Mixing
Binomial Coefficient Identity Conjecture
Clash Royale CLAN TAG#URR8PPP
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The following (conjectured) identity has come up in a research problem that I am working on:
for even $a$
$$sum_i=0^a-1 (-1)^a-ibinomai binom2m-i-2m-i-1=0;$$
and for odd $a$
$$sum_i=0^a-1 (-1)^a-ibinomai binom2m-i-2m-i-1=-2binom2m-a-2m-a-1,$$
where $a,m$ are positive integers with $1le ale m-2$.
I've verified the identity holds for small values of $a,m$.
The closest problem I have found is
Help with a Binomial Coefficient Identity. Any suggestion how to apply that identity or to find another proof?
combinatorics binomial-coefficients
asked Jul 24 at 5:48
Alex
263
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up vote
4
down vote
favorite
The following (conjectured) identity has come up in a research problem that I am working on:
for even $a$
$$sum_i=0^a-1 (-1)^a-ibinomai binom2m-i-2m-i-1=0;$$
and for odd $a$
$$sum_i=0^a-1 (-1)^a-ibinomai binom2m-i-2m-i-1=-2binom2m-a-2m-a-1,$$
where $a,m$ are positive integers with $1le ale m-2$.
I've verified the identity holds for small values of $a,m$.
The closest problem I have found is
Help with a Binomial Coefficient Identity. Any suggestion how to apply that identity or to find another proof?
combinatorics binomial-coefficients
asked Jul 24 at 5:48
Alex
263
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
The following (conjectured) identity has come up in a research problem that I am working on:
for even $a$
$$sum_i=0^a-1 (-1)^a-ibinomai binom2m-i-2m-i-1=0;$$
and for odd $a$
$$sum_i=0^a-1 (-1)^a-ibinomai binom2m-i-2m-i-1=-2binom2m-a-2m-a-1,$$
where $a,m$ are positive integers with $1le ale m-2$.
I've verified the identity holds for small values of $a,m$.
The closest problem I have found is
Help with a Binomial Coefficient Identity. Any suggestion how to apply that identity or to find another proof?
combinatorics binomial-coefficients
asked Jul 24 at 5:48
Alex
263
The following (conjectured) identity has come up in a research problem that I am working on:
for even $a$
$$sum_i=0^a-1 (-1)^a-ibinomai binom2m-i-2m-i-1=0;$$
and for odd $a$
$$sum_i=0^a-1 (-1)^a-ibinomai binom2m-i-2m-i-1=-2binom2m-a-2m-a-1,$$
where $a,m$ are positive integers with $1le ale m-2$.
I've verified the identity holds for small values of $a,m$.
The closest problem I have found is
Help with a Binomial Coefficient Identity. Any suggestion how to apply that identity or to find another proof?
combinatorics binomial-coefficients
asked Jul 24 at 5:48
Alex
263
asked Jul 24 at 5:48
Alex
263
asked Jul 24 at 5:48
Alex
263
asked Jul 24 at 5:48
Alex
263
263
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
10
down vote
accepted
If you extend the sum to $a$, you can combine the even and odd cases into
$$
sum_i=0^a(-1)^ibinom aibinom2n-in-i=binom2n-an;,
$$
with $n=m-1$. This is a double count using inclusion–exclusion of the number of ways of selecting $n$ from $2n$ elements such that $a$ particular elements are not included in the selection.
answered Jul 24 at 6:00
joriki
164k10180328
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up vote
2
down vote
For an algebraic proof of the re-formulated identity by @joriki we write
$$sum_q=0^a (-1)^q achoose q 2n-qchoose n-q
= sum_q=0^a (-1)^q achoose q [z^n-q] (1+z)^2n-q
\ = [z^n] (1+z)^2n
sum_q=0^a (-1)^q achoose q z^q (1+z)^-q
\ = [z^n] (1+z)^2n
left(1-fracz1+zright)^a
= [z^n] (1+z)^2n-a = 2n-achoose n.$$
answered Jul 25 at 17:49
Marko Riedel
36.4k333107
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
accepted
If you extend the sum to $a$, you can combine the even and odd cases into
$$
sum_i=0^a(-1)^ibinom aibinom2n-in-i=binom2n-an;,
$$
with $n=m-1$. This is a double count using inclusion–exclusion of the number of ways of selecting $n$ from $2n$ elements such that $a$ particular elements are not included in the selection.
answered Jul 24 at 6:00
joriki
164k10180328
add a comment |Â
up vote
10
down vote
accepted
If you extend the sum to $a$, you can combine the even and odd cases into
$$
sum_i=0^a(-1)^ibinom aibinom2n-in-i=binom2n-an;,
$$
with $n=m-1$. This is a double count using inclusion–exclusion of the number of ways of selecting $n$ from $2n$ elements such that $a$ particular elements are not included in the selection.
answered Jul 24 at 6:00
joriki
164k10180328
add a comment |Â
up vote
10
down vote
accepted
up vote
10
down vote
accepted
If you extend the sum to $a$, you can combine the even and odd cases into
$$
sum_i=0^a(-1)^ibinom aibinom2n-in-i=binom2n-an;,
$$
with $n=m-1$. This is a double count using inclusion–exclusion of the number of ways of selecting $n$ from $2n$ elements such that $a$ particular elements are not included in the selection.
answered Jul 24 at 6:00
joriki
164k10180328
If you extend the sum to $a$, you can combine the even and odd cases into
$$
sum_i=0^a(-1)^ibinom aibinom2n-in-i=binom2n-an;,
$$
with $n=m-1$. This is a double count using inclusion–exclusion of the number of ways of selecting $n$ from $2n$ elements such that $a$ particular elements are not included in the selection.
answered Jul 24 at 6:00
joriki
164k10180328
answered Jul 24 at 6:00
joriki
164k10180328
answered Jul 24 at 6:00
joriki
164k10180328
answered Jul 24 at 6:00
joriki
164k10180328
164k10180328
add a comment |Â
add a comment |Â
up vote
2
down vote
For an algebraic proof of the re-formulated identity by @joriki we write
$$sum_q=0^a (-1)^q achoose q 2n-qchoose n-q
= sum_q=0^a (-1)^q achoose q [z^n-q] (1+z)^2n-q
\ = [z^n] (1+z)^2n
sum_q=0^a (-1)^q achoose q z^q (1+z)^-q
\ = [z^n] (1+z)^2n
left(1-fracz1+zright)^a
= [z^n] (1+z)^2n-a = 2n-achoose n.$$
answered Jul 25 at 17:49
Marko Riedel
36.4k333107
add a comment |Â
up vote
2
down vote
For an algebraic proof of the re-formulated identity by @joriki we write
$$sum_q=0^a (-1)^q achoose q 2n-qchoose n-q
= sum_q=0^a (-1)^q achoose q [z^n-q] (1+z)^2n-q
\ = [z^n] (1+z)^2n
sum_q=0^a (-1)^q achoose q z^q (1+z)^-q
\ = [z^n] (1+z)^2n
left(1-fracz1+zright)^a
= [z^n] (1+z)^2n-a = 2n-achoose n.$$
answered Jul 25 at 17:49
Marko Riedel
36.4k333107
add a comment |Â
up vote
2
down vote
up vote
2
down vote
For an algebraic proof of the re-formulated identity by @joriki we write
$$sum_q=0^a (-1)^q achoose q 2n-qchoose n-q
= sum_q=0^a (-1)^q achoose q [z^n-q] (1+z)^2n-q
\ = [z^n] (1+z)^2n
sum_q=0^a (-1)^q achoose q z^q (1+z)^-q
\ = [z^n] (1+z)^2n
left(1-fracz1+zright)^a
= [z^n] (1+z)^2n-a = 2n-achoose n.$$
answered Jul 25 at 17:49
Marko Riedel
36.4k333107
For an algebraic proof of the re-formulated identity by @joriki we write
$$sum_q=0^a (-1)^q achoose q 2n-qchoose n-q
= sum_q=0^a (-1)^q achoose q [z^n-q] (1+z)^2n-q
\ = [z^n] (1+z)^2n
sum_q=0^a (-1)^q achoose q z^q (1+z)^-q
\ = [z^n] (1+z)^2n
left(1-fracz1+zright)^a
= [z^n] (1+z)^2n-a = 2n-achoose n.$$
answered Jul 25 at 17:49
Marko Riedel
36.4k333107
answered Jul 25 at 17:49
Marko Riedel
36.4k333107
answered Jul 25 at 17:49
Marko Riedel
36.4k333107
answered Jul 25 at 17:49
Marko Riedel
36.4k333107
36.4k333107
add a comment |Â
add a comment |Â
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