A079563 -id:A079563 - cp's OEIS frontend

A006256 a(n) = Sum_{k=0..n} binomial(3k,k)binomial(3n-3k,n-k).

1, 6, 39, 258, 1719, 11496, 77052, 517194, 3475071, 23366598, 157206519, 1058119992, 7124428836, 47983020624, 323240752272, 2177956129818, 14677216121871, 98923498131762, 666819212874501, 4495342330033938, 30308036621747679, 204356509814519712

Offset: 0

N. J. A. Sloane, Don Knuth

The right-hand sides of several of the "Ruehr identities". - N. J. A. Sloane, Feb 20 2020

Convolution of A005809 with itself. - Emeric Deutsch, May 22 2003

Allouche, J-P. "Two binomial identities of Ruehr Revisited." The American Mathematical Monthly 126.3 (2019): 217-225.
Alzer, Horst, and Helmut Prodinger. "On Ruehr's Identities." Ars Comb. 139 (2018): 247-254.
Bai, Mei, and Wenchang Chu. "Seven equivalent binomial sums." Discrete Mathematics 343.2 (2020): 111691.
M. Petkovsek et al., A=B, Peters, 1996, p. 165.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Jean-Paul Allouche, A generalization of an identity due to Kimura and Ruehr, arXiv preprint arXiv:1706.08929 [math.NT], 2017.
Jean-Paul Allouche, Two exercises of Comtet and two identities of Ruehr, arXiv preprint arXiv:1707.05751 [math.NT], 2017.
Rui Duarte and António Guedes de Oliveira, Short note on the convolution of binomial coefficients, arXiv:1302.2100 [math.CO], 2013 and J. Int. Seq. 16 (2013) #13.7.6.
Shalosh B. Ekhad, Doron Zeilberger, Some Remarks on a recent article by J.-P. Allouche, arXiv:1903.09511 [math.CO], 2019.
N, Kimura and O. G. Ruehr, Change of variable formula for definite integral. Problem E2765, Am. Math. Mnthly, 87, 1980, 307-308.
S. Meehan, A. Tefera, M. Weselcouch, A. Zeleke, Proofs of Ruehr's identities, Integers 14 (2014) A10.
D. Merlini, R. Sprugnoli and M. C. Verri, The tennis ball problem, J. Combin. Theory, A 99 (2002), 307-344 (Y_n for s=3).

Cf. A036829.
Cf. A005809, A160906, A183160, A226751.
Cf. A000302, A078995, A079563, A079678, A079679.
Cf. A141223, A385605, A385632.

a006256 n = a006256_list !! n
a006256_list = f (tail a005809_list) [1] where
   f (x:xs) zs = (sum $ zipWith (*) zs a005809_list) : f xs (x : zs)
-- Reinhard Zumkeller, Sep 21 2014

[&+[Binomial(3*k, k) *Binomial(3*n-3*k, n-k): k in [0..n]]:n in  [0..22]]; // Vincenzo Librandi, Feb 21 2020

a:= proc(n) option remember; `if`(n<2, 5*n+1,
      ((216*n^2-270*n+96) *a(n-1)
      -81*(3*n-2)*(3*n-4) *a(n-2)) /(n*(16*n-8)))
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Nov 07 2012

a[n_] := HypergeometricPFQ[{1/3, 2/3, 1/2-n, -n}, {1/2, 1/3-n, 2/3-n}, 1]*(3n)!/(n!*(2n)!); Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Jun 20 2012 *)
Table[Sum[Binomial[3k,k]Binomial[3n-3k,n-k],{k,0,n}],{n,0,30}] (* Harvey P. Dale, Oct 23 2013 *)

a(n)=sum(k=0,n, binomial(3*k,k)*binomial(3*n-3*k,n-k)) \\ Charles R Greathouse IV, Feb 07 2017

a = lambda n: binomial(3*n+1,n)*hypergeometric([1,-n],[2*n+2],-2)
[simplify(a(n)) for n in range(20)] # Peter Luschny, May 19 2015

a(n) = (3/4)*(27/4)^n*(1+c/sqrt(n)+o(n^(-1/2))) where c = (2/3)*sqrt(1/(3*Pi)) = 0.217156671956853298... More generally, a(n, m) = sum(k=0, n, C(m*k,k) *C(m*(n-k),n-k)) is asymptotic to (1/2)*m/(m-1)*(m^m/(m-1)^(m-1))^n. See A000302, A078995 for cases m=2 and 4. - Benoit Cloitre, Jan 26 2003, extended by Vaclav Kotesovec, Nov 06 2012
G.f.: 1/(1-3*z*g^2)^2, where g=g(z) is given by g=1+z*g^3, g(0)=1, i.e. (in Maple command) g := 2*sin(arcsin(3*sqrt(3*z)/2)/3)/sqrt(3*z). - Emeric Deutsch, May 22 2003
D-finite with recurrence: 6*(36*n^2-45*n+16)*a(n-1) - 81*(3*n-4)*(3*n-2)*a(n-2) - 8*n*(2*n-1)*a(n) = 0. - Vaclav Kotesovec, Oct 05 2012
From Rui Duarte and António Guedes de Oliveira, Feb 17 2013: (Start)
a(n) = sum(k=0, n, C(3*k+l,k)*C(3*(n-k)-l,n-k)) for every real number l.
a(n) = sum(k=0, n, 2^(n-k)*C(3n+1,k)).
a(n) = sum(k=0, n, 3^(n-k)*C(2n+k,k)).  (End)
From Akalu Tefera, Sean Meehan, Michael Weselcouch, and Aklilu Zeleke, May 11 2013: (Start)
a(n) = sum(k=0, 2n, (-3)^k*C(3n - k, n)).
a(n) = sum(k=0, 2n, (-4)^k*C(3n + 1, 2n - k)).
a(n) = sum(k=0, n, 3^k*C(3n - k, 2n)).
a(n) = sum(k=0, n, 2^k*C(3n + 1, n - k)).  (End)
a(n) = C(3*n+1,n)*Hyper2F1(1,-n,2*n+2,-2). - Peter Luschny, May 19 2015
a(n) = [x^n] 1/((1-3*x) * (1-x)^(2*n+1)). - Seiichi Manyama, Aug 03 2025
a(n) = Sum_{k=0..n} 3^k * (-2)^(n-k) * binomial(3*n+1,k) * binomial(3*n-k,n-k). - Seiichi Manyama, Aug 07 2025

A078995 a(n) = Sum_{k=0..n} C(4k,k)C(4*(n-k),n-k).

Original entry on oeis.org

1, 8, 72, 664, 6184, 57888, 543544, 5113872, 48180456, 454396000, 4288773152, 40503496536, 382701222296, 3617396099936, 34203591636048, 323492394385824, 3060238763412072, 28955508198895584, 274018698082833760, 2593539713410178528, 24550565251665845664

Offset: 0

N. J. A. Sloane, Jan 19 2003

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..200
D. Merlini, R. Sprugnoli and M. C. Verri, The tennis ball problem, J. Combin. Theory, A 99 (2002), 307-344 (Y_n for s=4).
Rui Duarte and António Guedes de Oliveira, Short note on the convolution of binomial coefficients, arXiv:1302.2100 [math.CO], 2013 and J. Int. Seq. 16 (2013) #13.7.6.

See A049235 for more information.
Cf. A006256, A079563, A079678, A079679.
Cf. A005810, A147855, A226733, A226761, A385605, A386811.

series(eval(g/(3*g-4), g=RootOf(g = 1+x*g^4,g))^2, x=0, 30); # Mark van Hoeij, May 06 2013

Table[Sum[Binomial[4*k, k]*Binomial[4*(n - k), n - k], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Nov 06 2012 *)

a(n) = sum(k=0, n, binomial(4*k, k)*binomial(4*(n-k), n-k)); \\ Michel Marcus, May 09 2020

a(n) = 2/3*(256/27)^n*(1+c/sqrt(n)+o(n^-1/2)) where c = 2/3*sqrt(2/(3*Pi)) = 0.307105910641187... More generally, a(n, m)=sum(k=0, n, binomial(m*k, k)*binomial(m*(n-k), n-k)) is asymptotic to 1/2*m/(m-1)*(m^m/(m-1)^(m-1))^n. See A000302, A006256 for cases m=2 and 3. - Benoit Cloitre, Jan 26 2003, corrected and extended by Vaclav Kotesovec, Nov 06 2012
243*n*(8*n - 17)*(3*n - 1)*(3*n - 4)*(3*n - 2)*(3*n - 5)*a(n) = 72*(3*n - 5)*(3*n - 4)*(6912*n^4 - 33120*n^3 + 58256*n^2 - 47798*n + 15309)*a(n - 1) - 3072*(2*n - 3)*(6912*n^5 - 55008*n^4 + 175696*n^3 - 282180*n^2 + 227825*n - 73710)*a(n - 2) + 262144*(n - 2)*(4*n - 7)*(2*n - 3)*(2*n - 5)*(4*n - 9)*(8*n - 9)*a(n - 3). - Vladeta Jovovic, Jul 16 2004
Shorter recurrence: 81*n*(3*n-2)*(3*n-1)*(8*n-11)*a(n) = 24*(4608*n^4-14400*n^3+15776*n^2-7346*n+1215)*a(n-1) - 2048*(2*n-3)*(4*n-5)*(4*n-3)*(8*n-3)*a(n-2). - Vaclav Kotesovec, Nov 06 2012
a(n) = Sum_{k=0..n} binomial(4*k+l,k) * binomial(4*(n-k)-l,n-k) for every real number l. - Rui Duarte and António Guedes de Oliveira, Feb 16 2013
From Rui Duarte and António Guedes de Oliveira, Feb 17 2013: (Start)
a(n) = Sum_{k=0..n} 3^(n-k) * binomial(4*n+1,k).
a(n) = Sum_{k=0..n} 4^(n-k) * binomial(3*n+k,k). (End)
G.f.: g^2/(3*g-4)^2 where g=ogf(A002293) satisfies g = 1+x*g^4. - Mark van Hoeij, May 06 2013
a(n) = [x^n] 1/((1-4*x) * (1-x)^(3*n+1)). - Seiichi Manyama, Aug 03 2025
a(n) = Sum_{k=0..n} 4^k * (-3)^(n-k) * binomial(4*n+1,k) * binomial(4*n-k,n-k). - Seiichi Manyama, Aug 15 2025

A079678 a(n) = a(n,m) = Sum_{k=0..n} binomial(mk,k)binomial(m*(n-k),n-k) for m=5.

Original entry on oeis.org

1, 10, 115, 1360, 16265, 195660, 2361925, 28577440, 346316645, 4201744870, 51023399190, 620022989200, 7538489480075, 91696845873760, 1115794688036920, 13581508654978560, 165357977228808925, 2013721466517360650, 24527742112263770425, 298805688708113438240, 3640695209795092874290

Offset: 0

Benoit Cloitre, Jan 26 2003

nonn

More generally : a(n,m)=sum(k=0,n,binomial(m*k,k)*binomial(m*(n-k),n-k)) is asymptotic to 1/2*m/(m-1)*(m^m/(m-1)^(m-1))^n. See A000302, A006256, A078995 for cases m=2,3 and 4.

Robert Israel, Table of n, a(n) for n = 0..828
D. Merlini, R. Sprugnoli and M. C. Verri, The tennis ball problem, J. Combin. Theory, A 99 (2002), 307-344.
Rui Duarte and António Guedes de Oliveira, Short note on the convolution of binomial coefficients, arXiv:1302.2100 [math.CO], 2013.

Cf. A006256, A078995, A079563, A079679.

Cf. A371753, A385632, A386812.

seq(add(binomial(5*k,k)*binomial(5*(n-k),n-k),k=0..n), n=0..30); # Robert Israel, Jul 16 2015

m = 5; Table[Sum[Binomial[m k, k] Binomial[m (n - k), n - k], {k, 0, n}], {n, 0, 17}] (* Michael De Vlieger, Sep 30 2015 *)

main(size)=my(k,n,m=5); concat(1,vector(size,n, sum(k=0,n, binomial(m*k,k)*binomial(m*(n-k),n-k)))) \\ Anders Hellström, Jul 16 2015

a(n) = sum(k=0,n,4^(n-k)*binomial(5*n+1,k));
vector(30, n, a(n-1)) \\ Altug Alkan, Sep 30 2015

a(n) = 5/8*(3125/256)^n*(1+c/sqrt(n)+o(n^-1/2)) where c=0.356...
c = sqrt(2)/sqrt(5*Pi) = 0.3568248232305542229... - Vaclav Kotesovec, May 25 2020
a(n) = Sum_{k=0..n} binomial(5*k+l,k) * binomial(5*(n-k)-l,n-k) for every real number l. - Rui Duarte and António Guedes de Oliveira, Feb 16 2013
From Rui Duarte and António Guedes de Oliveira, Feb 17 2013: (Start)
a(n) = Sum_{k=0..n} 4^(n-k) * binomial(5*n+1,k).
a(n) = Sum_{k=0..n} 5^(n-k) * binomial(4*n+k,k). (End)
G.f.: hypergeom([1/5, 2/5, 3/5, 4/5], [1/4, 1/2, 3/4], (3125/256)*x)^2 satisfies
((3125/2)*g^3*x^4-128*g^3*x^3)*g''''+((-3125*g^2*x^4+256*g^2*x^3)*g'+12500*g^3*x^3-576*g^3*x^2)*g'''+(-(9375/4)*g^2*x^4+192*g^2*x^3)*g''^2+(((28125/4)*g*x^4-576*g*x^3)*(g')^2+(-18750*g^2*x^3+864*g^2*x^2)*g'+22500*g^3*x^2-408*g^3*x)*g''+(-(46875/16)*x^4+240*x^3)*(g')^4+(9375*g*x^3-432*g*x^2)*(g')^3+(-11250*g^2*x^2+204*g^2*x)*(g')^2+(7500*g^3*x-12*g^3)*g'+120*g^4 = 0. - Robert Israel, Jul 16 2015
a(n) = [x^n] 1/((1-5*x) * (1-x)^(4*n+1)). - Seiichi Manyama, Aug 03 2025
From Seiichi Manyama, Aug 14 2025: (Start)
a(n) = Sum_{k=0..n} 5^k * (-4)^(n-k) * binomial(5*n+1,k) * binomial(5*n-k,n-k).
G.f.: g^2/(5-4*g)^2 where g = 1+x*g^5 is the g.f. of A002294. (End)

A079679 a(n) = a(n,m) = Sum_{k=0..n} binomial(mk,k)binomial(m*(n-k),n-k) for m=6.

Original entry on oeis.org

1, 12, 168, 2424, 35400, 520236, 7674144, 113482584, 1681028136, 24932533800, 370144424376, 5499182587416, 81748907485248, 1215834858032820, 18090048027643200, 269246037610828656, 4008495234662771688, 59692297399976544120, 889090275714779739120, 13245013739104555683600

Offset: 0

Benoit Cloitre, Jan 26 2003

nonn

More generally : a(n,m)=sum(k=0,n,binomial(m*k,k)*binomial(m*(n-k),n-k)) is asymptotic to 1/2*m/(m-1)*(m^m/(m-1)^(m-1))^n. See A000302, A006256, A078995 for cases m=2,3 and 4.

Seiichi Manyama, Table of n, a(n) for n = 0..851
D. Merlini, R. Sprugnoli and M. C. Verri, The tennis ball problem, J. Combin. Theory, A 99 (2002), 307-344.
Rui Duarte and António Guedes de Oliveira, Short note on the convolution of binomial coefficients, arXiv:1302.2100 [math.CO], 2013.

Cf. A000302, A006256, A078995, A079563, A079678.

Cf. A002295, A226705.

a(n) = sum(k=0,n,5^(n-k)*binomial(6*n+1,k));
vector(30, n, a(n-1)) \\  Altug Alkan, Sep 30 2015

a(n) = 3/5*(46656/3125)^n*(1+c/sqrt(n)+o(n^-1/2)) where c=0.388...
c = 8/(3*sqrt(15*Pi)) = 0.388461664210517... - Vaclav Kotesovec, May 25 2020
a(n) = Sum_{k=0..n} binomial(6*k+l,k)*binomial(6*(n-k)-l,n-k) for every real number l. - Rui Duarte and António Guedes de Oliveira, Feb 16 2013
From Rui Duarte and António Guedes de Oliveira, Feb 17 2013: (Start)
a(n) = Sum_{k=0..n} 5^(n-k) * binomial(6*n+1,k).
a(n) = Sum_{k=0..n} 6^(n-k) * binomial(5*n+k,k). (End)
G.f.: hypergeom([1/6, 1/3, 1/2, 2/3, 5/6],[1/5, 2/5, 3/5, 4/5],46656*x/3125)^2.  - Mark van Hoeij, Apr 19 2013
a(n) = [x^n] 1/((1-6*x) * (1-x)^(5*n+1)). - Seiichi Manyama, Aug 03 2025
From Seiichi Manyama, Aug 14 2025: (Start)
a(n) = Sum_{k=0..n} 6^k * (-5)^(n-k) * binomial(6*n+1,k) * binomial(6*n-k,n-k).
G.f.: g^2/(6-5*g)^2 where g = 1+x*g^6 is the g.f. of A002295. (End)

A358050 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) = Sum_{j=0..n} binomial(kj,j) binomial(k*(n-j),n-j).

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 4, 3, 0, 1, 6, 16, 4, 0, 1, 8, 39, 64, 5, 0, 1, 10, 72, 258, 256, 6, 0, 1, 12, 115, 664, 1719, 1024, 7, 0, 1, 14, 168, 1360, 6184, 11496, 4096, 8, 0, 1, 16, 231, 2424, 16265, 57888, 77052, 16384, 9, 0, 1, 18, 304, 3934, 35400, 195660, 543544, 517194, 65536, 10, 0

Offset: 0

Seiichi Manyama, Oct 31 2022

			Square array begins:
  1, 1,    1,     1,     1,      1, ...
  0, 2,    4,     6,     8,     10, ...
  0, 3,   16,    39,    72,    115, ...
  0, 4,   64,   258,   664,   1360, ...
  0, 5,  256,  1719,  6184,  16265, ...
  0, 6, 1024, 11496, 57888, 195660, ...

Column k=0-7 give: A000007, A001477(n+1), A000302, A006256, A078995, A079678, A079679, A079563.
Main diagonal gives A358145.
Cf. A358146.

T(n, k) = sum(j=0, n, binomial(k*j, j)*binomial(k*(n-j), n-j));

T(n, k) = sum(j=0, n, (k-1)^(n-j)*binomial(k*n+1, j));

T(n, k) = sum(j=0, n, k^(n-j)*binomial((k-1)*n+j, j));

T(n,k) = Sum_{j=0..n} (k-1)^(n-j) * binomial(k*n+1,j).

T(n,k) = Sum_{j=0..n} k^(n-j) * binomial((k-1)*n+j,j).

cp's OEIS Frontend

A006256 a(n) = Sum_{k=0..n} binomial(3*k,k)*binomial(3*n-3*k,n-k).

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A078995 a(n) = Sum_{k=0..n} C(4*k,k)*C(4*(n-k),n-k).

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A079678 a(n) = a(n,m) = Sum_{k=0..n} binomial(m*k,k)*binomial(m*(n-k),n-k) for m=5.

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A079679 a(n) = a(n,m) = Sum_{k=0..n} binomial(m*k,k)*binomial(m*(n-k),n-k) for m=6.

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A358050 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) = Sum_{j=0..n} binomial(k*j,j) * binomial(k*(n-j),n-j).

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A006256 a(n) = Sum_{k=0..n} binomial(3k,k)binomial(3n-3k,n-k).

A078995 a(n) = Sum_{k=0..n} C(4k,k)C(4*(n-k),n-k).

A079678 a(n) = a(n,m) = Sum_{k=0..n} binomial(mk,k)binomial(m*(n-k),n-k) for m=5.

A079679 a(n) = a(n,m) = Sum_{k=0..n} binomial(mk,k)binomial(m*(n-k),n-k) for m=6.

A358050 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) = Sum_{j=0..n} binomial(kj,j) binomial(k*(n-j),n-j).