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Evaluate: $sum_j=0^ksum_i=0^jk choose j^2k choose i$

Evaluate $sum_k=1^n (2k-1)n choose k$ using calculusShow that $sum_k=0^n(-1)^n+knchoose k(ak+b)_n=n!a^n$Show that $int_0^1ln(-lnx)cdotmathrm dxover 1+x^2=-sumlimits_n=0^infty1over 2n+1cdot2piover e^pi(2n+1)+1$ and evaluate itShowing that $sum_k=0^n(-1)^knchoose k1over k+1sum_j=0^kH_j+1over j+1=1over (n+1)^3$Is this sum $sum_k=0^nnchoose ksum_i=0^k(-1)^ikchoose iH_n+k-i=H_2n$ correct?Looking for the closed form of $sum_n=1^inftyzeta(2n+1)over (2n+1)2^4n$Evaluate $sumlimits_n=0^infty(-1)^nsumlimits_j=0^kk choose jfrac(-1)^j2n+2j+1$Concerning this sum $sum_n=0^inftyfrac14n+1left[frac14^n2n choose nright]^2=fracGamma^4left(frac14right)16pi^2$Looking at this sum $sum_n=0^inftyfrac(2n)!!(2n+1)!!frac2n choose n4^ng(n)$About this sum $sum_n=0^inftyfrac2n choose n^364^nf(n)$

5

0

$begingroup$

Can anybody help me to evaluate this sum $(1)$?

$$sum_j=0^ksum_i=0^jk choose j^2k choose itag1$$

I have manage to figure out:

$$sum_j=0^ksum_i=0^jk choose jk choose i=frac4^k+2k choose k2tag2$$
and

$$sum_j=0^ksum_i=0^jk+1 choose j+1^2k choose i=2^k2k choose kfrac2k+1k+1tag3$$

sequences-and-series binomial-theorem

share|cite|improve this question

asked Apr 1 at 15:50

coffeeecoffeee

18919

$endgroup$

add a comment | 

5

0

$begingroup$

Can anybody help me to evaluate this sum $(1)$?

$$sum_j=0^ksum_i=0^jk choose j^2k choose itag1$$

I have manage to figure out:

$$sum_j=0^ksum_i=0^jk choose jk choose i=frac4^k+2k choose k2tag2$$
and

$$sum_j=0^ksum_i=0^jk+1 choose j+1^2k choose i=2^k2k choose kfrac2k+1k+1tag3$$

sequences-and-series binomial-theorem

share|cite|improve this question

asked Apr 1 at 15:50

coffeeecoffeee

18919

$endgroup$

add a comment | 

$begingroup$

Can anybody help me to evaluate this sum $(1)$?

$$sum_j=0^ksum_i=0^jk choose j^2k choose itag1$$

I have manage to figure out:

$$sum_j=0^ksum_i=0^jk choose jk choose i=frac4^k+2k choose k2tag2$$
and

$$sum_j=0^ksum_i=0^jk+1 choose j+1^2k choose i=2^k2k choose kfrac2k+1k+1tag3$$

sequences-and-series binomial-theorem

share|cite|improve this question

asked Apr 1 at 15:50

coffeeecoffeee

18919

$endgroup$

share|cite|improve this question

asked Apr 1 at 15:50

coffeeecoffeee

18919

share|cite|improve this question

asked Apr 1 at 15:50

coffeeecoffeee

18919

asked Apr 1 at 15:50

coffeeecoffeee

18919

asked Apr 1 at 15:50

coffeeecoffeee

18919

1 Answer
1

active

oldest

votes

3

$begingroup$

We start with
beginalign*
colorbluesum_j=0^k&colorbluesum_i=0^kbinomkj^2binomki\
&=2^ksum_j=0^kbinomkjbinomkk-j\
&=2^ksum_j=0^kbinomkj[z^k-j](1+z)^ktag1\
&=2^k[z^k](1+z)^ksum_j=0^kbinomkjz^jtag2\
&=2^k[z^k](1+z)^2k\
&,,colorblue=2^kbinom2kktag3
endalign*

Comment:

• In (1) we use the coefficient of operator $[z^p]$ to denote the coefficient of $z^p$.




• In (2) we apply the rule $[z^p-q]A(z)=[z^p]z^qA(z)$.




• In (3) we select the coefficient of $z^k$.

We obtain from (3)

beginalign*
sum_j=0^k&sum_i=0^jbinomkj^2binomki=2^kbinom2kk-sum_j=0^ksum_i=j+1^kbinomkj^2binomkitag4
endalign*

The right-hand sum of (4) gives

beginalign*
colorbluesum_j=0^k&colorbluesum_i=j+1^kbinomkj^2binomki\
&=sum_j=0^ksum_i=k-j+1^kbinomkj^2binomkitag5\
&=sum_j=0^ksum_i=-j+1^0binomkj+k^2binomkitag6\
&=sum_j=0^ksum_i=0^j-1binomkk-j^2binomkitag7\
&=sum_j=0^ksum_i=0^j-1binomkj^2binomki\
&,,colorblue=sum_j=0^ksum_i=0^jbinomkj^2binomki-sum_j=0^kbinomkj^3tag8
endalign*

Comment:

• In (5) we change the order of the outer sum $jto k-j$.




• In (6) we shift the index $i$ by $k$.




• In (7) we replace $i$ with $-i$.

We conclude from (4) and (8)

beginalign*
colorbluesum_j=0^ksum_i=0^jbinomkj^2binomki=2^k-1binom2kk+frac12sum_j=0^kbinomkj^3
endalign*

with $sum_j=0^kbinomkj^3$ the Franel numbers stored as A000172 in OEIS.

share|cite|improve this answer

edited Apr 1 at 20:47

answered Apr 1 at 20:21

Markus ScheuerMarkus Scheuer

64k460152

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Interesting work. Verified. (+1).
  $endgroup$
  – Marko Riedel
  Apr 1 at 20:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MarkoRiedel: Many thanks for your friendly comment and the credit, Marko. :-)
  $endgroup$
  – Markus Scheuer
  Apr 1 at 20:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  An explanation to this answer is: let A be a matrix such that Aij = (kCi) * (kCj)^2, then the sum you want is the sum of lower triangular part of A = SL. Note that the sum of upper triangular part of A = SU is also equal to SL. Finally, we have SA=SU+SL-SD, where SA is the sum of all elements in A, and SD is the sum of Diagonal of A.
  $endgroup$
  – DragunityMAX
  Apr 1 at 22:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your wonderful answer. I have check the link on Franel number, I am still curious, does this sum $sum_j=0^kk choose j^3$ have a closed form?
  $endgroup$
  – coffeee
  Apr 2 at 5:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I manage to find a closed form for this sum (similar to Franel) $$sum_j=0^kleft[2k choose 2j^3-2k choose 2j-1^3right]=(-1)^k2k choose k3k choose k$$
  $endgroup$
  – coffeee
  Apr 2 at 5:31
  

  
  
  
  

 | 
show 2 more comments

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3

$begingroup$

We start with
beginalign*
colorbluesum_j=0^k&colorbluesum_i=0^kbinomkj^2binomki\
&=2^ksum_j=0^kbinomkjbinomkk-j\
&=2^ksum_j=0^kbinomkj[z^k-j](1+z)^ktag1\
&=2^k[z^k](1+z)^ksum_j=0^kbinomkjz^jtag2\
&=2^k[z^k](1+z)^2k\
&,,colorblue=2^kbinom2kktag3
endalign*

Comment:

• In (1) we use the coefficient of operator $[z^p]$ to denote the coefficient of $z^p$.




• In (2) we apply the rule $[z^p-q]A(z)=[z^p]z^qA(z)$.




• In (3) we select the coefficient of $z^k$.

We obtain from (3)

beginalign*
sum_j=0^k&sum_i=0^jbinomkj^2binomki=2^kbinom2kk-sum_j=0^ksum_i=j+1^kbinomkj^2binomkitag4
endalign*

The right-hand sum of (4) gives

beginalign*
colorbluesum_j=0^k&colorbluesum_i=j+1^kbinomkj^2binomki\
&=sum_j=0^ksum_i=k-j+1^kbinomkj^2binomkitag5\
&=sum_j=0^ksum_i=-j+1^0binomkj+k^2binomkitag6\
&=sum_j=0^ksum_i=0^j-1binomkk-j^2binomkitag7\
&=sum_j=0^ksum_i=0^j-1binomkj^2binomki\
&,,colorblue=sum_j=0^ksum_i=0^jbinomkj^2binomki-sum_j=0^kbinomkj^3tag8
endalign*

Comment:

• In (5) we change the order of the outer sum $jto k-j$.




• In (6) we shift the index $i$ by $k$.




• In (7) we replace $i$ with $-i$.

We conclude from (4) and (8)

beginalign*
colorbluesum_j=0^ksum_i=0^jbinomkj^2binomki=2^k-1binom2kk+frac12sum_j=0^kbinomkj^3
endalign*

with $sum_j=0^kbinomkj^3$ the Franel numbers stored as A000172 in OEIS.

share|cite|improve this answer

edited Apr 1 at 20:47

answered Apr 1 at 20:21

Markus ScheuerMarkus Scheuer

64k460152

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Interesting work. Verified. (+1).
  $endgroup$
  – Marko Riedel
  Apr 1 at 20:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MarkoRiedel: Many thanks for your friendly comment and the credit, Marko. :-)
  $endgroup$
  – Markus Scheuer
  Apr 1 at 20:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  An explanation to this answer is: let A be a matrix such that Aij = (kCi) * (kCj)^2, then the sum you want is the sum of lower triangular part of A = SL. Note that the sum of upper triangular part of A = SU is also equal to SL. Finally, we have SA=SU+SL-SD, where SA is the sum of all elements in A, and SD is the sum of Diagonal of A.
  $endgroup$
  – DragunityMAX
  Apr 1 at 22:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your wonderful answer. I have check the link on Franel number, I am still curious, does this sum $sum_j=0^kk choose j^3$ have a closed form?
  $endgroup$
  – coffeee
  Apr 2 at 5:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I manage to find a closed form for this sum (similar to Franel) $$sum_j=0^kleft[2k choose 2j^3-2k choose 2j-1^3right]=(-1)^k2k choose k3k choose k$$
  $endgroup$
  – coffeee
  Apr 2 at 5:31
  

  
  
  
  

 | 
show 2 more comments

3

$begingroup$

We start with
beginalign*
colorbluesum_j=0^k&colorbluesum_i=0^kbinomkj^2binomki\
&=2^ksum_j=0^kbinomkjbinomkk-j\
&=2^ksum_j=0^kbinomkj[z^k-j](1+z)^ktag1\
&=2^k[z^k](1+z)^ksum_j=0^kbinomkjz^jtag2\
&=2^k[z^k](1+z)^2k\
&,,colorblue=2^kbinom2kktag3
endalign*

Comment:

• In (1) we use the coefficient of operator $[z^p]$ to denote the coefficient of $z^p$.




• In (2) we apply the rule $[z^p-q]A(z)=[z^p]z^qA(z)$.




• In (3) we select the coefficient of $z^k$.

We obtain from (3)

beginalign*
sum_j=0^k&sum_i=0^jbinomkj^2binomki=2^kbinom2kk-sum_j=0^ksum_i=j+1^kbinomkj^2binomkitag4
endalign*

The right-hand sum of (4) gives

beginalign*
colorbluesum_j=0^k&colorbluesum_i=j+1^kbinomkj^2binomki\
&=sum_j=0^ksum_i=k-j+1^kbinomkj^2binomkitag5\
&=sum_j=0^ksum_i=-j+1^0binomkj+k^2binomkitag6\
&=sum_j=0^ksum_i=0^j-1binomkk-j^2binomkitag7\
&=sum_j=0^ksum_i=0^j-1binomkj^2binomki\
&,,colorblue=sum_j=0^ksum_i=0^jbinomkj^2binomki-sum_j=0^kbinomkj^3tag8
endalign*

Comment:

• In (5) we change the order of the outer sum $jto k-j$.




• In (6) we shift the index $i$ by $k$.




• In (7) we replace $i$ with $-i$.

We conclude from (4) and (8)

beginalign*
colorbluesum_j=0^ksum_i=0^jbinomkj^2binomki=2^k-1binom2kk+frac12sum_j=0^kbinomkj^3
endalign*

with $sum_j=0^kbinomkj^3$ the Franel numbers stored as A000172 in OEIS.

share|cite|improve this answer

edited Apr 1 at 20:47

answered Apr 1 at 20:21

Markus ScheuerMarkus Scheuer

64k460152

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Interesting work. Verified. (+1).
  $endgroup$
  – Marko Riedel
  Apr 1 at 20:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MarkoRiedel: Many thanks for your friendly comment and the credit, Marko. :-)
  $endgroup$
  – Markus Scheuer
  Apr 1 at 20:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  An explanation to this answer is: let A be a matrix such that Aij = (kCi) * (kCj)^2, then the sum you want is the sum of lower triangular part of A = SL. Note that the sum of upper triangular part of A = SU is also equal to SL. Finally, we have SA=SU+SL-SD, where SA is the sum of all elements in A, and SD is the sum of Diagonal of A.
  $endgroup$
  – DragunityMAX
  Apr 1 at 22:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your wonderful answer. I have check the link on Franel number, I am still curious, does this sum $sum_j=0^kk choose j^3$ have a closed form?
  $endgroup$
  – coffeee
  Apr 2 at 5:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I manage to find a closed form for this sum (similar to Franel) $$sum_j=0^kleft[2k choose 2j^3-2k choose 2j-1^3right]=(-1)^k2k choose k3k choose k$$
  $endgroup$
  – coffeee
  Apr 2 at 5:31
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

We start with
beginalign*
colorbluesum_j=0^k&colorbluesum_i=0^kbinomkj^2binomki\
&=2^ksum_j=0^kbinomkjbinomkk-j\
&=2^ksum_j=0^kbinomkj[z^k-j](1+z)^ktag1\
&=2^k[z^k](1+z)^ksum_j=0^kbinomkjz^jtag2\
&=2^k[z^k](1+z)^2k\
&,,colorblue=2^kbinom2kktag3
endalign*

Comment:

• In (1) we use the coefficient of operator $[z^p]$ to denote the coefficient of $z^p$.




• In (2) we apply the rule $[z^p-q]A(z)=[z^p]z^qA(z)$.




• In (3) we select the coefficient of $z^k$.

We obtain from (3)

beginalign*
sum_j=0^k&sum_i=0^jbinomkj^2binomki=2^kbinom2kk-sum_j=0^ksum_i=j+1^kbinomkj^2binomkitag4
endalign*

The right-hand sum of (4) gives

beginalign*
colorbluesum_j=0^k&colorbluesum_i=j+1^kbinomkj^2binomki\
&=sum_j=0^ksum_i=k-j+1^kbinomkj^2binomkitag5\
&=sum_j=0^ksum_i=-j+1^0binomkj+k^2binomkitag6\
&=sum_j=0^ksum_i=0^j-1binomkk-j^2binomkitag7\
&=sum_j=0^ksum_i=0^j-1binomkj^2binomki\
&,,colorblue=sum_j=0^ksum_i=0^jbinomkj^2binomki-sum_j=0^kbinomkj^3tag8
endalign*

Comment:

• In (5) we change the order of the outer sum $jto k-j$.




• In (6) we shift the index $i$ by $k$.




• In (7) we replace $i$ with $-i$.

We conclude from (4) and (8)

beginalign*
colorbluesum_j=0^ksum_i=0^jbinomkj^2binomki=2^k-1binom2kk+frac12sum_j=0^kbinomkj^3
endalign*

with $sum_j=0^kbinomkj^3$ the Franel numbers stored as A000172 in OEIS.

share|cite|improve this answer

edited Apr 1 at 20:47

answered Apr 1 at 20:21

Markus ScheuerMarkus Scheuer

64k460152

$endgroup$

share|cite|improve this answer

edited Apr 1 at 20:47

answered Apr 1 at 20:21

Markus ScheuerMarkus Scheuer

64k460152

answered Apr 1 at 20:21

Markus ScheuerMarkus Scheuer

64k460152

answered Apr 1 at 20:21

Markus ScheuerMarkus Scheuer

64k460152