add a comment |Â
Did you tried to use Faulhaber's formula (en.wikipedia.org/wiki/Faulhaber%27s_formula) to verify your general expression?
– mrtaurho
Jul 25 at 10:34
@mrtaurho No, I wasn't aware of Faulhaber's formula, but looking at that wikipedia page, I'm not seeing how it applies. As far as I can tell, $sum_k=1^nk^p$ doesn't seem related. How do you think it's applicable?
– John Forkosh
Jul 25 at 11:04
Oh, I have mistaken one formula. I am sorry. After I saw your expressions for $H_n^3$ and $H_n^4$ I did not realized that the one factor in you product for $H_n^4$ is $(n+2)$ and not as I assumed $(2n+1)$.
– mrtaurho
Jul 25 at 11:10
Did you tried to use Faulhaber's formula (en.wikipedia.org/wiki/Faulhaber%27s_formula) to verify your general expression?
– mrtaurho
Jul 25 at 10:34
Did you tried to use Faulhaber's formula (en.wikipedia.org/wiki/Faulhaber%27s_formula) to verify your general expression?
– mrtaurho
Jul 25 at 10:34
@mrtaurho No, I wasn't aware of Faulhaber's formula, but looking at that wikipedia page, I'm not seeing how it applies. As far as I can tell, $sum_k=1^nk^p$ doesn't seem related. How do you think it's applicable?
– John Forkosh
Jul 25 at 11:04
@mrtaurho No, I wasn't aware of Faulhaber's formula, but looking at that wikipedia page, I'm not seeing how it applies. As far as I can tell, $sum_k=1^nk^p$ doesn't seem related. How do you think it's applicable?
– John Forkosh
Jul 25 at 11:04
Oh, I have mistaken one formula. I am sorry. After I saw your expressions for $H_n^3$ and $H_n^4$ I did not realized that the one factor in you product for $H_n^4$ is $(n+2)$ and not as I assumed $(2n+1)$.
– mrtaurho
Jul 25 at 11:10
Oh, I have mistaken one formula. I am sorry. After I saw your expressions for $H_n^3$ and $H_n^4$ I did not realized that the one factor in you product for $H_n^4$ is $(n+2)$ and not as I assumed $(2n+1)$.
– mrtaurho
Jul 25 at 11:10
add a comment |Â
2 Answers
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This is a well-known fact, called the Hockey stick identity.
https://en.wikipedia.org/wiki/Hockey-stick_identity
answered Jul 25 at 14:15
Yves Daoust
111k665203
1
Thanks, Yves. That indeed seems closely related, and I wasn't aware of it. But note that their combinatorial proof construction $binomn+k-1k-1$ is "one off", so to speak, from $H^m_n=binomn+m-2m-1$. So, e.g., that means the $H^m_n=H^n_m$ symmetry is lost (as illustrated in the "stars and bars" link from that page). And maybe more fundamentally, these binomial coefficients are merely the "answer". It's that generating-function-like approach/definition of $H^m_n$ which gives rise to that answer which would be the underlying interesting "thing". At least that's what I've been thinking.
– John Forkosh
Jul 25 at 14:55
@JohnForkosh: I understand your frustration.
– Yves Daoust
Jul 25 at 14:58
add a comment |Â
up vote
1
down vote
I really tried to play around with your two "magic formulas" but I did not found anything useful. Nevertheless I can prove your explicit formula for $H^m_n$ using a proof I did on a different post (which you can find here : Help on induction, couldn't make both side the same value). The relation there is
$$sum_k=1^n k(k+1)~cdots (k+p-1)~=~ fracn(n+1)cdots (n+p)p+1$$
which I will use later on. First of all we write $H_n^m+1$ in terms of $H_n^m$ with the equation you defined
$$H^m+1_n~=~sum_k=1^n H_k^m $$
Now using your derived indentity
$$H_n^m~=~binomn+m-2m-1~~textand respectively~~H_n^m+1~=~binomn+m-1m$$
we obtain
$$beginalign
H^m+1_n~&=~sum_k=1^n H_k^m\
binomn+m-1m~&=~sum_k=1^n binomk+m-2m-1\
frac(n+m-1)!(n-1)!~m!~&=~sum_k=1^n frac(k+m-2)!(k-1)!~(m-1)!\
frac(n+m-1)!(n-1)!~m~&=~sum_k=1^n frac(k+m-2)!(k-1)!\
fracn(n+1)cdots (n+m-1)m~&=~sum_k=1^n k(k+1)~cdots (k+m-2)
endalign$$
which equals my given equation by setting $m=p+1$.
Maybe you can go further from there on but atleast you got a proof as a verification for your solution by examination.
My try on the first "magic formula"
$$beginalign
H_m^m+2-sum_k=1^m-1 k H_m^m-k+1~&=~m\
H_m^m+2-sum_k=1^m-1 k H_m^m-k+1+mH_m^1-mH_m^1~&=~m\
H_m^m+2-sum_k=1^m k H_m^m-k+1+mH_m^1~&=~m\
H_m^m+2-sum_k=1^m k H_m^m-k+1+mcdot (1)~&=~m\
H_m^m+2~&=~sum_k=1^m k H_m^m-k+1\
sum_k=1^m H_k^m+1~&=~sum_k=1^m k H_m^m-k+1
endalign$$
so therefore I guess it should be somehow possible to show
$$H_k^m+1~=~k H_m^m-k+1$$
But till now it leads me to nowhere
answered Jul 25 at 12:12
mrtaurho
740219
Thanks @mrtaurho. Actually I'd algebraically derived the binomial coefficient $H^m_n=binomn+m-2n-1$, but just didn't bother displaying the algebra. And I apparently didn't make it clear that it had already been derived. The numerical tests at the bottom were merely a double-check confirmation. It's only the "magic formulas" I couldn't derive. As for "usefulness", that's from my earlier question math.stackexchange.com/questions/2860403 where the $H^m_n$'s appear in an iterative definition of something else. And that other thing simplified nicely only if those magic formulas hold.
– John Forkosh
Jul 25 at 13:07
Ah, okay. I guess from now one you can be quite sure that this binomial coefficient is right. I tried some different ways, only concerning the first formula, involving some kind of induction and by taking advantage of the definitions but it always ended in some very strange binomial coefficients together with weird sums. For me the main problem which occurs is the extra $k$-factor infront of the $H_m^m+1-k$. I just looked over your first question and I am not sure how exactly you derived it there. Could you maybe add it to this question since you are asking here for these "magic formulas"?
– mrtaurho
Jul 25 at 13:17
Sorry, "..how exactly you derived it there"? What "it" are we referring to? If you're referring to that original product form for $H^m_n$, then I mis-spoke in the preceding comment -- I solved three equations in three unknowns to get the general form for $H^4_n$, and then inferred $H^m>4_n$ without algebraic proof (but with lots of numerical verification). And from there, the simplification from products to factorials to binomial coefficients was pretty straightforward.
– John Forkosh
Jul 25 at 13:45
I am sorry. My comment had ran out of characters. I was speaking about your first "magic formula". I do not exactly understand what you have done there, in your first post, to get this special relation. I would appreciate if you could add the process you have gone through to end up with this formula to this post. Or, if you do not mind, just explain it to me further than in your original question. Hence I am quite fascinated by this formula aswell I will work on it a while. Maybe I can find something useful, who knows.
– mrtaurho
Jul 25 at 13:52
Oh, right. I mentioned that above -- for both formulas. I >>did<< absolutely nothing. Rather, it was that other question math.stackexchange.com/questions/2860403 where I noticed that I >>needed<< those formulas to be true (or else the stuff that I felt should simplify wouldn't simplify). So I >>assumed<< them true, and just did numerous numerical tests (for more values than you can shake a stick at). Then I indeed tried to prove them, but got absolutely nowhere (and it wasn't "nowhere fast", but "nowhere very slowly":)
– John Forkosh
Jul 25 at 14:01
 |Â
show 3 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
This is a well-known fact, called the Hockey stick identity.
https://en.wikipedia.org/wiki/Hockey-stick_identity
answered Jul 25 at 14:15
Yves Daoust
111k665203
1
Thanks, Yves. That indeed seems closely related, and I wasn't aware of it. But note that their combinatorial proof construction $binomn+k-1k-1$ is "one off", so to speak, from $H^m_n=binomn+m-2m-1$. So, e.g., that means the $H^m_n=H^n_m$ symmetry is lost (as illustrated in the "stars and bars" link from that page). And maybe more fundamentally, these binomial coefficients are merely the "answer". It's that generating-function-like approach/definition of $H^m_n$ which gives rise to that answer which would be the underlying interesting "thing". At least that's what I've been thinking.
– John Forkosh
Jul 25 at 14:55
@JohnForkosh: I understand your frustration.
– Yves Daoust
Jul 25 at 14:58
add a comment |Â
up vote
2
down vote
This is a well-known fact, called the Hockey stick identity.
https://en.wikipedia.org/wiki/Hockey-stick_identity
answered Jul 25 at 14:15
Yves Daoust
111k665203
1
Thanks, Yves. That indeed seems closely related, and I wasn't aware of it. But note that their combinatorial proof construction $binomn+k-1k-1$ is "one off", so to speak, from $H^m_n=binomn+m-2m-1$. So, e.g., that means the $H^m_n=H^n_m$ symmetry is lost (as illustrated in the "stars and bars" link from that page). And maybe more fundamentally, these binomial coefficients are merely the "answer". It's that generating-function-like approach/definition of $H^m_n$ which gives rise to that answer which would be the underlying interesting "thing". At least that's what I've been thinking.
– John Forkosh
Jul 25 at 14:55
@JohnForkosh: I understand your frustration.
– Yves Daoust
Jul 25 at 14:58
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This is a well-known fact, called the Hockey stick identity.
https://en.wikipedia.org/wiki/Hockey-stick_identity
answered Jul 25 at 14:15
Yves Daoust
111k665203
This is a well-known fact, called the Hockey stick identity.
https://en.wikipedia.org/wiki/Hockey-stick_identity
answered Jul 25 at 14:15
Yves Daoust
111k665203
answered Jul 25 at 14:15
Yves Daoust
111k665203
answered Jul 25 at 14:15
Yves Daoust
111k665203
answered Jul 25 at 14:15
Yves Daoust
111k665203
111k665203
1
Thanks, Yves. That indeed seems closely related, and I wasn't aware of it. But note that their combinatorial proof construction $binomn+k-1k-1$ is "one off", so to speak, from $H^m_n=binomn+m-2m-1$. So, e.g., that means the $H^m_n=H^n_m$ symmetry is lost (as illustrated in the "stars and bars" link from that page). And maybe more fundamentally, these binomial coefficients are merely the "answer". It's that generating-function-like approach/definition of $H^m_n$ which gives rise to that answer which would be the underlying interesting "thing". At least that's what I've been thinking.
– John Forkosh
Jul 25 at 14:55
@JohnForkosh: I understand your frustration.
– Yves Daoust
Jul 25 at 14:58
add a comment |Â
1
Thanks, Yves. That indeed seems closely related, and I wasn't aware of it. But note that their combinatorial proof construction $binomn+k-1k-1$ is "one off", so to speak, from $H^m_n=binomn+m-2m-1$. So, e.g., that means the $H^m_n=H^n_m$ symmetry is lost (as illustrated in the "stars and bars" link from that page). And maybe more fundamentally, these binomial coefficients are merely the "answer". It's that generating-function-like approach/definition of $H^m_n$ which gives rise to that answer which would be the underlying interesting "thing". At least that's what I've been thinking.
– John Forkosh
Jul 25 at 14:55
@JohnForkosh: I understand your frustration.
– Yves Daoust
Jul 25 at 14:58
1
1
Thanks, Yves. That indeed seems closely related, and I wasn't aware of it. But note that their combinatorial proof construction $binomn+k-1k-1$ is "one off", so to speak, from $H^m_n=binomn+m-2m-1$. So, e.g., that means the $H^m_n=H^n_m$ symmetry is lost (as illustrated in the "stars and bars" link from that page). And maybe more fundamentally, these binomial coefficients are merely the "answer". It's that generating-function-like approach/definition of $H^m_n$ which gives rise to that answer which would be the underlying interesting "thing". At least that's what I've been thinking.
– John Forkosh
Jul 25 at 14:55
Thanks, Yves. That indeed seems closely related, and I wasn't aware of it. But note that their combinatorial proof construction $binomn+k-1k-1$ is "one off", so to speak, from $H^m_n=binomn+m-2m-1$. So, e.g., that means the $H^m_n=H^n_m$ symmetry is lost (as illustrated in the "stars and bars" link from that page). And maybe more fundamentally, these binomial coefficients are merely the "answer". It's that generating-function-like approach/definition of $H^m_n$ which gives rise to that answer which would be the underlying interesting "thing". At least that's what I've been thinking.
– John Forkosh
Jul 25 at 14:55
@JohnForkosh: I understand your frustration.
– Yves Daoust
Jul 25 at 14:58
@JohnForkosh: I understand your frustration.
– Yves Daoust
Jul 25 at 14:58
add a comment |Â
up vote
1
down vote
I really tried to play around with your two "magic formulas" but I did not found anything useful. Nevertheless I can prove your explicit formula for $H^m_n$ using a proof I did on a different post (which you can find here : Help on induction, couldn't make both side the same value). The relation there is
$$sum_k=1^n k(k+1)~cdots (k+p-1)~=~ fracn(n+1)cdots (n+p)p+1$$
which I will use later on. First of all we write $H_n^m+1$ in terms of $H_n^m$ with the equation you defined
$$H^m+1_n~=~sum_k=1^n H_k^m $$
Now using your derived indentity
$$H_n^m~=~binomn+m-2m-1~~textand respectively~~H_n^m+1~=~binomn+m-1m$$
we obtain
$$beginalign
H^m+1_n~&=~sum_k=1^n H_k^m\
binomn+m-1m~&=~sum_k=1^n binomk+m-2m-1\
frac(n+m-1)!(n-1)!~m!~&=~sum_k=1^n frac(k+m-2)!(k-1)!~(m-1)!\
frac(n+m-1)!(n-1)!~m~&=~sum_k=1^n frac(k+m-2)!(k-1)!\
fracn(n+1)cdots (n+m-1)m~&=~sum_k=1^n k(k+1)~cdots (k+m-2)
endalign$$
which equals my given equation by setting $m=p+1$.
Maybe you can go further from there on but atleast you got a proof as a verification for your solution by examination.
My try on the first "magic formula"
$$beginalign
H_m^m+2-sum_k=1^m-1 k H_m^m-k+1~&=~m\
H_m^m+2-sum_k=1^m-1 k H_m^m-k+1+mH_m^1-mH_m^1~&=~m\
H_m^m+2-sum_k=1^m k H_m^m-k+1+mH_m^1~&=~m\
H_m^m+2-sum_k=1^m k H_m^m-k+1+mcdot (1)~&=~m\
H_m^m+2~&=~sum_k=1^m k H_m^m-k+1\
sum_k=1^m H_k^m+1~&=~sum_k=1^m k H_m^m-k+1
endalign$$
so therefore I guess it should be somehow possible to show
$$H_k^m+1~=~k H_m^m-k+1$$
But till now it leads me to nowhere
answered Jul 25 at 12:12
mrtaurho
740219
Thanks @mrtaurho. Actually I'd algebraically derived the binomial coefficient $H^m_n=binomn+m-2n-1$, but just didn't bother displaying the algebra. And I apparently didn't make it clear that it had already been derived. The numerical tests at the bottom were merely a double-check confirmation. It's only the "magic formulas" I couldn't derive. As for "usefulness", that's from my earlier question math.stackexchange.com/questions/2860403 where the $H^m_n$'s appear in an iterative definition of something else. And that other thing simplified nicely only if those magic formulas hold.
– John Forkosh
Jul 25 at 13:07
Ah, okay. I guess from now one you can be quite sure that this binomial coefficient is right. I tried some different ways, only concerning the first formula, involving some kind of induction and by taking advantage of the definitions but it always ended in some very strange binomial coefficients together with weird sums. For me the main problem which occurs is the extra $k$-factor infront of the $H_m^m+1-k$. I just looked over your first question and I am not sure how exactly you derived it there. Could you maybe add it to this question since you are asking here for these "magic formulas"?
– mrtaurho
Jul 25 at 13:17
Sorry, "..how exactly you derived it there"? What "it" are we referring to? If you're referring to that original product form for $H^m_n$, then I mis-spoke in the preceding comment -- I solved three equations in three unknowns to get the general form for $H^4_n$, and then inferred $H^m>4_n$ without algebraic proof (but with lots of numerical verification). And from there, the simplification from products to factorials to binomial coefficients was pretty straightforward.
– John Forkosh
Jul 25 at 13:45
I am sorry. My comment had ran out of characters. I was speaking about your first "magic formula". I do not exactly understand what you have done there, in your first post, to get this special relation. I would appreciate if you could add the process you have gone through to end up with this formula to this post. Or, if you do not mind, just explain it to me further than in your original question. Hence I am quite fascinated by this formula aswell I will work on it a while. Maybe I can find something useful, who knows.
– mrtaurho
Jul 25 at 13:52
Oh, right. I mentioned that above -- for both formulas. I >>did<< absolutely nothing. Rather, it was that other question math.stackexchange.com/questions/2860403 where I noticed that I >>needed<< those formulas to be true (or else the stuff that I felt should simplify wouldn't simplify). So I >>assumed<< them true, and just did numerous numerical tests (for more values than you can shake a stick at). Then I indeed tried to prove them, but got absolutely nowhere (and it wasn't "nowhere fast", but "nowhere very slowly":)
– John Forkosh
Jul 25 at 14:01
 |Â
show 3 more comments
up vote
1
down vote
I really tried to play around with your two "magic formulas" but I did not found anything useful. Nevertheless I can prove your explicit formula for $H^m_n$ using a proof I did on a different post (which you can find here : Help on induction, couldn't make both side the same value). The relation there is
$$sum_k=1^n k(k+1)~cdots (k+p-1)~=~ fracn(n+1)cdots (n+p)p+1$$
which I will use later on. First of all we write $H_n^m+1$ in terms of $H_n^m$ with the equation you defined
$$H^m+1_n~=~sum_k=1^n H_k^m $$
Now using your derived indentity
$$H_n^m~=~binomn+m-2m-1~~textand respectively~~H_n^m+1~=~binomn+m-1m$$
we obtain
$$beginalign
H^m+1_n~&=~sum_k=1^n H_k^m\
binomn+m-1m~&=~sum_k=1^n binomk+m-2m-1\
frac(n+m-1)!(n-1)!~m!~&=~sum_k=1^n frac(k+m-2)!(k-1)!~(m-1)!\
frac(n+m-1)!(n-1)!~m~&=~sum_k=1^n frac(k+m-2)!(k-1)!\
fracn(n+1)cdots (n+m-1)m~&=~sum_k=1^n k(k+1)~cdots (k+m-2)
endalign$$
which equals my given equation by setting $m=p+1$.
Maybe you can go further from there on but atleast you got a proof as a verification for your solution by examination.
My try on the first "magic formula"
$$beginalign
H_m^m+2-sum_k=1^m-1 k H_m^m-k+1~&=~m\
H_m^m+2-sum_k=1^m-1 k H_m^m-k+1+mH_m^1-mH_m^1~&=~m\
H_m^m+2-sum_k=1^m k H_m^m-k+1+mH_m^1~&=~m\
H_m^m+2-sum_k=1^m k H_m^m-k+1+mcdot (1)~&=~m\
H_m^m+2~&=~sum_k=1^m k H_m^m-k+1\
sum_k=1^m H_k^m+1~&=~sum_k=1^m k H_m^m-k+1
endalign$$
so therefore I guess it should be somehow possible to show
$$H_k^m+1~=~k H_m^m-k+1$$
But till now it leads me to nowhere
answered Jul 25 at 12:12
mrtaurho
740219
Thanks @mrtaurho. Actually I'd algebraically derived the binomial coefficient $H^m_n=binomn+m-2n-1$, but just didn't bother displaying the algebra. And I apparently didn't make it clear that it had already been derived. The numerical tests at the bottom were merely a double-check confirmation. It's only the "magic formulas" I couldn't derive. As for "usefulness", that's from my earlier question math.stackexchange.com/questions/2860403 where the $H^m_n$'s appear in an iterative definition of something else. And that other thing simplified nicely only if those magic formulas hold.
– John Forkosh
Jul 25 at 13:07
Ah, okay. I guess from now one you can be quite sure that this binomial coefficient is right. I tried some different ways, only concerning the first formula, involving some kind of induction and by taking advantage of the definitions but it always ended in some very strange binomial coefficients together with weird sums. For me the main problem which occurs is the extra $k$-factor infront of the $H_m^m+1-k$. I just looked over your first question and I am not sure how exactly you derived it there. Could you maybe add it to this question since you are asking here for these "magic formulas"?
– mrtaurho
Jul 25 at 13:17
Sorry, "..how exactly you derived it there"? What "it" are we referring to? If you're referring to that original product form for $H^m_n$, then I mis-spoke in the preceding comment -- I solved three equations in three unknowns to get the general form for $H^4_n$, and then inferred $H^m>4_n$ without algebraic proof (but with lots of numerical verification). And from there, the simplification from products to factorials to binomial coefficients was pretty straightforward.
– John Forkosh
Jul 25 at 13:45
I am sorry. My comment had ran out of characters. I was speaking about your first "magic formula". I do not exactly understand what you have done there, in your first post, to get this special relation. I would appreciate if you could add the process you have gone through to end up with this formula to this post. Or, if you do not mind, just explain it to me further than in your original question. Hence I am quite fascinated by this formula aswell I will work on it a while. Maybe I can find something useful, who knows.
– mrtaurho
Jul 25 at 13:52
Oh, right. I mentioned that above -- for both formulas. I >>did<< absolutely nothing. Rather, it was that other question math.stackexchange.com/questions/2860403 where I noticed that I >>needed<< those formulas to be true (or else the stuff that I felt should simplify wouldn't simplify). So I >>assumed<< them true, and just did numerous numerical tests (for more values than you can shake a stick at). Then I indeed tried to prove them, but got absolutely nowhere (and it wasn't "nowhere fast", but "nowhere very slowly":)
– John Forkosh
Jul 25 at 14:01
 |Â
show 3 more comments
up vote
1
down vote
up vote
1
down vote
I really tried to play around with your two "magic formulas" but I did not found anything useful. Nevertheless I can prove your explicit formula for $H^m_n$ using a proof I did on a different post (which you can find here : Help on induction, couldn't make both side the same value). The relation there is
$$sum_k=1^n k(k+1)~cdots (k+p-1)~=~ fracn(n+1)cdots (n+p)p+1$$
which I will use later on. First of all we write $H_n^m+1$ in terms of $H_n^m$ with the equation you defined
$$H^m+1_n~=~sum_k=1^n H_k^m $$
Now using your derived indentity
$$H_n^m~=~binomn+m-2m-1~~textand respectively~~H_n^m+1~=~binomn+m-1m$$
we obtain
$$beginalign
H^m+1_n~&=~sum_k=1^n H_k^m\
binomn+m-1m~&=~sum_k=1^n binomk+m-2m-1\
frac(n+m-1)!(n-1)!~m!~&=~sum_k=1^n frac(k+m-2)!(k-1)!~(m-1)!\
frac(n+m-1)!(n-1)!~m~&=~sum_k=1^n frac(k+m-2)!(k-1)!\
fracn(n+1)cdots (n+m-1)m~&=~sum_k=1^n k(k+1)~cdots (k+m-2)
endalign$$
which equals my given equation by setting $m=p+1$.
Maybe you can go further from there on but atleast you got a proof as a verification for your solution by examination.
My try on the first "magic formula"
$$beginalign
H_m^m+2-sum_k=1^m-1 k H_m^m-k+1~&=~m\
H_m^m+2-sum_k=1^m-1 k H_m^m-k+1+mH_m^1-mH_m^1~&=~m\
H_m^m+2-sum_k=1^m k H_m^m-k+1+mH_m^1~&=~m\
H_m^m+2-sum_k=1^m k H_m^m-k+1+mcdot (1)~&=~m\
H_m^m+2~&=~sum_k=1^m k H_m^m-k+1\
sum_k=1^m H_k^m+1~&=~sum_k=1^m k H_m^m-k+1
endalign$$
so therefore I guess it should be somehow possible to show
$$H_k^m+1~=~k H_m^m-k+1$$
But till now it leads me to nowhere
answered Jul 25 at 12:12
mrtaurho
740219
I really tried to play around with your two "magic formulas" but I did not found anything useful. Nevertheless I can prove your explicit formula for $H^m_n$ using a proof I did on a different post (which you can find here : Help on induction, couldn't make both side the same value). The relation there is
$$sum_k=1^n k(k+1)~cdots (k+p-1)~=~ fracn(n+1)cdots (n+p)p+1$$
which I will use later on. First of all we write $H_n^m+1$ in terms of $H_n^m$ with the equation you defined
$$H^m+1_n~=~sum_k=1^n H_k^m $$
Now using your derived indentity
$$H_n^m~=~binomn+m-2m-1~~textand respectively~~H_n^m+1~=~binomn+m-1m$$
we obtain
$$beginalign
H^m+1_n~&=~sum_k=1^n H_k^m\
binomn+m-1m~&=~sum_k=1^n binomk+m-2m-1\
frac(n+m-1)!(n-1)!~m!~&=~sum_k=1^n frac(k+m-2)!(k-1)!~(m-1)!\
frac(n+m-1)!(n-1)!~m~&=~sum_k=1^n frac(k+m-2)!(k-1)!\
fracn(n+1)cdots (n+m-1)m~&=~sum_k=1^n k(k+1)~cdots (k+m-2)
endalign$$
which equals my given equation by setting $m=p+1$.
Maybe you can go further from there on but atleast you got a proof as a verification for your solution by examination.
My try on the first "magic formula"
$$beginalign
H_m^m+2-sum_k=1^m-1 k H_m^m-k+1~&=~m\
H_m^m+2-sum_k=1^m-1 k H_m^m-k+1+mH_m^1-mH_m^1~&=~m\
H_m^m+2-sum_k=1^m k H_m^m-k+1+mH_m^1~&=~m\
H_m^m+2-sum_k=1^m k H_m^m-k+1+mcdot (1)~&=~m\
H_m^m+2~&=~sum_k=1^m k H_m^m-k+1\
sum_k=1^m H_k^m+1~&=~sum_k=1^m k H_m^m-k+1
endalign$$
so therefore I guess it should be somehow possible to show
$$H_k^m+1~=~k H_m^m-k+1$$
But till now it leads me to nowhere
answered Jul 25 at 12:12
mrtaurho
740219
edited Jul 25 at 14:15
answered Jul 25 at 12:12
mrtaurho
740219
answered Jul 25 at 12:12
mrtaurho
740219
answered Jul 25 at 12:12
mrtaurho
740219
740219
Thanks @mrtaurho. Actually I'd algebraically derived the binomial coefficient $H^m_n=binomn+m-2n-1$, but just didn't bother displaying the algebra. And I apparently didn't make it clear that it had already been derived. The numerical tests at the bottom were merely a double-check confirmation. It's only the "magic formulas" I couldn't derive. As for "usefulness", that's from my earlier question math.stackexchange.com/questions/2860403 where the $H^m_n$'s appear in an iterative definition of something else. And that other thing simplified nicely only if those magic formulas hold.
– John Forkosh
Jul 25 at 13:07
Ah, okay. I guess from now one you can be quite sure that this binomial coefficient is right. I tried some different ways, only concerning the first formula, involving some kind of induction and by taking advantage of the definitions but it always ended in some very strange binomial coefficients together with weird sums. For me the main problem which occurs is the extra $k$-factor infront of the $H_m^m+1-k$. I just looked over your first question and I am not sure how exactly you derived it there. Could you maybe add it to this question since you are asking here for these "magic formulas"?
– mrtaurho
Jul 25 at 13:17
Sorry, "..how exactly you derived it there"? What "it" are we referring to? If you're referring to that original product form for $H^m_n$, then I mis-spoke in the preceding comment -- I solved three equations in three unknowns to get the general form for $H^4_n$, and then inferred $H^m>4_n$ without algebraic proof (but with lots of numerical verification). And from there, the simplification from products to factorials to binomial coefficients was pretty straightforward.
– John Forkosh
Jul 25 at 13:45
I am sorry. My comment had ran out of characters. I was speaking about your first "magic formula". I do not exactly understand what you have done there, in your first post, to get this special relation. I would appreciate if you could add the process you have gone through to end up with this formula to this post. Or, if you do not mind, just explain it to me further than in your original question. Hence I am quite fascinated by this formula aswell I will work on it a while. Maybe I can find something useful, who knows.
– mrtaurho
Jul 25 at 13:52
Oh, right. I mentioned that above -- for both formulas. I >>did<< absolutely nothing. Rather, it was that other question math.stackexchange.com/questions/2860403 where I noticed that I >>needed<< those formulas to be true (or else the stuff that I felt should simplify wouldn't simplify). So I >>assumed<< them true, and just did numerous numerical tests (for more values than you can shake a stick at). Then I indeed tried to prove them, but got absolutely nowhere (and it wasn't "nowhere fast", but "nowhere very slowly":)
– John Forkosh
Jul 25 at 14:01
 |Â
show 3 more comments
Thanks @mrtaurho. Actually I'd algebraically derived the binomial coefficient $H^m_n=binomn+m-2n-1$, but just didn't bother displaying the algebra. And I apparently didn't make it clear that it had already been derived. The numerical tests at the bottom were merely a double-check confirmation. It's only the "magic formulas" I couldn't derive. As for "usefulness", that's from my earlier question math.stackexchange.com/questions/2860403 where the $H^m_n$'s appear in an iterative definition of something else. And that other thing simplified nicely only if those magic formulas hold.
– John Forkosh
Jul 25 at 13:07
Ah, okay. I guess from now one you can be quite sure that this binomial coefficient is right. I tried some different ways, only concerning the first formula, involving some kind of induction and by taking advantage of the definitions but it always ended in some very strange binomial coefficients together with weird sums. For me the main problem which occurs is the extra $k$-factor infront of the $H_m^m+1-k$. I just looked over your first question and I am not sure how exactly you derived it there. Could you maybe add it to this question since you are asking here for these "magic formulas"?
– mrtaurho
Jul 25 at 13:17
Sorry, "..how exactly you derived it there"? What "it" are we referring to? If you're referring to that original product form for $H^m_n$, then I mis-spoke in the preceding comment -- I solved three equations in three unknowns to get the general form for $H^4_n$, and then inferred $H^m>4_n$ without algebraic proof (but with lots of numerical verification). And from there, the simplification from products to factorials to binomial coefficients was pretty straightforward.
– John Forkosh
Jul 25 at 13:45
I am sorry. My comment had ran out of characters. I was speaking about your first "magic formula". I do not exactly understand what you have done there, in your first post, to get this special relation. I would appreciate if you could add the process you have gone through to end up with this formula to this post. Or, if you do not mind, just explain it to me further than in your original question. Hence I am quite fascinated by this formula aswell I will work on it a while. Maybe I can find something useful, who knows.
– mrtaurho
Jul 25 at 13:52
Oh, right. I mentioned that above -- for both formulas. I >>did<< absolutely nothing. Rather, it was that other question math.stackexchange.com/questions/2860403 where I noticed that I >>needed<< those formulas to be true (or else the stuff that I felt should simplify wouldn't simplify). So I >>assumed<< them true, and just did numerous numerical tests (for more values than you can shake a stick at). Then I indeed tried to prove them, but got absolutely nowhere (and it wasn't "nowhere fast", but "nowhere very slowly":)
– John Forkosh
Jul 25 at 14:01
Thanks @mrtaurho. Actually I'd algebraically derived the binomial coefficient $H^m_n=binomn+m-2n-1$, but just didn't bother displaying the algebra. And I apparently didn't make it clear that it had already been derived. The numerical tests at the bottom were merely a double-check confirmation. It's only the "magic formulas" I couldn't derive. As for "usefulness", that's from my earlier question math.stackexchange.com/questions/2860403 where the $H^m_n$'s appear in an iterative definition of something else. And that other thing simplified nicely only if those magic formulas hold.
– John Forkosh
Jul 25 at 13:07
Thanks @mrtaurho. Actually I'd algebraically derived the binomial coefficient $H^m_n=binomn+m-2n-1$, but just didn't bother displaying the algebra. And I apparently didn't make it clear that it had already been derived. The numerical tests at the bottom were merely a double-check confirmation. It's only the "magic formulas" I couldn't derive. As for "usefulness", that's from my earlier question math.stackexchange.com/questions/2860403 where the $H^m_n$'s appear in an iterative definition of something else. And that other thing simplified nicely only if those magic formulas hold.
– John Forkosh
Jul 25 at 13:07
Ah, okay. I guess from now one you can be quite sure that this binomial coefficient is right. I tried some different ways, only concerning the first formula, involving some kind of induction and by taking advantage of the definitions but it always ended in some very strange binomial coefficients together with weird sums. For me the main problem which occurs is the extra $k$-factor infront of the $H_m^m+1-k$. I just looked over your first question and I am not sure how exactly you derived it there. Could you maybe add it to this question since you are asking here for these "magic formulas"?
– mrtaurho
Jul 25 at 13:17
Ah, okay. I guess from now one you can be quite sure that this binomial coefficient is right. I tried some different ways, only concerning the first formula, involving some kind of induction and by taking advantage of the definitions but it always ended in some very strange binomial coefficients together with weird sums. For me the main problem which occurs is the extra $k$-factor infront of the $H_m^m+1-k$. I just looked over your first question and I am not sure how exactly you derived it there. Could you maybe add it to this question since you are asking here for these "magic formulas"?
– mrtaurho
Jul 25 at 13:17
Sorry, "..how exactly you derived it there"? What "it" are we referring to? If you're referring to that original product form for $H^m_n$, then I mis-spoke in the preceding comment -- I solved three equations in three unknowns to get the general form for $H^4_n$, and then inferred $H^m>4_n$ without algebraic proof (but with lots of numerical verification). And from there, the simplification from products to factorials to binomial coefficients was pretty straightforward.
– John Forkosh
Jul 25 at 13:45
Sorry, "..how exactly you derived it there"? What "it" are we referring to? If you're referring to that original product form for $H^m_n$, then I mis-spoke in the preceding comment -- I solved three equations in three unknowns to get the general form for $H^4_n$, and then inferred $H^m>4_n$ without algebraic proof (but with lots of numerical verification). And from there, the simplification from products to factorials to binomial coefficients was pretty straightforward.
– John Forkosh
Jul 25 at 13:45
I am sorry. My comment had ran out of characters. I was speaking about your first "magic formula". I do not exactly understand what you have done there, in your first post, to get this special relation. I would appreciate if you could add the process you have gone through to end up with this formula to this post. Or, if you do not mind, just explain it to me further than in your original question. Hence I am quite fascinated by this formula aswell I will work on it a while. Maybe I can find something useful, who knows.
– mrtaurho
Jul 25 at 13:52
I am sorry. My comment had ran out of characters. I was speaking about your first "magic formula". I do not exactly understand what you have done there, in your first post, to get this special relation. I would appreciate if you could add the process you have gone through to end up with this formula to this post. Or, if you do not mind, just explain it to me further than in your original question. Hence I am quite fascinated by this formula aswell I will work on it a while. Maybe I can find something useful, who knows.
– mrtaurho
Jul 25 at 13:52
Oh, right. I mentioned that above -- for both formulas. I >>did<< absolutely nothing. Rather, it was that other question math.stackexchange.com/questions/2860403 where I noticed that I >>needed<< those formulas to be true (or else the stuff that I felt should simplify wouldn't simplify). So I >>assumed<< them true, and just did numerous numerical tests (for more values than you can shake a stick at). Then I indeed tried to prove them, but got absolutely nowhere (and it wasn't "nowhere fast", but "nowhere very slowly":)
– John Forkosh
Jul 25 at 14:01
Oh, right. I mentioned that above -- for both formulas. I >>did<< absolutely nothing. Rather, it was that other question math.stackexchange.com/questions/2860403 where I noticed that I >>needed<< those formulas to be true (or else the stuff that I felt should simplify wouldn't simplify). So I >>assumed<< them true, and just did numerous numerical tests (for more values than you can shake a stick at). Then I indeed tried to prove them, but got absolutely nowhere (and it wasn't "nowhere fast", but "nowhere very slowly":)
– John Forkosh
Jul 25 at 14:01
 |Â
show 3 more comments
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Did you tried to use Faulhaber's formula (en.wikipedia.org/wiki/Faulhaber%27s_formula) to verify your general expression?
– mrtaurho
Jul 25 at 10:34
@mrtaurho No, I wasn't aware of Faulhaber's formula, but looking at that wikipedia page, I'm not seeing how it applies. As far as I can tell, $sum_k=1^nk^p$ doesn't seem related. How do you think it's applicable?
– John Forkosh
Jul 25 at 11:04
Oh, I have mistaken one formula. I am sorry. After I saw your expressions for $H_n^3$ and $H_n^4$ I did not realized that the one factor in you product for $H_n^4$ is $(n+2)$ and not as I assumed $(2n+1)$.
– mrtaurho
Jul 25 at 11:10