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How to rewrite this nested summation as a single sum?
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$Q$ and $p$ are natural numbers. $E$ equals the $p$-nested sum of products with $p+1$ factors
$$beginalignE&=sum _large k_1=2^Q sum _large k_2=k_1+1^Qldotssum _large k_p-1 =\large k_p-2+1^Qsum _large k_p =\large k_p-1+1^Q a_large k_1-1,1 a_large k_2-1,k_1ldots a_large k_p-1,k_p-1 a_large Q,k_p\\&=sum _large k_1=2^Q-p+1sum _large k_2=k_1+1^Q-p+2ldotssum _large k_p-1 =\large k_p-2+1^Q-1sum _large k_p =\large k_p-1+1^Q a_large k_1-1,1 a_large k_2-1,k_1ldots a_large k_p-1,k_p-1 a_large Q,k_pendalign$$
E.g. when $Q=7$ and $p=4$ the products to be sumed are the $binomQ-1p$ columns
$$beginarrayccccccccccccccc
a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_2,1 & a_2,1 & a_2,1 & a_2,1 & a_3,1 \
a_2,2 & a_2,2 & a_2,2 & a_2,2 & a_2,2 & a_2,2 & a_3,2 & a_3,2 & a_3,2 & a_4,2 & a_3,3 & a_3,3 & a_3,3 & a_4,3 & a_4,4 \
a_3,3 & a_3,3 & a_3,3 & a_4,3 & a_4,3 & a_5,3 & a_4,4 & a_4,4 & a_5,4 & a_5,5 & a_4,4 & a_4,4 & a_5,4 & a_5,5 & a_5,5 \
a_4,4 & a_5,4 & a_6,4 & a_5,5 & a_6,5 & a_6,6 & a_5,5 & a_6,5 & a_6,6 & a_6,6 & a_5,5 & a_6,5 & a_6,6 & a_6,6 & a_6,6 \
a_7,5 & a_7,6 & a_7,7 & a_7,6 & a_7,7 & a_7,7 & a_7,6 & a_7,7 & a_7,7 & a_7,7 & a_7,6 & a_7,7 & a_7,7 & a_7,7 & a_7,7 \
endarray$$
How to define functions $f_1$ and $f_2$ that compute the indices of the $k$'th term's $j$'th factor?
I.e. such that
$$E=sum_k=1^binomQ-1pprod_j=1^p+1a_large f_1(Q,p,k,j),f_2(Q,p,k,j)$$
(that is, the order of terms as well as factors doesn't matter)
combinatorics discrete-mathematics summation
asked Jul 18 at 11:13
MeMyselfI
540216
add a comment |Â
up vote
0
down vote
favorite
$Q$ and $p$ are natural numbers. $E$ equals the $p$-nested sum of products with $p+1$ factors
$$beginalignE&=sum _large k_1=2^Q sum _large k_2=k_1+1^Qldotssum _large k_p-1 =\large k_p-2+1^Qsum _large k_p =\large k_p-1+1^Q a_large k_1-1,1 a_large k_2-1,k_1ldots a_large k_p-1,k_p-1 a_large Q,k_p\\&=sum _large k_1=2^Q-p+1sum _large k_2=k_1+1^Q-p+2ldotssum _large k_p-1 =\large k_p-2+1^Q-1sum _large k_p =\large k_p-1+1^Q a_large k_1-1,1 a_large k_2-1,k_1ldots a_large k_p-1,k_p-1 a_large Q,k_pendalign$$
E.g. when $Q=7$ and $p=4$ the products to be sumed are the $binomQ-1p$ columns
$$beginarrayccccccccccccccc
a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_2,1 & a_2,1 & a_2,1 & a_2,1 & a_3,1 \
a_2,2 & a_2,2 & a_2,2 & a_2,2 & a_2,2 & a_2,2 & a_3,2 & a_3,2 & a_3,2 & a_4,2 & a_3,3 & a_3,3 & a_3,3 & a_4,3 & a_4,4 \
a_3,3 & a_3,3 & a_3,3 & a_4,3 & a_4,3 & a_5,3 & a_4,4 & a_4,4 & a_5,4 & a_5,5 & a_4,4 & a_4,4 & a_5,4 & a_5,5 & a_5,5 \
a_4,4 & a_5,4 & a_6,4 & a_5,5 & a_6,5 & a_6,6 & a_5,5 & a_6,5 & a_6,6 & a_6,6 & a_5,5 & a_6,5 & a_6,6 & a_6,6 & a_6,6 \
a_7,5 & a_7,6 & a_7,7 & a_7,6 & a_7,7 & a_7,7 & a_7,6 & a_7,7 & a_7,7 & a_7,7 & a_7,6 & a_7,7 & a_7,7 & a_7,7 & a_7,7 \
endarray$$
How to define functions $f_1$ and $f_2$ that compute the indices of the $k$'th term's $j$'th factor?
I.e. such that
$$E=sum_k=1^binomQ-1pprod_j=1^p+1a_large f_1(Q,p,k,j),f_2(Q,p,k,j)$$
(that is, the order of terms as well as factors doesn't matter)
combinatorics discrete-mathematics summation
asked Jul 18 at 11:13
MeMyselfI
540216
1
IMO the original sum is easier to read as $$sum_large 1 = k_0 < k_1 < cdots < k_p < k_p+1 = Q+1; prod_large j=1^p+1 a_k_j - 1, k_j-1$$ Also the question can be simplified by noting that $f_1(Q,p,k,j) = f_2(Q,p,k,j+1)-1$ requiring only the addition of a special case $f_2(Q,p,k,p+2)=Q+1$.
– Peter Taylor
Jul 18 at 13:02
Replace the k's with their actual values - 2,3,4,...
– William Elliot
Jul 18 at 19:50
Instead of $f_1$ and $f_2$, you can write a few lines of code to find the NEXT set, based on which factor gets its index increased by $1$.
– Empy2
Jul 21 at 0:49
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$Q$ and $p$ are natural numbers. $E$ equals the $p$-nested sum of products with $p+1$ factors
$$beginalignE&=sum _large k_1=2^Q sum _large k_2=k_1+1^Qldotssum _large k_p-1 =\large k_p-2+1^Qsum _large k_p =\large k_p-1+1^Q a_large k_1-1,1 a_large k_2-1,k_1ldots a_large k_p-1,k_p-1 a_large Q,k_p\\&=sum _large k_1=2^Q-p+1sum _large k_2=k_1+1^Q-p+2ldotssum _large k_p-1 =\large k_p-2+1^Q-1sum _large k_p =\large k_p-1+1^Q a_large k_1-1,1 a_large k_2-1,k_1ldots a_large k_p-1,k_p-1 a_large Q,k_pendalign$$
E.g. when $Q=7$ and $p=4$ the products to be sumed are the $binomQ-1p$ columns
$$beginarrayccccccccccccccc
a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_2,1 & a_2,1 & a_2,1 & a_2,1 & a_3,1 \
a_2,2 & a_2,2 & a_2,2 & a_2,2 & a_2,2 & a_2,2 & a_3,2 & a_3,2 & a_3,2 & a_4,2 & a_3,3 & a_3,3 & a_3,3 & a_4,3 & a_4,4 \
a_3,3 & a_3,3 & a_3,3 & a_4,3 & a_4,3 & a_5,3 & a_4,4 & a_4,4 & a_5,4 & a_5,5 & a_4,4 & a_4,4 & a_5,4 & a_5,5 & a_5,5 \
a_4,4 & a_5,4 & a_6,4 & a_5,5 & a_6,5 & a_6,6 & a_5,5 & a_6,5 & a_6,6 & a_6,6 & a_5,5 & a_6,5 & a_6,6 & a_6,6 & a_6,6 \
a_7,5 & a_7,6 & a_7,7 & a_7,6 & a_7,7 & a_7,7 & a_7,6 & a_7,7 & a_7,7 & a_7,7 & a_7,6 & a_7,7 & a_7,7 & a_7,7 & a_7,7 \
endarray$$
How to define functions $f_1$ and $f_2$ that compute the indices of the $k$'th term's $j$'th factor?
I.e. such that
$$E=sum_k=1^binomQ-1pprod_j=1^p+1a_large f_1(Q,p,k,j),f_2(Q,p,k,j)$$
(that is, the order of terms as well as factors doesn't matter)
combinatorics discrete-mathematics summation
asked Jul 18 at 11:13
MeMyselfI
540216
$Q$ and $p$ are natural numbers. $E$ equals the $p$-nested sum of products with $p+1$ factors
$$beginalignE&=sum _large k_1=2^Q sum _large k_2=k_1+1^Qldotssum _large k_p-1 =\large k_p-2+1^Qsum _large k_p =\large k_p-1+1^Q a_large k_1-1,1 a_large k_2-1,k_1ldots a_large k_p-1,k_p-1 a_large Q,k_p\\&=sum _large k_1=2^Q-p+1sum _large k_2=k_1+1^Q-p+2ldotssum _large k_p-1 =\large k_p-2+1^Q-1sum _large k_p =\large k_p-1+1^Q a_large k_1-1,1 a_large k_2-1,k_1ldots a_large k_p-1,k_p-1 a_large Q,k_pendalign$$
E.g. when $Q=7$ and $p=4$ the products to be sumed are the $binomQ-1p$ columns
$$beginarrayccccccccccccccc
a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_1,1 & a_2,1 & a_2,1 & a_2,1 & a_2,1 & a_3,1 \
a_2,2 & a_2,2 & a_2,2 & a_2,2 & a_2,2 & a_2,2 & a_3,2 & a_3,2 & a_3,2 & a_4,2 & a_3,3 & a_3,3 & a_3,3 & a_4,3 & a_4,4 \
a_3,3 & a_3,3 & a_3,3 & a_4,3 & a_4,3 & a_5,3 & a_4,4 & a_4,4 & a_5,4 & a_5,5 & a_4,4 & a_4,4 & a_5,4 & a_5,5 & a_5,5 \
a_4,4 & a_5,4 & a_6,4 & a_5,5 & a_6,5 & a_6,6 & a_5,5 & a_6,5 & a_6,6 & a_6,6 & a_5,5 & a_6,5 & a_6,6 & a_6,6 & a_6,6 \
a_7,5 & a_7,6 & a_7,7 & a_7,6 & a_7,7 & a_7,7 & a_7,6 & a_7,7 & a_7,7 & a_7,7 & a_7,6 & a_7,7 & a_7,7 & a_7,7 & a_7,7 \
endarray$$
How to define functions $f_1$ and $f_2$ that compute the indices of the $k$'th term's $j$'th factor?
I.e. such that
$$E=sum_k=1^binomQ-1pprod_j=1^p+1a_large f_1(Q,p,k,j),f_2(Q,p,k,j)$$
(that is, the order of terms as well as factors doesn't matter)
combinatorics discrete-mathematics summation
asked Jul 18 at 11:13
MeMyselfI
540216
edited Jul 19 at 7:58
asked Jul 18 at 11:13
MeMyselfI
540216
asked Jul 18 at 11:13
MeMyselfI
540216
asked Jul 18 at 11:13
MeMyselfI
540216
540216
1
IMO the original sum is easier to read as $$sum_large 1 = k_0 < k_1 < cdots < k_p < k_p+1 = Q+1; prod_large j=1^p+1 a_k_j - 1, k_j-1$$ Also the question can be simplified by noting that $f_1(Q,p,k,j) = f_2(Q,p,k,j+1)-1$ requiring only the addition of a special case $f_2(Q,p,k,p+2)=Q+1$.
– Peter Taylor
Jul 18 at 13:02
Replace the k's with their actual values - 2,3,4,...
– William Elliot
Jul 18 at 19:50
Instead of $f_1$ and $f_2$, you can write a few lines of code to find the NEXT set, based on which factor gets its index increased by $1$.
– Empy2
Jul 21 at 0:49
add a comment |Â
1
IMO the original sum is easier to read as $$sum_large 1 = k_0 < k_1 < cdots < k_p < k_p+1 = Q+1; prod_large j=1^p+1 a_k_j - 1, k_j-1$$ Also the question can be simplified by noting that $f_1(Q,p,k,j) = f_2(Q,p,k,j+1)-1$ requiring only the addition of a special case $f_2(Q,p,k,p+2)=Q+1$.
– Peter Taylor
Jul 18 at 13:02
Replace the k's with their actual values - 2,3,4,...
– William Elliot
Jul 18 at 19:50
Instead of $f_1$ and $f_2$, you can write a few lines of code to find the NEXT set, based on which factor gets its index increased by $1$.
– Empy2
Jul 21 at 0:49
1
1
IMO the original sum is easier to read as $$sum_large 1 = k_0 < k_1 < cdots < k_p < k_p+1 = Q+1; prod_large j=1^p+1 a_k_j - 1, k_j-1$$ Also the question can be simplified by noting that $f_1(Q,p,k,j) = f_2(Q,p,k,j+1)-1$ requiring only the addition of a special case $f_2(Q,p,k,p+2)=Q+1$.
– Peter Taylor
Jul 18 at 13:02
IMO the original sum is easier to read as $$sum_large 1 = k_0 < k_1 < cdots < k_p < k_p+1 = Q+1; prod_large j=1^p+1 a_k_j - 1, k_j-1$$ Also the question can be simplified by noting that $f_1(Q,p,k,j) = f_2(Q,p,k,j+1)-1$ requiring only the addition of a special case $f_2(Q,p,k,p+2)=Q+1$.
– Peter Taylor
Jul 18 at 13:02
Replace the k's with their actual values - 2,3,4,...
– William Elliot
Jul 18 at 19:50
Replace the k's with their actual values - 2,3,4,...
– William Elliot
Jul 18 at 19:50
Instead of $f_1$ and $f_2$, you can write a few lines of code to find the NEXT set, based on which factor gets its index increased by $1$.
– Empy2
Jul 21 at 0:49
Instead of $f_1$ and $f_2$, you can write a few lines of code to find the NEXT set, based on which factor gets its index increased by $1$.
– Empy2
Jul 21 at 0:49
add a comment |Â
1 Answer
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Reading from right to left, the last $a_3,1,a_2,1$ and $a_1,1$ are in columns $1,5,15$ which are $4choose4,5choose4$ and $6choose4$. So you want the inverse function of $$f(n)=nchoose4$$ Within each of those, the second factor comes from similar $nchoose3$ and so on. I wonder how effective is $$g(m)=sqrt[4]4!f(n)+frac32$$
Lookup tables would be effective.
answered Jul 19 at 13:29
Empy2
31.9k12059
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Reading from right to left, the last $a_3,1,a_2,1$ and $a_1,1$ are in columns $1,5,15$ which are $4choose4,5choose4$ and $6choose4$. So you want the inverse function of $$f(n)=nchoose4$$ Within each of those, the second factor comes from similar $nchoose3$ and so on. I wonder how effective is $$g(m)=sqrt[4]4!f(n)+frac32$$
Lookup tables would be effective.
answered Jul 19 at 13:29
Empy2
31.9k12059
add a comment |Â
up vote
0
down vote
Reading from right to left, the last $a_3,1,a_2,1$ and $a_1,1$ are in columns $1,5,15$ which are $4choose4,5choose4$ and $6choose4$. So you want the inverse function of $$f(n)=nchoose4$$ Within each of those, the second factor comes from similar $nchoose3$ and so on. I wonder how effective is $$g(m)=sqrt[4]4!f(n)+frac32$$
Lookup tables would be effective.
answered Jul 19 at 13:29
Empy2
31.9k12059
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Reading from right to left, the last $a_3,1,a_2,1$ and $a_1,1$ are in columns $1,5,15$ which are $4choose4,5choose4$ and $6choose4$. So you want the inverse function of $$f(n)=nchoose4$$ Within each of those, the second factor comes from similar $nchoose3$ and so on. I wonder how effective is $$g(m)=sqrt[4]4!f(n)+frac32$$
Lookup tables would be effective.
answered Jul 19 at 13:29
Empy2
31.9k12059
Reading from right to left, the last $a_3,1,a_2,1$ and $a_1,1$ are in columns $1,5,15$ which are $4choose4,5choose4$ and $6choose4$. So you want the inverse function of $$f(n)=nchoose4$$ Within each of those, the second factor comes from similar $nchoose3$ and so on. I wonder how effective is $$g(m)=sqrt[4]4!f(n)+frac32$$
Lookup tables would be effective.
answered Jul 19 at 13:29
Empy2
31.9k12059
edited Jul 19 at 14:01
answered Jul 19 at 13:29
Empy2
31.9k12059
answered Jul 19 at 13:29
Empy2
31.9k12059
answered Jul 19 at 13:29
Empy2
31.9k12059
31.9k12059
add a comment |Â
add a comment |Â
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1
IMO the original sum is easier to read as $$sum_large 1 = k_0 < k_1 < cdots < k_p < k_p+1 = Q+1; prod_large j=1^p+1 a_k_j - 1, k_j-1$$ Also the question can be simplified by noting that $f_1(Q,p,k,j) = f_2(Q,p,k,j+1)-1$ requiring only the addition of a special case $f_2(Q,p,k,p+2)=Q+1$.
– Peter Taylor
Jul 18 at 13:02
Replace the k's with their actual values - 2,3,4,...
– William Elliot
Jul 18 at 19:50
Instead of $f_1$ and $f_2$, you can write a few lines of code to find the NEXT set, based on which factor gets its index increased by $1$.
– Empy2
Jul 21 at 0:49