A087127 -id:A087127 - cp's OEIS frontend

A236463 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(4n+1,i) binomial(k+4-i,4)^n, 0 <= k <= 4*(n-1).

1, 1, 16, 36, 16, 1, 1, 112, 1828, 8464, 13840, 8464, 1828, 112, 1, 1, 608, 40136, 724320, 4961755, 15018688, 21571984, 15018688, 4961755, 724320, 40136, 608, 1, 1, 3104, 693960, 37229920, 733059110, 6501577152, 29066972368, 69830127680, 93200908410, 69830127680

Offset: 1

Yahia Kahloune, Feb 01 2014

In general, define b(k,e,p) = Sum_{i=0..k} (-1)^i*binomial(e*p+1,i)*binomial(k+e-i,e)^p. Then T(n,k) = b(k,4,n).
Using these coefficients we can obtain formulas for binomial(n,e)^p and for Sum_{i=1..n} binomial(e-1+i,e)^p.
In particular:
  binomial(n, e)^p = Sum_{k=0..e*(p-1)} b(k,e,p) * binomial(n+k, e*p).
  Sum_{i=1..n} binomial(e-1+i, e)^p = Sum_{k=0..e*(p-1)} b(k,e,p) * binomial(n+e+k, e*p+1).
T(n,k) is the number of permutations of 4 indistinguishable copies of 1..n with exactly k descents. A descent is a pair of adjacent elements with the second element less than the first. - Andrew Howroyd, May 08 2020

			T(n,0) = 1;
T(n,1) = 5^n - (4*n+1);
T(n,2) = 15^n - (4*n+1)*5^n + C(4*n+1,2);
T(n,3) = 35^n - (4*n+1)*15^n + C(4*n+1,2)*5^n - C(4*n+1,3);
T(n,4) = 70^n - (4*n+1)*35^n + C(4*n+1,2)*15^n - C(4*n+1,3)*5^n + C(4*n+1,4).
Triangle T(n,k) begins:
1,
1, 16, 36, 16, 1;
1, 112, 1828, 8464, 13840, 8464, 1828, 112, 1;
1, 608, 40136, 724320, 4961755, 15018688, 21571984, 15018688, 4961755, 724320, 40136, 608, 1;
1, 3104, 693960, 37229920, 733059110, 6501577152, 29066972368, 69830127680, 93200908410, 69830127680, 29066972368, 6501577152, 733059110, 37229920, 693960, 3104, 1;
1, 15600, 11000300, 1558185200, 75073622025, 1585757994496, 16938467955200, 99825129369600, 342907451401150, 710228619472800, 903546399077256, 710228619472800, 342907451401150, 99825129369600, 16938467955200, 1585757994496, 75073622025, 1558185200, 11000300, 15600, 1;
  ...
Example:
Sum_{i=1..n} C(3+i,4)^3 = C(n+4,13) + 112*C(n+5,13) + 1828*C(n+6,13) + 8464*C(n+7,13) + 13840*C(n+8,13) + 8464*C(n+9,13) + 1828*C(n+10,13) + 112*C(n+11,13) + C(+12,13).
C(n,4)^3 = C(n,12) + 112*C(n+1,12) + 1828*C(n+2,12) + 8464*C(n+3,12) + 13840*C(n+4,12) + 8464*C(n+5,12) + 1828*C(n+6,12) + 112*C(n+7,12) + C(n+8,12).

Vincenzo Librandi, Table of n, a(n) for n = 1..780
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 12.

Columns k=2..8 are A151640, A151641, A151642, A151643, A151644, A151645, A151646.
Row sums are A014608.
Similar triangles for e=1..6: A173018 (or A008292), A154283, A174266, this sequence, A237202, A237252.
Sum_{i=1..n} binomial(3+i,4)^p for p=2..3 gives: A086023, A086024.
Cf. A087127, A087107, A087108, A087109, A087110, A087111, A181544.

b[k_, 4, p_] := Sum[(-1)^i*Binomial[4*p+1, i]*Binomial[k-i, 4]^p /. k -> 4+i, {i, 0, k-4}]; row[p_] := Table[b[k, 4, p], {k, 4, 4*p}]; Table[row[p], {p, 1, 6}] // Flatten (* Jean-François Alcover, Feb 05 2014 *)

T(n,k)={sum(i=0, k, (-1)^i*binomial(4*n+1, i)*binomial(k+4-i, 4)^n)} \\ Andrew Howroyd, May 08 2020

Sum_{i=1..n} binomial(3+i,4)^p = Sum{k=0..4*(p-1)} T(p,k) * binomial(n+4+k, 4*p+1).

binomial(n,4)^p = Sum_{k=0..4*(p-1)} T(p,k) * binomial(n+k, 4*p).

a(36) corrected by Vincenzo Librandi, Feb 14 2014

Edited by Andrew Howroyd, May 08 2020

A087107 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of tetrahedral numbers. The p-th row (p>=1) contains a(i,p) for i=1 to 3p-2, where a(i,p) satisfies Sum_{i=1..n} C(i+2,3)^p = 4 C(n+3,4) * Sum_{i=1..3p-2} a(i,p) C(n-1,i-1)/(i+3).

Original entry on oeis.org

1, 1, 3, 3, 1, 1, 15, 69, 147, 162, 90, 20, 1, 63, 873, 5191, 16620, 31560, 36750, 25830, 10080, 1680, 1, 255, 9489, 130767, 919602, 3832650, 10238000, 18244380, 21990360, 17745000, 9198000, 2772000, 369600, 1, 1023, 97953, 2903071, 40317780

Offset: 1

André F. Labossière, Aug 11 2003

Let s_n denote the sequence (1, 4^n, 10^n, 20^n, ...) regarded as an infinite column vector, where 1, 4, 10, 20, ... is the sequence of tetrahedral numbers A000292. It appears that the n-th row of this table is determined by the matrix product P^(-1)s_n, where P denotes Pascal's triangle A007318. - Peter Bala, Nov 26 2017
From Peter Bala, Mar 11 2018: (Start)
The observation above is correct.
The table entries T(n,k) are the coefficients when expressing the polynomial C(x+3,3)^p of degree 3*p in terms of falling factorials: C(x+3,3)^p = Sum_{k = 0..3*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+3,3)^p = Sum_{k = 0..3*p} T(p,k)*C(n,k+1).
The sum of the p-th powers of the tetrahedral numbers is also given by Sum_{i = 0..n-1} C(i+3,3)^p = Sum_{k = 3..3*p} A299041(p,k)*C(n+3,k+1) for p >= 1. (End)

			Row 3 contains 1,15,69,147,162,90,20, so Sum_{i=1..n} C(i+2,3)^3 = 4 * C(n+3,4) * [ a(1,3)/4 + a(2,3)*C(n-1,1)/5 + a(3,3)*C(n-1,2)/6 + ... + a(7,3)*C(n-1,6)/10 ] = 4 * C(n+3,4) * [ 1/4 + 15*C(n-1,1)/5 + 69*C(n-1,2)/6 + 147*C(n-1,3)/7 + 162*C(n-1,4)/8 + 90*C(n-1,5)/9 + 20*C(n-1,6)/10 ]. Cf. A086021 for more details.
From _Peter Bala_, Mar 11 2018: (Start)
Table begins
n=0 | 1
n=1 | 1  3   3    1
n=2 | 1 15  69  147   162    90    20
n=3 | 1 63 873 5191 16620 31560 36750 25830 10080 1680
...
Row 2: C(i+3,3)^2 = C(i,0) + 15*C(i,1) + 69*C(i,2) + 147*C(i,3) + 162*C(i,4) + 90*C(i,5) + 20*C(i,6). Hence, Sum_{i = 0..n-1} C(i+3,3)^2 =  C(n,1) + 15*C(n,2) + 69*C(n,3) + 147*C(n,4) + 162*C(n,5) + 90*C(n,6) + 20*C(n,7). (End)

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Cf. A000292, A024166, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A087127, A087110, A174266, A299041.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+3, 3)^n, i= 0..k), k = 0..3*n), n = 0..8); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 4, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 3, i - 2*k]^(p - 1), {k, 1, (2*i + 1 + (-1)^(i - 1))/4}]; Table[If[p == 1, 1, a[i, p]], {p, 1, 10}, {i, 1, 3*p - 2}]//Flatten (* G. C. Greubel, Nov 23 2017 *)

{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 4, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 3, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 3*p-2, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+4, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+3, i-2*k)^(p-1) ].
From Peter Bala, Nov 26 2017: (Start)
Conjectural formula for table entries: T(n,k) = Sum_{j = 0..k} (-1)^(k+j)*binomial(k,j)*binomial(j+3,3)^n.
Conjecturally, the n-th row polynomial R(n,x) = 1/(1 + x)*Sum_{i >= 0} binomial(i+3,3)^n *(x/(1 + x))^n. (End)
From Peter Bala, Mar 11 2018: (Start)
The conjectures above are correct.
The following remarks assume the row and column indices start at 0.
T(n+1,k) = C(k+3,3)*T(n,k) + 3*C(k+2,3)*T(n,k-1) + 3*C(k+1,3)*T(n,k-2) + C(k,3)*T(n,k-3) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 3*n.
Sum_{k = 0..3*n} T(n,k)*binomial(x,k) = (binomial(x+3,3))^n.
x^3*R(n,x) = (1 + x)^3 * the n-th row polynomial of A299041.
R(n+1,x) = 1/3!*(1 + x)^3*(d/dx)^3 (x^3*R(n,x)).
(1 - x)^(3*n)*R(n,x/(1 - x)) gives the n-th row polynomial of A174266.
R(n,x) = (1 + x)^3 o (1 + x)^3 o ... o (1 + x)^3 (n factors), where o denotes the black diamond product of power series defined in Dukes and White. Note the polynomial x^3 o ... o x^3 (n factors) is the n-th row polynomial of A299041. (End)

Edited by Dean Hickerson, Aug 16 2003

A087111 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of binomial coefficients C(n,7). The p-th row (p>=1) contains a(i,p) for i=1 to 7p-6, where a(i,p) satisfies Sum_{i=1..n} C(i+6,7)^p = 8 C(n+7,8) * Sum_{i=1..7p-6} a(i,p) C(n-1,i-1)/(i+7).

Original entry on oeis.org

1, 1, 7, 21, 35, 35, 21, 7, 1, 1, 63, 1169, 10703, 58821, 214123, 545629, 1004307, 1356194, 1347318, 974862, 500346, 172788, 36036, 3432, 1, 511, 45633, 1589567, 29302889, 333924087, 2577462937, 14287393351, 59159005164, 188008120188

Offset: 1

André F. Labossière, Aug 11 2003

From Peter Bala, Mar 11 2018: (Start)

The table entries T(n,k) are the coefficients when expressing the polynomial C(x+7,7)^p of degree 7*p in terms of falling factorials: C(x+7,7)^p = Sum_{k = 0..7*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+7,7)^p = Sum_{k = 0..7*p} T(p,k)*C(n,k+1). (End)

			Row 3 contains 1,63,1169,...,3432, so Sum_{i=1..n} C(i+6,7)^3 = 8 * C(n+7,8) * [ a(1,3)/8 + a(2,3)*C(n-1,1)/9 + a(3,3)*C(n-1,2)/10 + ... + a(15,3)*C(n-1,14)/22 ] = 8 * C(n+7,8) * [ 1/8 + 63*C(n-1,1)/9 + 1169*C(n-1,2)/10 + ... + 3432*C(n-1,14)/22 ]. Cf. A086030 for more details.

G. C. Greubel, Table of n, a(n) for the first 40 rows, flattebed
Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Cf. A000292, A024166, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A087107, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A027555, A086029, A086030, A087127.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+7, 7)^n, i = 0..k), k = 0..7*n), n = 0..4); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 8, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 7, i - 2*k]^(p - 1), {k, 1, (2*i + 1 + (-1)^(i - 1))/4}]; Table[If[p == 1, 1, a[i, p]], {p, 1, 10}, {i, 1, 7*p - 6}]//Flatten (* G. C. Greubel, Nov 23 2017 *)

{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 8, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 7, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 7*p-6, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+8, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+7, i-2*k)^(p-1) ]
From Peter Bala, Mar 11 2018: (Start)
The following remarks assume the row and column indices start at 0.
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i) * binomial(i+7,7)^n. Equivalently, let v_n denote the sequence (1, 8^n, 36^n, 120^n, ...) regarded as an infinite column vector, where 1, 8, 36, 120, ... is the sequence binomial(n+7,7) - see A000580. Then the n-th row of this table is determined by the matrix product P^(-1)*v_n, where P denotes Pascal's triangle A007318.
Recurrence: T(n+1,k) = Sum_{i = 0..7} C(7,i)*C(k+7-i,7)*T(n,k-i) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 7*n.
n-th row polynomial R(n,x) = (1 + x)^7 o (1 + x)^7 o ... o (1 + x)^7 (n factors), where o denotes the black diamond product of power series defined in Dukes and White.
R(n+1,x) = 1/7!*(1 + x)^7 * (d/dx)^7(x^7*R(n,x)).
R(n,x) = Sum_{i >= 0} binomial(i+7,7)^n*x^i/(1 + x)^(i+1).
(End)

Edited by Dean Hickerson, Aug 16 2003

A237252 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(6n+1,i) binomial(k+6-i,6)^n, 0 <= k <= 6*(n-1).

Original entry on oeis.org

1, 1, 36, 225, 400, 225, 36, 1, 1, 324, 15606, 233300, 1424925, 4050864, 5703096, 4050864, 1424925, 233300, 15606, 324, 1, 1, 2376, 554931, 35138736, 879018750, 10490842656, 66555527346, 239677178256, 509723668476, 654019630000, 509723668476, 239677178256, 66555527346, 10490842656, 879018750, 35138736, 554931, 2376, 1

Offset: 1

Yahia Kahloune, Feb 05 2014

In general, define b(k,e,p) = Sum_{i=0..k} (-1)^i*binomial(e*p+1,i)*binomial(k+e-i,e)^p. Then T(n,k) = b(k,6,n).
Using these coefficients we can obtain formulas for binomial(n,e)^p and for Sum_{i=1..n} binomial(e-1+i,e)^p.
In particular:
  binomial(n, e)^p = Sum_{k=0..e*(p-1)} b(k,e p) * binomial(n+k, e*p).
  Sum_{i=1..n} binomial(e-1+i, e)^p = Sum_{k=0..e*(p-1)} b(k,e,p) * binomial(n+e+k, e*p+1).
T(n,k) is the number of permutations of 6 indistinguishable copies of 1..n with exactly k descents. A descent is a pair of adjacent elements with the second element less than the first. - Andrew Howroyd, May 06 2020

			For example :
  T(n,0) = 1;
  T(n,1) = 7^n - (6*n+1);
  T(n,2) = 28^n - (6*n+1)*7^n + C(6*n+1,2);
  T(n,3) = 84^n - (6*n+1)*28^n + C(6*n+1,2)*7^n + C(6*n+1,3);
  T(n,4) = 210^n - (6*n+1)*84^n + C(6*n+1,2)*28^n - C(6*n+1,3)*7^n + C(6*n+1,4).
Triangle T(n,k) begins:
 1;
 1, 36, 225, 400, 225, 36, 1;
 1, 324, 15606, 233300, 1424925, 4050864, 5703096, 4050864, 1424925, 233300, 15606, 324, 1;
 1, 2376, 554931, 35138736, 879018750, 10490842656, 66555527346, 239677178256, 509723668476, 654019630000, 509723668476, 239677178256, 66555527346, 10490842656, 879018750, 35138736, 554931, 2376, 1;
 1, 16776, 16689816, 3656408776, 286691702976, 10255094095176, 192698692565176, 2080037792142216, 13690633212385551, 57229721552316976, 156200093827061616, 283397584598631216, 345271537321293856, 283397584598631216, 156200093827061616, 57229721552316976,13690633212385551, 2080037792142216, 192698692565176, 10255094095176, 286691702976, 3656408776, 16689816, 16776, 1;
...
Example:
Sum_{i=1..n} C(5+i,6)^2 = A086027(n) = C(n+6,13) + 36*C(n+7,13) + 225*C(n+8,13) + 400*C(n+9,13) + 225*C(n+10,13) + 36*C(n+11,13) + C(n+12,13).
binomial(n,6)^2 = C(n,12) + 36*C(n+1,12) + 225*C(n+2,12) + 400*C(n+3,12) + 225*C(n+4,12) + 36*C(n+5,12) + C(n+6,12).

G. C. Greubel, Table of n, a(n) for the first 25 rows, flattened
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 12.

Columns k=2..6 are A151651, A151652, A151653, A151654, A151655.
Row sums are A248814.
Similar triangles for e=1..5: A173018 (or A008292), A154283, A174266, A236463, A237202.
Sum_{i=1..n} binomial(5+i,6)^p for p=1..3 gives: A000580, A086027, A086028.
Cf. A087127, A086023, A086024, A086025, A087107, A087108, A087109, A087110, A087111, A181544.

b[k_, 6, p_] := Sum[(-1)^i*Binomial[6*p+1, i]*Binomial[k-i, 6]^p /. k -> 6+i, {i, 0, k-6}]; row[p_] := Table[b[k, 6, p], {k, 6, 6*p}]; Table[row[p], {p, 1, 5}] // Flatten (* Jean-François Alcover, Feb 05 2014 *)

T(n,k)={sum(i=0, k, (-1)^i*binomial(6*n+1, i)*binomial(k+6-i, 6)^n)} \\ Andrew Howroyd, May 06 2020

Sum_{i=1..n} binomial(5+i,6)^p = Sum{k=0..6*(p-1)} T(p,k) * binomial(n+6+k, 6*p+1).

binomial(n,6)^p = Sum_{k=0..6*(p-1)} T(p,k) * binomial(n+k, 6*p).

Edited by Andrew Howroyd, May 06 2020

A087108 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of binomial coefficients C(n,4). The p-th row (p>=1) contains a(i,p) for i=1 to 4p-3, where a(i,p) satisfies Sum_{i=1..n} C(i+3,4)^p = 5 C(n+4,5) * Sum_{i=1..4p-3} a(i,p) C(n-1,i-1)/(i+4).

Original entry on oeis.org

1, 1, 4, 6, 4, 1, 1, 24, 176, 624, 1251, 1500, 1070, 420, 70, 1, 124, 3126, 33124, 191251, 681000, 1596120, 2543520, 2780820, 2058000, 987000, 277200, 34650, 1, 624, 49376, 1350624, 18308751, 146500500, 763418870, 2749648020, 7101675070, 13440210000

Offset: 1

André F. Labossière, Aug 11 2003

From Peter Bala, Mar 11 2018: (Start)

The table entries T(n,k) are the coefficients when expressing the polynomial C(x+4,4)^p of degree 4*p in terms of falling factorials: C(x+4,4)^p = Sum_{k = 0..4*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+4,4)^p = Sum_{k = 0..4*p} T(p,k)*C(n,k+1). (End)

			Row 3 contains 1,24,176,...,70, so Sum_{i=1..n} C(i+3,4)^3 = 5 * C(n+4,5) * [ a(1,3)/5 + a(2,3)*C(n-1,1)/6 + a(3,3)*C(n-1,2)/7 + ... + a(9,3)*C(n-1,8)/13 ] = 5 * C(n+4,5) * [ 1/5 + 24*C(n-1,1)/6 + 176*C(n-1,2)/7 + ... + 70*C(n-1,8)/13 ]. Cf. A086024 for more details.
From _Peter Bala_, Mar 11 2018: (Start)
Table begins
  n = 0 | 1
  n = 1 | 1   4    6     4      1
  n = 2 | 1  24  176   624   1251   1500    1070  420  70
  n = 3 | 1 124 3126 33124 191251 681000 1596120 ...
  ...
Row 2: C(i+4,4)^2 = C(i,0) + 24*C(i,1) + 176*C(i,2) + 624*C(i,3) + 1251*C(i,4) + 1500*C(i,5) + 1070*C(i,6) + 420*C(i,7) + 70*C(i,8). Hence, Sum_{i = 0..n-1} C(i+4,4)^2 =  C(n,1) + 24*C(n,2) + 176*C(n,3) + 624*C(n,4) + 1251*C(n,5) + 1500*C(n,6) + 1070*C(n,7) + 420*C(n,8) + 70*C(n,9) .(End)

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Cf. A000292, A024166, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A087107, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A087127, A236463.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+4, 4)^n, i = 0..k), k = 0..4*n), n = 0..6); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 5, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 4, i - 2*k]^(p - 1), {k, 1, (2*i + 1 + (-1)^(i - 1))/4}]; Table[If[p == 1, 1, a[i, p]], {p, 1, 10}, {i, 1, 4*p - 3}]//Flatten (* G. C. Greubel, Nov 23 2017 *)

{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 5, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 4, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 4*p-3, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+5, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+4, i-2*k)^(p-1) ]
From Peter Bala, Mar 11 2018: (Start)
The following remarks assume the row and column indices start at 0.
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i) * binomial(i+4,4)^n. Equivalently, let v_n denote the sequence (1, 5^n, 15^n, 35^n, ...) regarded as an infinite column vector, where 1, 5, 15, 35, ... is the sequence binomial(n+4,4) - see A000332. Then the n-th row of this table is determined by the matrix product P^(-1)*v_n, where P denotes Pascal's triangle A007318.
Recurrence: T(n+1,k) = C(k+4,4)*T(n,k) + 4*C(k+3,4)*T(n,k-1) + 6*C(k+2,4)*T(n,k-2) + 4*C(k+1,4)*T(n,k-3) + C(k,4)*T(n,k-4) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 4*n.
n-th row polynomial R(n,x) = (1 + x)^4 o (1 + x)^4 o ... o (1 + x)^4 (n factors), where o denotes the black diamond product of power series defined in Dukes and White.
R(n,x) = Sum_{i >= 0} binomial(i+4,4)^n*x^i/(1 + x)^(i+1).
R(n+1,x) = 1/4! * (1 + x)^4 * (d/dx)^4(x^4*R(n,x)).
(1 - x)^(4*n)*R(n,x/(1 - x)) appears to equal the n-th row polynomial of A236463. (End)

Edited by Dean Hickerson, Aug 16 2003

A087109 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of binomial coefficients C(n,5). The p-th row (p>=1) contains a(i,p) for i=1 to 5p-4, where a(i,p) satisfies Sum_{i=1..n} C(i+4,5)^p = 6 C(n+5,6) * Sum_{i=1..5p-4} a(i,p) C(n-1,i-1)/(i+5).

Original entry on oeis.org

1, 1, 5, 10, 10, 5, 1, 1, 35, 370, 1920, 5835, 11253, 14240, 11830, 6230, 1890, 252, 1, 215, 8830, 148480, 1352615, 7665757, 29224020, 78518790, 152794740, 218270220, 229279512, 175227360, 94864770, 34504470, 7567560, 756756, 1, 1295, 191890

Offset: 1

André F. Labossière, Aug 11 2003

From Peter Bala, Mar 11 2018: (Start)

The table entries T(n,k) are the coefficients when expressing the polynomial C(x+5,5)^p of degree 5*p in terms of falling factorials: C(x+5,5)^p = Sum_{k = 0..5*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+5,5)^p = Sum_{k = 0..5*p} T(p,k)*C(n,k+1). (End)

			Row 3 contains 1,35,370,...,252, so Sum_{i=1..n} C(i+4,5)^3 = 6 * C(n+5,6) * [ a(1,3)/6 + a(2,3)*C(n-1,1)/7 + a(3,3)*C(n-1,2)/8 + ... + a(11,3)*C(n-1,10)/16 ] = 6 * C(n+5,6) * [ 1/6 + 35*C(n-1,1)/7 + 370*C(n-1,2)/8 + ... + 252*C(n-1,10)/16 ]. Cf. A086026 for more details.
From _Peter Bala_, Mar 11 2018: (Start)
Table begins
1
1  5  10   10    5     1
1 35 370 1920 5835 11253 14240 11830 6230 1890 252
...
Row 2: C(i+5,5)^2 = C(i,0) + 35*C(i,1) + 370*C(i,2) + 1920*C(i,3) + 5835*C(i,4) + 11253*C(i,5) + 14240*C(i,6) + 11830*C(i,7) + 6230*C(i,8) + 1890*C(i,9) + 252*C(i,10). Hence, Sum_{i = 0..n-1} C(i+5,5)^2 = C(n,1) + 35*C(n,2) + 370*C(n,3) + 1920*C(n,4) + 5835*C(n,5) + 11253*C(n,6) + 14240*C(n,7) + 11830*C(n,8) + 6230*C(n,9) + 1890*C(n,10) + 252*C(n,11). (End)

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Cf. A000292, A024166, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A087107, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A087127, A237202.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+5, 5)^n, i = 0..k), k = 0..5*n), n = 0..5); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 6, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 5, i - 2*k]^(p - 1), {k, 1, (2*i + 1 + (-1)^(i - 1))/4}]; Table[If[p == 1, 1, a[i, p]], {p, 1, 10}, {i, 1, 5*p - 4}]//Flatten (* G. C. Greubel, Nov 23 2017 *)

{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 6, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 5, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 5*p-4, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+6, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+5, i-2*k)^(p-1) ]
From Peter Bala, Mar 11 2018: (Start)
The following remarks assume the row and column indices start at 0.
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i) * binomial(i+5,5)^n. Equivalently, let v_n denote the sequence (1, 6^n, 21^n, 56^n, ...) regarded as an infinite column vector, where 1, 6, 21, 56, ... is the sequence binomial(n+5,5) - see A000389. Then the n-th row of this table is determined by the matrix product P^(-1)*v_n, where P denotes Pascal's triangle A007318.
Recurrence: T(n+1,k) = Sum_{i = 0..5} C(5,i)*C(k+5-i,5)*T(n,k-i) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 5*n.
n-th row polynomial R(n,x) = (1 + x)^5 o (1 + x)^5 o ... o (1 + x)^5 (n factors), where o denotes the black diamond product of power series defined in Dukes and White.
R(n+1,x) = 1/5!*(1 + x)^5 * (d/dx)^5(x^5*R(n,x)).
R(n,x) = Sum_{i >= 0} binomial(i+5,5)^n*x^i/(1 + x)^(i+1).
(1 - x)^(5*n)*R(n,x/(1 - x)) appears to equal the n-th row polynomial of A237202. (End)

Edited by Dean Hickerson, Aug 16 2003

A087110 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of binomial coefficients C(n,6). The p-th row (p>=1) contains a(i,p) for i=1 to 6p-5, where a(i,p) satisfies Sum_{i=1..n} C(i+5,6)^p = 7 C(n+6,7) * Sum_{i=1..6p-5} a(i,p) C(n-1,i-1)/(i+6).

Original entry on oeis.org

1, 1, 6, 15, 20, 15, 6, 1, 1, 48, 687, 4850, 20385, 55908, 104959, 137886, 127050, 80640, 33642, 8316, 924, 1, 342, 21267, 527876, 7020525, 58015362, 324610399, 1297791264, 3839203452, 8595153000, 14760228672, 19560928464, 19987430694

Offset: 1

André F. Labossière, Aug 11 2003

From Peter Bala, Mar 11 2018: (Start)

The table entries T(n,k) are the coefficients when expressing the polynomial C(x+6,6)^p of degree 6*p in terms of falling factorials: C(x+6,6)^p = Sum_{k = 0..6*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+6,6)^p = Sum_{k = 0..6*p} T(p,k)*C(n,k+1). (End)

			Row 3 contains 1,48,687,...,924, so Sum_{i=1..n} C(i+5,6)^3 = 7 * C(n+6,7) * [ a(1,3)/7 + a(2,3)*C(n-1,1)/8 + a(3,3)*C(n-1,2)/9 + ... + a(13,3)*C(n-1,12)/19 ] = 7 * C(n+6,7) * [ 1/7 + 48*C(n-1,1)/8 + 687*C(n-1,2)/9 + ... + 924*C(n-1,12)/19 ]. Cf. A086028 for more details.

G. C. Greubel, Table of n, a(n) for the first 40 rows, flattened
Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Cf. A000292, A024166, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A087107, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A087127, A237252.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+6, 6)^n, i = 0..k), k = 0..6*n), n = 0..5); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 7, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 6, i - 2*k]^(p - 1), {k, 1, (2*i + 1 + (-1)^(i - 1))/4}]; Table[If[p == 1, 1, a[i, p]], {p, 1, 10}, {i, 1, 6*p - 5}]//Flatten (* G. C. Greubel, Nov 23 2017 *)

{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 7, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 6, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 6*p-5, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+7, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+6, i-2*k)^(p-1) ]
From Peter Bala, Mar 11 2018: (Start)
The following remarks assume the row and column indices start at 0.
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i) * binomial(i+6,6)^n. Equivalently, let v_n denote the sequence (1, 7^n, 28^n, 84^n, ...) regarded as an infinite column vector, where 1, 7, 28, 84, ... is the sequence binomial(n+6,6) - see A000579. Then the n-th row of this table is determined by the matrix product P^(-1)*v_n, where P denotes Pascal's triangle A007318.
Recurrence: T(n+1,k) = Sum_{i = 0..6} C(6,i)*C(k+6-i,6)*T(n,k-i) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 6*n.
n-th row polynomial R(n,x) = (1 + x)^6 o (1 + x)^6 o ... o (1 + x)^6 (n factors), where o denotes the black diamond product of power series defined in Dukes and White.
R(n+1,x) = 1/6!*(1 + x)^6 * (d/dx)^6(x^6*R(n,x)).
R(n,x) = Sum_{i >= 0} binomial(i+6,6)^n*x^i/(1 + x)^(i+1).
(1 - x)^(6*n)*R(n,x/(1 - x)) appears to equal the n-th row polynomial of A237252. (End)

Edited by Dean Hickerson, Aug 16 2003

A122193 Triangle T(n,k) of number of loopless multigraphs with n labeled edges and k labeled vertices and without isolated vertices, n >= 1; 2 <= k <= 2*n.

Original entry on oeis.org

1, 1, 6, 6, 1, 24, 114, 180, 90, 1, 78, 978, 4320, 8460, 7560, 2520, 1, 240, 6810, 63540, 271170, 604800, 730800, 453600, 113400, 1, 726, 43746, 774000, 6075900, 25424280, 61923960, 90720000, 78813000, 37422000, 7484400

Offset: 1

Vladeta Jovovic, Aug 24 2006

T(n,k) equals the number of arrangements on a line of n (nondegenerate) finite closed intervals having k distinct endpoints. See the 'IBM Ponder This' link. An example is given below. - Peter Bala, Jan 28 2018

T(n,k) equals the number of alignments of length k of n strings each of length 2. See Slowinski. Cf. A131689 (alignments of strings of length 1) and A299041 (alignments of strings of length 3). - Peter Bala, Feb 04 2018

			Triangle begins:
  1;
  1,  6,   6;
  1, 24, 114,  180,   90;
  1, 78, 978, 4320, 8460, 7560, 2520;
  ...
From _Francisco Santos_, Nov 17 2017: (Start)
For n=3 edges and k=4 vertices there are three loopless multigraphs without isolated vertices: a path, a Y-graph, and the multigraph {12, 34, 34}. The number of labelings in each is 3!4!/a, where a is the number of automorphisms. This gives respectively 3!4!/2 = 72, 3!4!/6 = 24 and 3!4!/8 = 18, adding up to 72 + 24 + 18 = 114. (End)
From _Peter Bala_, Jan 28 2018: (Start)
T(2,3) = 6: Consider 2 (nondegenerate) finite closed intervals [a, b] and [c, d]. There are 6 arrangements of these two intervals with 3 distinct endpoints:
  ...a--b--d....  a < b = c < d
  ...a...c--b...  a < c < b = d
  ...a--d...b...  a = c < d < b
  ...a--b...d...  a = c < b < d
  ...c...a--d...  c < a < b = d
  ...c--a--b....  c < a = d < b
T(2,4) = 6: There are 6 arrangements of the two intervals with 4 distinct endpoints:
  ...a--b...c--d.....  no intersection a < b < c < d
  ...a...c...b...d...  a < c < b < d
  ...a...c--d...b....  [c,d] is a proper subset of [a,b]
  ...c...a...d...b...  c < a < d < b
  ...c...a--b...d... [a,b] is a proper subset of [c,d]
  ...c--d...a--b.....  no intersection c < d < a < b.
Sums of powers of triangular numbers:
Row 2: Sum_{i = 2..n-1} C(i,2)^2 = C(n,3) + 6*C(n,4) + 6*C(n,5);
Row 3: Sum_{i = 2..n-1} C(i,2)^3 = C(n,3) + 24*C(n,4) + 114*C(n,5) + 180*C(n,6) + 90*C(n,7). See A024166 and A085438.
exp( Sum_{n >= 1} R(n,2)*x^n/n ) = (1 + x + 19*x^2 + 1147*x^3 + 145606*x^4 + 31784062*x^5 + ... )^4
exp( Sum_{n >= 1} R(n,3)*x^n/n ) = (1 + x + 37*x^2 + 4453*x^3 + 1126375*x^4 + 489185863*x^5 + ... )^9
exp( Sum_{n >= 1} R(n,4)*x^n/n ) = (1 + x + 61*x^2 + 12221*x^3 + 5144411*x^4 + 3715840571*x^5 + ... )^16 (End)
From _Peter Bala_, Feb 04 2018: (Start)
T(3,3) = 24 alignments of length 3 of 3 strings each of length 2. Examples include
  (i) A B -    (ii) A - B
      - C D         - C D
      - E F         E F -
There are 18 alignments of type (i) with two gap characters in one of the columns (3 ways of putting 2 gap characters in a column x 2 ways to place the other letter in the row which doesn't yet have a gap character x 3 columns: there are 6 alignments of type (ii) with a single gap character in each column (3 ways to put a single gap character in the first column x 2 ways to then place a single gap character in the second column). (End)

Michael De Vlieger, Table of n, a(n) for n = 1..10000 (rows 1 <= n <= 100).
Peter Bala, Deformations of the Hadamard product of power series
Peter Bala, Notes on A122193
S. Eger, On the Number of Many-to-Many Alignments of N Sequences, arXiv:1511.00622 [math.CO], 2015.
J. Engbers and C. Stocker, Two Combinatorial Proofs of Identities Involving Sums of Powers of Binomial Coefficients, Integers 16 (2016), #A58.
M. Dukes and C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.
IBM Ponder This, Jan 01 2001
J. B. Slowinski, The Number of Multiple Alignments, Molecular Phylogenetics and Evolution 10:2 (1998), 264-266. doi:10.1006/mpev.1998.0522

Row sums give A055203.
Cf. A078739, A005799, A019538, A131689, A154283, A299041, A087127.
For Sum_{i = 2..n} C(i,2)^k see A024166 (k = 2), A085438 - A085442 ( k = 3 thru 7).

# Note that the function implements the full triangle because it can be
# much better reused and referenced in this form.
A122193 := (n,k) -> A078739(n,k)*k!/2^n:
# Displays the truncated triangle from the definition:
seq(print(seq(A122193(n,k),k=2..2*n)),n=1..6); # Peter Luschny, Mar 25 2011

t[n_, k_] := Sum[(-1)^(n - r) Binomial[n, r] StirlingS2[n + r, k], {r, 0, n}]; Table[t[n, k] k!/2^n, {n, 6}, {k, 2, 2 n}] // Flatten (* Michael De Vlieger, Nov 18 2017, after Jean-François Alcover at A078739 *)

Double e.g.f.: exp(-x)*Sum_{n>=0} exp(binomial(n,2)*y)*x^n/n!.
T(n,k) = S_{2,2}(n,k)*k!/2^n; S_{2,2} the generalized Stirling numbers A078739. - Peter Luschny, Mar 25 2011
From Peter Bala, Jan 28 2018: (Start)
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i)*(i*(i-1)/2)^n.
T(n,k) = k*(k-1)/2*( T(n-1,k) + 2*T(n-1,k-1) + T(n-1,k-2) ) for 2 < k <= 2*n with boundary conditions T(n,2) = 1 for n >= 1 and T(n,k) = 0 if (k < 2) or (k > 2*n).
n-th row polynomial R(n,x) = Sum_{i >= 2} (i*(i-1)/2)^n * x^i/(1+x)^(i+1) for n >= 1.
1/(1-x)*R(n,x/(1-x)) = Sum_{i >= 2} (i*(i-1)/2)^n*x^i for n >= 1.
R(n,x) = 1/2^n*Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*F(n+k,x), where F(n,x) = Sum_{k = 0..n} k!*Stirling2(n,k)*x^k is the n-th Fubini polynomial, the n-th row polynomial of A131689.
R(n,x) = x/(1+x)*1/2^n*Sum_{k = 0..n} binomial(n,k)*F(n+k,x) for n >= 1.
The polynomials Sum_{k = 2..2*n} T(n,k)*x^(k-2)*(1-x)^(2*n-k) are the row polynomials of A154283.
A154283 * A007318 equals the row reverse of this array.
Sum_{k = 2..2*n} T(n,k)*binomial(x,k) = ( binomial(x,2) )^n. Equivalently, Sum_{k = 2..2*n} (-1)^k*T(n,k)*binomial(x+k,k) = ( binomial(x+2,2) )^n. Cf. the Worpitzky-type identity Sum_{k = 1..n} A019538(n,k)*binomial(x,k) = x^n.
Sum_{i = 2..n-1} (i*(i-1)/2)^m = Sum_{k = 2..2*m} T(m,k) * binomial(n,k+1) for m >= 1. See Examples below.
R(n,x) = x^2 o x^2 o ... o x^2 (n factors), where o is the black diamond product of power series defined in Dukes and White. Note the polynomial x o x o ... o x (n factors) is the n-th row polynomial of A019538.
x^2*R(n,-1-x) = (1+x)^2*R(n,x) for n >= 1.
R(n+1,x) = 1/2*x^2*(d/dx)^2 ((1+x)^2*R(n,x)).
The zeros of R(n,x) belong to the interval [-1, 0].
Alternating row sums equal 1, that is R(n,-1) = 1.
R(n,-2) = 4*R(n,1) = 4*A055203(n).
4^n*Sum_{k = 2..2*n} T(n,k)*(-1/2)^k appears to equal (-1)^(n+1)*A005799(n) for n >= 1.
For k a nonzero integer, the power series A(k,x) := exp( Sum_{n >= 1} 1/k^2*R(n,k)*x^n/n ) appear to have integer coefficients. See the Example section.
Sum_{k = 2..2*n} T(n,k)*binomial(x,k-2) = binomial(x,2)^n - 2*binomial(x+1,2)^n + binomial(x+2,2)^n. These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane (the corresponding property also holds for the row polynomials of A019538 with a factor of x removed). (End)
From Peter Bala, Mar 08 2018: (Start)
n-th row polynomial R(n,x) = coefficient of (z_1)^2 * ... * (z_n)^2 in the expansion of the rational function 1/(1 + x - x*(1 + z_1)*...*(1 + z_n)).
The n-th row of the table is given by the matrix product P^(-1)*v_n, where P denotes Pascal's triangle A007318 and v_n is the sequence (0, 0, 1, 3^n, 6^n, 10^n, ...) regarded as an infinite column vector, where 1, 3, 6, 10, ... is the sequence of triangular numbers A000217. Cf. A087127. (End)

Definition corrected by Francisco Santos, Nov 17 2017

A237202 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(5n+1,i) binomial(k+5-i,5)^n, 0 <= k <= 5*(n-1).

Original entry on oeis.org

1, 1, 25, 100, 100, 25, 1, 1, 200, 5925, 52800, 182700, 273504, 182700, 52800, 5925, 200, 1, 1, 1275, 167475, 6021225, 84646275, 554083761, 1858142825, 3363309675, 3363309675, 1858142825, 554083761, 84646275

Offset: 1

Yahia Kahloune, Feb 05 2014

In general, define b(k,e,p) = Sum_{i=0..k} (-1)^i*binomial(e*p+1,i)*binomial(k+e-i,e)^p. Then T(n,k) = b(k,5,n).
Using these coefficients we can obtain formulas for binomial(n,e)^p and for Sum_{i=1..n} binomial(e-1+i,e)^p.
In particular:
  binomial(n, e)^p = Sum_{k=0..e*(p-1)} b(k,e,p) * binomial(n+k, e*p).
  Sum_{i=1..n} binomial(e-1+i, e)^p = Sum_{k=0..e*(p-1)} b(k,e,p) * binomial(n+e+k, e*p+1).
T(n,k) is the number of permutations of 5 indistinguishable copies of 1..n with exactly k descents. A descent is a pair of adjacent elements with the second element less than the first. - Andrew Howroyd, May 08 2020

			T(n,0) = 1;
T(n,1) = 6^n - (5*n+1);
T(n,2) = 21^n - (5*n+1)*6^n + C(5*n+1,2);
T(n,3) = 56^n - (5*n+1)*21^n + C(5*n+1,2)*6^n - C(5*n+1,3) ;
T(n,4) = 126^n - (5*n+1)*56^n + C(5*n+1,2)*21^n - C(5*n+1,3)*6^n  + C(5*n+1,4).
Triangle T(n,k) begins:
1;
1, 25, 100, 100, 25, 1;
1, 200, 5925, 52800, 182700, 273504, 182700, 52800, 5925, 200, 1;
1, 1275, 167475, 6021225, 84646275, 554083761, 1858142825, 3363309675, 3363309675, 1858142825, 554083761, 84646275, 6021225, 167475, 125, 1;
1, 7750, 3882250, 447069750, 18746073375, 359033166276, 3575306548500, 20052364456500, 66640122159000, 135424590593500, 171219515211316, 135424590593500, 66640122159000, 20052364456500, 3575306548500, 359033166276, 18746073375, 447069750, 3882250, 7750, 1;
...
Example:
Sum_{i=1..n} C(4+i,5)^3 = C(n+5,16) + 200*C(n+6,16) + 5925*(n+7,16) + 52800*C(n+8,16) + 182700*C(n+9,16) + 273504*C(n+10,16) + 182700*C(n+11,16) + 52800*C(n+12,16) + 5925*C(n+13,16) + 200*C(n+14,16) + C(n+15,16).
C(n,5)^3 = C(n,15) + 200*C(n+1,15) + 5925*C(n+2,15) + 52800*C(n+3,15) + 182700*C(n+4,15) + 273504*C(n+5,15) + 182700*C(n+6,15) + 52800*C(n+7,15) + 5925*C(n+8,15) + 200*C(n+9,15) + C(n+10,15).

G. C. Greubel, Table of n, a(n) for the first 25 rows, flattened
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 12.

Columns k=2..5 are A151647, A151648, A151649, A151650.
Row sums are A014609.
Similar triangles for e=1..6: A173018 (or A008292), A154283, A174266, A236463, this sequence, A237252.
Sum_{i=1..n} binomial(4+i,5)^p for p=2..3 gives: A086025, A086026.
Cf. A087127, A087107, A087108, A087109, A087110, A087111, A181544.

b[k_, 5, p_] := Sum[(-1)^i*Binomial[5*p+1, i]*Binomial[k-i, 5]^p /. k -> 5+i, {i, 0, k-5}]; row[p_] := Table[b[k, 5, p], {k, 5, 5*p}]; Table[row[p], {p, 1, 5}] // Flatten (* Jean-François Alcover, Feb 05 2014 *)

T(n,k)={sum(i=0, k, (-1)^i*binomial(5*n+1, i)*binomial(k+5-i, 5)^n)} \\ Andrew Howroyd, May 08 2020

Sum_{i=1..n} binomial(4+i,5)^p = Sum{k=0..5*(p-1)} T(p,k) * binomial(n+5+k, 5*p+1).

binomial(n,5)^p = Sum_{k=0..5*(p-1)} T(p,k) * binomial(n+k, 5*p).

Edited by Andrew Howroyd, May 08 2020

cp's OEIS Frontend

A236463 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(4*n+1,i) * binomial(k+4-i,4)^n, 0 <= k <= 4*(n-1).

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A237252 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(6*n+1,i) * binomial(k+6-i,6)^n, 0 <= k <= 6*(n-1).

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A236463 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(4n+1,i) binomial(k+4-i,4)^n, 0 <= k <= 4*(n-1).

A237252 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(6n+1,i) binomial(k+6-i,6)^n, 0 <= k <= 6*(n-1).

A237202 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(5n+1,i) binomial(k+5-i,5)^n, 0 <= k <= 5*(n-1).