A005260 -id:A005260 - cp's OEIS frontend

A182421 a(n) = Sum_{k = 0..n} C(n,k)^7.

1, 2, 130, 4376, 312706, 20156252, 1622278624, 132282417920, 11716609750402, 1067553850832372, 101275413131018380, 9844149854624122160, 980597565209377223200, 99518148302583383833280, 10272819477206557916630080, 1075608762378043981968997376

Offset: 0

Vaclav Kotesovec, Apr 28 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Vaclav Kotesovec, Recurrence (of order 4)
M. A. Perlstadt, Some Recurrences for Sums of Powers of Binomial Coefficients, Journal of Number Theory 27 (1987), pp. 304-309.
Jin Yuan, Zhi-Juan Lu, Asmus L. Schmidt, On recurrences for sums of powers of binomial coefficients, J. Numb. Theory 128 (2008) 2784-2794

Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.

a := n -> hypergeom([seq(-n, i=1..7)],[seq(1, i=1..6)],-1):
seq(simplify(a(n)),n=0..15); # Peter Luschny, Jul 27 2016

Table[Total[Binomial[n, Range[0, n]]^7], {n, 0, 20}] (* T. D. Noe, Apr 28 2012 *)

a(n) = sum(k=0, n, binomial(n,k)^7); \\ Michel Marcus, Jul 17 2020

Asymptotic (p = 7): a(n) ~ 2^(p*n)/sqrt(p)*(2/(Pi*n))^((p - 1)/2)*( 1 - (p - 1)^2/(4*p*n) + O(1/n^2) ).
For r a nonnegative integer, Sum_{k = r..n} C(k,r)^7*C(n,k)^7 = C(n,r)^7*a(n-r), where we take a(n) = 0 for n < 0. - Peter Bala, Jul 27 2016
Sum_{n>=0} a(n) * x^n / (n!)^7 = (Sum_{n>=0} x^n / (n!)^7)^2. - Ilya Gutkovskiy, Jul 17 2020

A182422 a(n) = Sum_{k = 0..n} C(n,k)^8.

Original entry on oeis.org

1, 2, 258, 13124, 1810690, 200781252, 30729140484, 4579408029576, 770670360699138, 132018919625044100, 23913739057463037508, 4433505541977804821256, 848185646293853978499844, 165563367990287610967653512, 32993144260428865295508700680

Offset: 0

Vaclav Kotesovec, Apr 28 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Vaclav Kotesovec, Recurrence (of order 4)
M. A. Perlstadt, Some Recurrences for Sums of Powers of Binomial Coefficients, Journal of Number Theory 27 (1987), pp. 304-309.

Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.

a := n -> hypergeom([seq(-n, i=1..8)],[seq(1, i=1..7)],1):
seq(simplify(a(n)),n=0..14); # Peter Luschny, Jul 27 2016

Table[Total[Binomial[n, Range[0, n]]^8], {n, 0, 20}] (* T. D. Noe, Apr 28 2012 *)

a(n) = sum(k=0, n, binomial(n,k)^8); \\ Michel Marcus, Jul 17 2020

Asymptotic (p = 8): a(n) ~ 2^(p*n)/sqrt(p)*(2/(Pi*n))^((p - 1)/2)*( 1 - (p - 1)^2/(4*p*n) + O(1/n^2) ).
For r a nonnegative integer, Sum_{k = r..n} C(k,r)^8*C(n,k)^8 = C(n,r)^8*a(n-r), where we take a(n) = 0 for n < 0. - Peter Bala, Jul 27 2016
Sum_{n>=0} a(n) * x^n / (n!)^8 = (Sum_{n>=0} x^n / (n!)^8)^2. - Ilya Gutkovskiy, Jul 17 2020

A182446 a(n) = Sum_{k = 0..n} C(n,k)^9.

Original entry on oeis.org

1, 2, 514, 39368, 10601986, 2003906252, 588906874144, 159219918144128, 51207103076632066, 16425660314368351892, 5697191745563573732764, 2010823973962863400708688, 739753103704422167184400096, 277511604090132008416695054272, 106814999715696983804826836579584

Offset: 0

Vaclav Kotesovec, Apr 29 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Vaclav Kotesovec, Recurrence (of order 5)
M. A. Perlstadt, Some Recurrences for Sums of Powers of Binomial Coefficients, Journal of Number Theory 27 (1987), pp. 304-309.

Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.

a := n -> hypergeom([seq(-n, i=1..9)],[seq(1, i=1..8)],-1):
seq(simplify(a(n)),n=0..14); # Peter Luschny, Jul 27 2016

Table[Sum[Binomial[n, k]^9, {k, 0, n}], {n, 0, 25}]

a(n) = sum(k=0, n, binomial(n,k)^9); \\ Michel Marcus, Jul 17 2020

Asymptotic (p = 9): a(n) ~ 2^(p*n)/sqrt(p)*(2/(Pi*n))^((p - 1)/2)*( 1 - (p - 1)^2/(4*p*n) + O(1/n^2) ).
For r a nonnegative integer, Sum_{k = r..n} C(k,r)^9*C(n,k)^9 = C(n,r)^9*a(n-r), where we take a(n) = 0 for n < 0. - Peter Bala, Jul 27 2016
Sum_{n>=0} a(n) * x^n / (n!)^9 = (Sum_{n>=0} x^n / (n!)^9)^2. - Ilya Gutkovskiy, Jul 17 2020

A182447 a(n) = Sum_{k = 0..n} C(n,k)^10.

Original entry on oeis.org

1, 2, 1026, 118100, 62563330, 20019531252, 11393421713604, 5550455033938152, 3431955863873102850, 2052124795850957537060, 1367610300690018553312276, 916694195766256069610158152, 649630217578404016288230718276, 467800319852823195772146025385000

Offset: 0

Vaclav Kotesovec, Apr 29 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Vaclav Kotesovec, Recurrence (of order 5)
M. A. Perlstadt, Some Recurrences for Sums of Powers of Binomial Coefficients, Journal of Number Theory 27 (1987), pp. 304-309.

Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.

a := n -> hypergeom([seq(-n, i=1..10)],[seq(1, i=1..9)],1):
seq(simplify(a(n)),n=0..13); # Peter Luschny, Jul 27 2016

Table[Sum[Binomial[n, k]^10, {k, 0, n}], {n, 0, 25}]

a(n) = sum(k=0, n, binomial(n,k)^10); \\ Michel Marcus, Jul 17 2020

Asymptotic (p = 10): a(n) ~ 2^(p*n)/sqrt(p)*(2/(Pi*n))^((p - 1)/2)*( 1 - (p - 1)^2/(4*p*n) + O(1/n^2) ).
For r a nonnegative integer, Sum_{k = r..n} C(k,r)^10*C(n,k)^10 = C(n,r)^10*a(n-r), where we take a(n) = 0 for n < 0. - Peter Bala, Jul 27 2016
Sum_{n>=0} a(n) * x^n / (n!)^10 = (Sum_{n>=0} x^n / (n!)^10)^2. - Ilya Gutkovskiy, Jul 17 2020

A309010 Square array A(n, k) = Sum_{j=0..n} binomial(n,j)^k, n >= 0, k >= 0, read by antidiagonals.

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 1, 2, 4, 4, 1, 2, 6, 8, 5, 1, 2, 10, 20, 16, 6, 1, 2, 18, 56, 70, 32, 7, 1, 2, 34, 164, 346, 252, 64, 8, 1, 2, 66, 488, 1810, 2252, 924, 128, 9, 1, 2, 130, 1460, 9826, 21252, 15184, 3432, 256, 10, 1, 2, 258, 4376, 54850, 206252, 263844, 104960, 12870, 512, 11

Offset: 0

Seiichi Manyama, Jul 06 2019

A(n,k) is the constant term in the expansion of (Product_{j=1..k-1} (1 + x_j) + Product_{j=1..k-1} (1 + 1/x_j))^n for k > 0. - Seiichi Manyama, Oct 27 2019
Let B_k be the binomial poset containing all k-tuples of equinumerous subsets of {1,2,...} ordered by inclusion componentwise (described in Stanley reference below).  Then A(k,n) is the number of elements in any n-interval of B_k. - Geoffrey Critzer, Apr 16 2020
Column k is the diagonal of the rational function 1 / (Product_{j=1..k} (1-x_j) - Product_{j=1..k} x_j) for k>0. - Seiichi Manyama, Jul 11 2020

			Square array, A(n, k), begins:
   1,  1,   1,    1,     1,      1, ... A000012;
   2,  2,   2,    2,     2,      2, ... A007395;
   3,  4,   6,   10,    18,     34, ... A052548;
   4,  8,  20,   56,   164,    488, ... A115099;
   5, 16,  70,  346,  1810,   9826, ...
   6, 32, 252, 2252, 21252, 206252, ...
Antidiagonals, T(n, k), begin:
  1;
  1,  2;
  1,  2,   3;
  1,  2,   4,    4;
  1,  2,   6,    8,    5;
  1,  2,  10,   20,   16,     6;
  1,  2,  18,   56,   70,    32,     7;
  1,  2,  34,  164,  346,   252,    64,    8;
  1,  2,  66,  488, 1810,  2252,   924,  128,   9;
  1,  2, 130, 1460, 9826, 21252, 15184, 3432, 256,  10;

R. P. Stanley, Enumerative Combinatorics Vol I, Second Edition, Cambridge, 2011, Example 3.18.3 d, page 366.

Seiichi Manyama, Antidiagonals n = 0..100, flattened

Columns k=0..12 give A000027(n+1), A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.
Main diagonal gives A167010.
Cf. A328747, A328748, A328807, A328812.
Cf. A000012, A007395, A052548, A115099.

[(&+[Binomial(k,j)^(n-k): j in [0..k]]): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 26 2022

nn = 8; Table[ek[x_] := Sum[x^n/n!^k, {n, 0, nn}];Range[0, nn]!^k CoefficientList[Series[ek[x]^2, {x, 0, nn}],x], {k, 0, nn}] // Transpose // Grid (* Geoffrey Critzer, Apr 17 2020 *)

A(n, k) = sum(j=0, n, binomial(n, j)^k); \\ Seiichi Manyama, Jan 08 2022

flatten([[sum(binomial(k,j)^(n-k) for j in (0..k)) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Aug 26 2022

A(n, k) = Sum_{j=0..n} binomial(n,j)^k (array).
A(n, n+1) = A328812(n).
A(n, n) = A167010(n).
T(n, k) = A(k, n-k) (antidiagonals).
T(n, n) = A000027(n+1).
T(n, n-1) = A000079(n-1).
T(n, n-2) = A000984(n-2).
T(n, n-3) = A000172(n-3).
T(n, n-4) = A005260(n-4).
T(n, n-5) = A005261(n-5).
T(n, n-6) = A069865(n-6).
T(n, n-7) = A182421(n-7).
T(n, n-8) = A182422(n-8).
T(n, n-9) = A182446(n-9).
T(n, n-10) = A182447(n-10).
T(n, n-11) = A342294(n-11).
T(n, n-12) = A342295(n-12).
Sum_{n>=0} A(n,k) x^n/(n!^k) = (Sum_{n>=0} x^n/(n!^k))^2. - Geoffrey Critzer, Apr 17 2020

A342294 a(n) = Sum_{k = 0..n} binomial(n,k)^11.

Original entry on oeis.org

1, 2, 2050, 354296, 371185666, 200097656252, 222100237312864, 193798873701831680, 231719476114879600642, 257097895846251291074612, 330463219813679264204224300, 419460465362069257397304825200, 573863850341313751827291703127200

Offset: 0

N. J. A. Sloane, Mar 27 2021

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..306
M. A. Perlstadt, Some Recurrences for Sums of Powers of Binomial Coefficients, Journal of Number Theory 27 (1987), pp. 304-309.

Column 11 of A309010.

Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.

Table[Sum[Binomial[n,k]^11,{k,0,n}],{n,0,15}] (* Harvey P. Dale, May 08 2025 *)

a(n) = sum(k=0, n, binomial(n, k)^11); \\ Michel Marcus, Mar 27 2021

a(n) ~ 2^(p*n)/sqrt(p) * (2/(Pi*n))^((p-1)/2) * (1 - (p-1)^2/(4*p*n)), set p=11. - Vaclav Kotesovec, Aug 04 2022

A342295 a(n) = Sum_{k = 0..n} binomial(n,k)^12.

Original entry on oeis.org

1, 2, 4098, 1062884, 2210336770, 2000488281252, 4355497029345924, 6773152698818628936, 15744083665278445490178, 32270900877696351763796420, 80314784333143089874093429348, 192454957455454582636391397662856, 509571049488109525160616367158261124

Offset: 0

N. J. A. Sloane, Mar 27 2021

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..281
M. A. Perlstadt, Some Recurrences for Sums of Powers of Binomial Coefficients, Journal of Number Theory 27 (1987), pp. 304-309.

Column 12 of A309010.

Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.

a(n) = sum(k=0, n, binomial(n, k)^12); \\ Michel Marcus, Mar 27 2021

a(n) ~ 2^(p*n)/sqrt(p) * (2/(Pi*n))^((p-1)/2) * (1 - (p-1)^2/(4*p*n)), set p=12. - Vaclav Kotesovec, Aug 04 2022

A198256 Row sums of A197653.

Original entry on oeis.org

1, 5, 46, 485, 5626, 69062, 882540, 11614437, 156343330, 2142556130, 29791689148, 419260001030, 5960334608788, 85469709312860, 1234797737654296, 17955907741675749, 262607675818816050, 3860239468267647914, 57002176852356800700, 845159480056345448610

Offset: 0

Susanne Wienand, Oct 22 2011

nonn

Number of meanders of length (n+1)*4 which are composed by arcs of equal length and a central angle of 90 degrees.
Definition of a meander:
A binary curve C is a triple (m, S, dir) such that
(a) S is a list with values in {L,R} which starts with an L,
(b) dir is a list of m different values, each value of S being allocated a value of dir,
(c) consecutive Ls increment the index of dir,
(d) consecutive Rs decrement the index of dir,
(e) the integer m > 0 divides the length of S and
(f) C is a meander if each value of dir occurs length(S)/m times.
For this sequence, m = 4.
The terms are proved by brute force for 0 <= n <= 8, but not yet in general. - Susanne Wienand, Oct 29 2011

			Some examples of list S and allocated values of dir if n = 4:
Length(S) = (4+1)*4 = 20.
  S: L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L
dir: 1,2,3,0,1,2,3,0,1,2,3,0,1,2,3,0,1,2,3,0
  S: L,L,L,L,R,L,R,R,L,R,R,R,L,R,L,L,R,L,L,L
dir: 1,2,3,0,0,0,0,3,3,3,2,1,1,1,1,2,2,2,3,0
  S: L,L,R,L,L,L,L,R,R,L,R,R,L,R,L,L,L,L,R,R
dir: 1,2,2,2,3,0,1,1,0,0,0,3,3,3,3,0,1,2,2,1
Each value of dir occurs 20/4 = 5 times.

Peter Luschny, Meanders and walks on the circle.

Cf. A197653, A198060, A198257, A198258.
Cf. A005260, A181069.
Cf. A181067, A374614.

A198256[n_] := Sum[Sum[Sum[(-1)^(j + i)* Binomial[i, j]*Binomial[n, k]^4*(n + 1)^j*(k + 1)^(3 - j)/(k + 1)^3, {i, 0, 3}], {j, 0, 3}], {k, 0, n}]; Table[A198256[n], {n, 0, 16}] (* Peter Luschny, Nov 02 2011 *)

A198256(n) = {sum(k=0,n,if(n == 1+2*k,4,(1+k)*(1-((n-k)/(1+k))^4)/(1+2*k-n))*binomial(n,k)^4)} \\ Peter Luschny, Nov 24 2011

from mpmath import mp, hyper
def A198256(n) : return hyper([1-n, 1-n, 1-n, 1-n], [3, 3, 3], 1)*(n^4-n^6)/4 + hyper([-n, -n, -n, -n], [2, 2, 2], 1)*(1+n+n^2+n^3) + hyper([2, 1-n, 1-n, 1-n, 1-n], [1, 3, 3, 3], 1)*(n^4+n^5)/4
mp.dps = 32
for n in (0..19) : print(int(A198256(n)))  # Peter Luschny, Oct 24 2011

a(n) = Sum_{k=0..n} Sum_{j=0..3} Sum_{i=0..3} (-1)^(j+i)*C(i,j)*C(n,k)^4*(n+1)^j*(k+1)^(3-j)/(k+1)^3. - Peter Luschny, Nov 02 2011
a(n) = Sum_{k=0..n} h(n,k)*binomial(n,k)^4, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^4)/(1+2*k-n) if 1+2*k-n <> 0 else h(n,k) = 4. - Peter Luschny, Nov 24 2011
From Peter Bala, Mar 21 2023: (Start)
Conjecture 1: a(n) = Sum_{k = 0..n} binomial(n+1,k)^2*binomial(n,k)^2.
If true, then we have the third-order recurrence equation
n^2*(n + 1)^3*P(n-1)*a(n) = 2*n^2*(400*n^8 - 1260*n^7 + 20*n^6 + 3020*n^5 - 1646*n^4 - 1951*n^3 + 1142*n^2 + 465*n - 290)*a(n-1) + 4*(n - 1)*(2800*n^9 - 15420*n^8 + 30620*n^7 - 23710*n^6 + 808*n^5 + 6863*n^4 - 1309*n^3 - 1218*n^2 + 496*n - 60)*a(n-2) + 8*(n - 2)^2*(2*n - 3)*(4*n - 5)*(4*n - 7)*P(n)*a(n-3) with a(0) = 1, a(1) = 5 and a(2) = 46 and where P(n) = 100*n^5 - 65*n^4 - 35*n^3 + 25*n^2 + 6*n - 5.
Conjecture 2: working with offset 1, that is, a(1) = 1, a(2) = 5, ..., then the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for positive integers n and r and all primes p >= 5. (End)
a(n) ~ 2^(4*n + 5/2) / (Pi*n)^(3/2). - Vaclav Kotesovec, Apr 17 2023
Peter Bala's conjecture 1 can equivalently written: a(n) = hypergeom([-n - 1, -n - 1, -n, -n], [1, 1, 1], 1). - Detlef Meya, May 28 2024
a(n) = Sum_{k=0..n+1} (k/(n+1))^2 * binomial(n+1,k)^4. - Seiichi Manyama, Jul 14 2024

A328725 Constant term in the expansion of (1 + x + y + z + 1/x + 1/y + 1/z + xy + yz + zx + 1/(xy) + 1/(yz) + 1/(zx) + xyz + 1/(xyz))^n.

Original entry on oeis.org

1, 1, 15, 115, 1255, 13671, 160461, 1936425, 24071895, 305313415, 3939158905, 51521082405, 681635916325, 9105864515125, 122657982366375, 1664151758259915, 22720725637684215, 311933068664333175, 4303704125389134825, 59640225721889127525, 829774531966386480705

Offset: 0

Seiichi Manyama, Oct 26 2019

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..854

Sum_{i=0..n} (-1)^(n-i)*binomial(n,i)*Sum_{j=0..i} binomial(i,j)^m: A002426 (m=2), A172634 (m=3), this sequence (m=4), A328750 (m=5).

Cf. A002899, A005260, A328713.

{a(n) = polcoef(polcoef(polcoef((-1+(1+x)*(1+y)*(1+z)+(1+1/x)*(1+1/y)*(1+1/z))^n, 0), 0), 0)}

{a(n) = sum(i=0, n, (-1)^(n-i)*binomial(n,i)*sum(j=0, i, binomial(i, j)^4))}

a(n) = Sum_{i=0..n} (-1)^(n-i)*binomial(n,i)*Sum_{j=0..i} binomial(i,j)^4.
From Vaclav Kotesovec, Oct 28 2019: (Start)
Recurrence: n^3*a(n) = (2*n - 1)^3*a(n-1) + (n-1)*(94*n^2 - 188*n + 93)*a(n-2) + 80*(n-2)*(n-1)*(2*n - 3)*a(n-3) + 75*(n-3)*(n-2)*(n-1)*a(n-4).
a(n) ~ 15^(n + 3/2) / (2^(11/2) * Pi^(3/2) * n^(3/2)). (End)

A181069 Expansion of l.g.f. Sum_{n>=1} [ Sum_{k>=0} C(n+k-1,k)^4 x^k ] x^n/n.

Original entry on oeis.org

1, 3, 28, 275, 3126, 37632, 475056, 6192531, 82754650, 1127504378, 15603575208, 218727171104, 3099183987004, 44315462038200, 638663235342528, 9267264584278419, 135279095477748642, 1985221072388231742

Offset: 1

Paul D. Hanna, Oct 08 2010

nonn

			L.g.f.: L(x) = x + 3*x^2/2 + 28*x^3/3 + 275*x^4/4 + 3126*x^5/5 +...
which equals the series:
  L(x) = (1 + x + x^2 + x^3 + x^4 + x^5 + x^6 +...)*x
  + (1 + 2^4*x +  3^4*x^2 +  4^4*x^3 +   5^4*x^4 +   6^4*x^5 + ...)*x^2/2
  + (1 + 3^4*x +  6^4*x^2 + 10^4*x^3 +  15^4*x^4 +  21^4*x^5 + ...)*x^3/3
  + (1 + 4^4*x + 10^4*x^2 + 20^4*x^3 +  35^4*x^4 +  56^4*x^5 + ...)*x^4/4
  + (1 + 5^4*x + 15^4*x^2 + 35^4*x^3 +  70^4*x^4 + 126^4*x^5 + ...)*x^5/5
  + (1 + 6^4*x + 21^4*x^2 + 56^4*x^3 + 126^4*x^4 + 252^4*x^5 + ...)*x^6/6
  + (1 + 7^4*x + 28^4*x^2 + 84^4*x^3 + 210^4*x^4 + 462^4*x^5 + ...)*x^7/7 + ...
Exponentiation yields the g.f. of A181068:
  exp(L(x)) = 1 + x + 2*x^2 + 11*x^3 + 80*x^4 + 714*x^5 + 7095*x^6 +...

G. C. Greubel, Table of n, a(n) for n = 1..500

Cf. A181067 (variant), A181068.

Cf. A005260, A198256.

[(&+[Binomial(n,k)*Binomial(n-1, k)^3: k in [0..n-1]]): n in [1..20]]; // G. C. Greubel, Apr 05 2021

A181069:= n-> add( binomial(n,k)*binomial(n-1,k)^3, k=0..n-1); seq(A181069(n), n=1..20); # G. C. Greubel, Apr 05 2021

Table[Sum[Binomial[n-1,k]^3 * Binomial[n,k],{k,0,n-1}],{n,1,20}] (* Vaclav Kotesovec, Mar 06 2014 *)
a[n_] := HypergeometricPFQ[{-n + 1, -n + 1, -n + 1, -n}, {1, 1, 1}, 1];Table[a[n], {n, 1, 18}] (* Detlef Meya, May 28 2024 *)

{a(n)=sum(k=0, n-1, binomial(n-1, k)^4*n/(n-k))}

{a(n)=n*polcoeff(sum(m=1, n, sum(k=0, n, binomial(m+k-1,k)^4*x^k)*x^m/m)+x*O(x^n), n)}
for(n=1,20,print1(a(n),", "))

[sum( binomial(n,k)*binomial(n-1,k)^3 for k in (0..n-1) ) for n in (1..20)] # G. C. Greubel, Apr 05 2021

a(n) = Sum_{k=0..n-1} binomial(n-1,k)^3 * binomial(n,k).
From Vaclav Kotesovec, Mar 06 2014: (Start)
Recurrence: (n-1)^2*n^3*(10*n^2 - 25*n + 16)*a(n) = 2*(n-1)^2*(60*n^5 - 240*n^4 + 341*n^3 - 225*n^2 + 90*n - 16)*a(n-1) + 4*(n-2)^2*n*(4*n - 7)*(4*n - 5)*(10*n^2 - 5*n + 1)*a(n-2).
a(n) ~ 2^(4*n-5/2) / (Pi*n)^(3/2). (End)
a(n) = hypergeom([-n + 1, -n + 1, -n + 1, -n], [1, 1, 1], 1). - Detlef Meya, May 28 2024
a(n) = Sum_{k=0..n} (k/n)^3 * binomial(n,k)^4. - Seiichi Manyama, Jul 14 2024

cp's OEIS Frontend

A182421 a(n) = Sum_{k = 0..n} C(n,k)^7.

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A182422 a(n) = Sum_{k = 0..n} C(n,k)^8.

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A182446 a(n) = Sum_{k = 0..n} C(n,k)^9.

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A182447 a(n) = Sum_{k = 0..n} C(n,k)^10.

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A309010 Square array A(n, k) = Sum_{j=0..n} binomial(n,j)^k, n >= 0, k >= 0, read by antidiagonals.

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A342294 a(n) = Sum_{k = 0..n} binomial(n,k)^11.

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A342295 a(n) = Sum_{k = 0..n} binomial(n,k)^12.

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A198256 Row sums of A197653.

A328725 Constant term in the expansion of (1 + x + y + z + 1/x + 1/y + 1/z + xy + yz + zx + 1/(xy) + 1/(yz) + 1/(zx) + xyz + 1/(xyz))^n.

A181069 Expansion of l.g.f. Sum_{n>=1} [ Sum_{k>=0} C(n+k-1,k)^4 x^k ] x^n/n.