A174266 -id:A174266 - cp's OEIS frontend

A154283 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(2n+1,i) binomial(k+2-i,2)^n, 0 <= k <= 2*(n-1).

1, 1, 4, 1, 1, 20, 48, 20, 1, 1, 72, 603, 1168, 603, 72, 1, 1, 232, 5158, 27664, 47290, 27664, 5158, 232, 1, 1, 716, 37257, 450048, 1822014, 2864328, 1822014, 450048, 37257, 716, 1, 1, 2172, 247236, 6030140, 49258935, 163809288, 242384856, 163809288, 49258935, 6030140, 247236, 2172, 1

Offset: 1

Roger L. Bagula, Jan 06 2009

From Yahia Kahloune, Jan 30 2014: (Start)
In general, let 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,2,n).
With these coefficients we can calculate: Sum_{i=1..n} binomial(i+e-1,e)^p = Sum_{k=0..e*(p-1)} b(k,e,p)*binomial(n+e+k,e*p+k).
For example, A085438(n) = Sum_{i=1..n} binomial(1+i,2)^3 = T(3,0)*binomial(2+n,7) + T(3,1)*binomial(3+n,7) + T(3,2)*binomial(4+n,7) + T(3,3)*binomial(5+n,7) + T(3,4)*binomial(6+n,7) = (1/5040)*(90*n^7 + 630*n^6 + 1638*n^5 + 1890*n^4 + 840*n^3 - 48*n).
(End)
T(n,k) is the number of permutations of 2 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

			Triangle begins:
  1;
  1,     4,       1;
  1,    20,      48,        20,           1;
  1,    72,     603,      1168,         603,           72,           1;
  1,   232,    5158,     27664,       47290,        27664,        5158,  232, 1;
  1,   716,   37257,    450048,     1822014,      2864328,     1822014, ...;
  1,  2172,  247236,   6030140,    49258935,    163809288,   242384856, ...;
  1,  6544, 1568215,  72338144,  1086859301,   6727188848, 19323413187, ...;
  1, 19664, 9703890, 811888600, 21147576440, 225167210712, ... ;
  ...
The T(2,1) = 4 permutations of 1122 with 1 descent are 1212, 1221, 2112, 2211. - _Andrew Howroyd_, May 15 2020

Andrew Howroyd, Table of n, a(n) for n = 1..1600 (rows 1..40)
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.
Helmut Prodinger, On Touchard's continued fraction and extensions: combinatorics-free, self-contained proofs , arXiv:1102.5186 [math.CO], 2011.

Columns k=0..9 are A000012, A061981, A151624, A151625, A151626, A151627, A151628, A151629, A151630, A151631.
Row sums are A000680.
Similar triangles for e=1..6: A173018 (or A008292), this sequence, A174266, A236463, A237202, A237252.
Cf. A000680, A005799, A085438, A334781.

[(&+[(-1)^j*Binomial(2*n+1,j)*Binomial(k-j+2,2)^n: j in [0..k]]): k in [0..2*n-2], n in [1..12]]; // G. C. Greubel, Jun 13 2022

A154283 := proc(n,k)
        (1-x)^(2*n+1)*add( (l*(l+1)/2)^n*x^(l-1),l=0..k+1) ;
        coeftayl(%,x=0,k) ;
end proc: # R. J. Mathar, Feb 01 2013

p[x_, n_]= (1-x)^(2*n+1)*Sum[(k*(k+1)/2)^n*x^k, {k, 0, Infinity}]/x;
Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n,10}]//Flatten

T(n,k)={sum(i=0, k, (-1)^i*binomial(2*n+1, i)*binomial(k+2-i, 2)^n)} \\ Andrew Howroyd, May 09 2020

def A154283(n,k): return sum((-1)^j*binomial(2*n+1, j)*binomial(k-j+2, 2)^n for j in (0..k))
flatten([[A154283(n,k) for k in (0..2*n-2)] for n in (1..12)]) # G. C. Greubel, Jun 13 2022

T(n,k) = (-1) times coefficient of x^k in (x-1)^(2*n+1) * Sum_{k>=0} (k*(k+1)/2)^n *x^(k-1).
From Yahia Kahloune, Jan 29 2014: (Start)
Sum_{i=1..n} binomial(1+i,2)^p = Sum_{k=0..2*p-2} T(p,k)*binomial(n+2+k,2*p+1).
binomial(n,2)^p = Sum_{k=0..2*p-2} T(p,k)*binomial(n+k,2*p). (End)
From Peter Bala, Dec 21 2019: (Start)
E.g.f. as a continued fraction: (1-x)/(1-x + ( 1-exp((1-x)^2*t))*x/(1-x + (1-exp(2*(1-x)^2*t))*x/(1-x + (1-exp(3*(1-x)^2*t))*x/(1-x + ... )))) =  1 + x*t + x*(x^2 + 4*x + 1)*t^2/2! + x*(x^4 + 20*x^3 + 48*x^2 + 20*x + 1)*t^3/3! + ... (use Prodinger equation 1.1).
The sequence of alternating row sums (unsigned) [1, 1, 2, 10, 104, 1816,...] appears to be A005799. (End)

Edited by N. J. A. Sloane, Jan 30 2014 following suggestions from Yahia Kahloune (among other things, the signs of all terms have been reversed).

Edited by Andrew Howroyd, May 09 2020

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).

Original entry on oeis.org

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

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

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

A299041 Irregular table: T(n,k) equals the number of alignments of length k of n strings each of length 3.

Original entry on oeis.org

1, 1, 12, 30, 20, 1, 60, 690, 2940, 5670, 5040, 1680, 1, 252, 8730, 103820, 581700, 1767360, 3087000, 3099600, 1663200, 369600, 1, 1020, 94890, 2615340, 32186070, 214628400, 859992000, 2189325600, 3628409400, 3903900000, 2630628000, 1009008000, 168168000, 1, 4092, 979530, 58061420, 1411122300

Offset: 1

Peter Bala, Feb 02 2018

An alignment of n strings of various lengths is a way of inserting blank characters into the n strings so that the resulting strings all have the same length. We don't allow insertion of a blank character into the same position in each of the n strings.
In this case, let s_1,...,s_n be n strings each of length 3 over an alphabet A. Let - be a gap symbol not in A and let A' = union of A and {-}. An alignment of the n strings is an n-tuple (s_1',...,s_n') of strings each of length >= 3 over the alphabet A' such that
(a) the strings s_i', 1 <= i <= n, have the same length. This common length is called the length of the alignment.
(b) deleting the gap symbols from s_i' yields the string s_i for 1 <= i <= n
(c) there is no value j such that all the strings s_i', 1 <= i <= n have a gap symbol at position j.
By writing the strings s_i' one under another we can consider an alignment of n strings as an n X L matrix, where L, the length of the alignment, ranges from a minimum value of 3 to a maximum value of 3*n. Each row of the matrix has 3 characters from the alphabet A and (L - 3) gap characters.
For example,
  s_1' = ABC------
  s_2' = ---DEF---
  s_3' = ------GHI
is an alignment (of maximum length L = 9) of three strings s_1 = ABC, s_2 = DEF and s_3 = GHI each of length 3.
For the number of alignments of length k of n strings of length 1 (resp. 2) see A131689 (resp. A122193).

			Table begins
n\k| 3   4     5       6      7      8        9      10
- - - - - - - - - - - - - - - - - - - - - - - - - - - -
  1| 1
  2| 1  12    30      20
  3| 1  60   690    2940   5670    5040    1680
  4| 1 252  8730  103820 581700 1767360 3087000 3099600 ...
...
T(2,5) = 30: An alignment of length 5 will have two gap symbols on each line. There are C(5,2) = 10 ways of choosing the 2 positions to insert the gap symbols in the first string. The second string in the alignment must then have nongap symbols at these two positions leaving three positions in which to insert the remaining 1 nongap symbol, giving in total 10 x 3 = 30 possible alignments of 2 strings of 3 characters. Some examples are
  ABC--   ABC--   ABC--
  D--EF   -D-EF   --DEF
Row 2: Sum_{i = 3..n-1} C(i,3)^2 = C(n,4) + 12*C(n,5) + 30*C(n,6) + 20*C(n,7).
Row 3: Sum_{i = 3..n-1} C(i,3)^3 = C(n,4) + 60*C(n,5) + 690*C(n,6) + 2940*C(n,7) + 5670*C(n,8)+ 5040*C(n,9)+ 1680*C(n,10).
exp( Sum_{n >= 1} R(n,2)*x^n/n ) = (1 + x + 153*x^2 + 128793*x^3 + 319155321*x^4 + 1744213657689*x^5 + ....)^8
exp( Sum_{n >= 1} R(n,3)*x^n/n ) = (1 + x + 424*x^2 + 998584*x^3 + 6925040260*x^4 + 105920615923684*x^5 + ....)^27.

P. Bala, Notes on A299041
M. Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.
J. Engbers and C. Stocker, Two Combinatorial Proofs of Identities Involving Sums of Powers of Binomial Coefficients, Integers 16 (2016), #A58.
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 A062208. Cf. A014606, A019538, A078741, A087127, A122193, A131689, A174266.

Cf. A086020, A086021, A086022.

seq(seq(add( (-1)^(k-i) *binomial(k,i)*binomial(i,3)^n, i = 0..k ), k = 3..3*n), n = 1..6);

nmax = 6; T[n_, k_] := Sum[(-1)^(k-i) Binomial[k, i] Binomial[i, 3]^n, {i, 0, k}]; Table[T[n, k], {n, 1, nmax}, {k, 3, 3n}] // Flatten (* Jean-François Alcover, Feb 20 2018 *)

T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i)*binomial(i,3)^n.
T(n,3) = 1; T(n,3*n) = (3*n)!/6^n = A014606(n)
T(n,k) = binomial(k,3)*( T(n-1,k) + 3*T(n-1,k-1) + 3*T(n-1,k-2) + T(n-1,k-3) ) for 3 <= k <= 3*n with boundary conditions T(n,3) = 1 for n >= 1 and T(n,k) = 0 if (k < 3) or (k > 3*n).
Double e.g.f.: exp(-x)*Sum_{n >= 0} exp(binomial(n,3)*y)*x^n/n! = 1 + (x^3/3!)*y + (x^3/3! + 12*x^4/4! + 30*x^5/5! + 20*x^6/6!)*y^2/2! + ....
n-th row polynomial R(n,x) = Sum_{i >= 3} binomial(i,3)^n*x^i/(1 + x)^(i+1) for n >= 1.
1/(1 - x)*R(n,x/(1 - x)) = Sum_{i >= 3} binomial(i,3)^n*x^i for n >= 1.
R(n,x) = x^3 o x^3 o ... o x^3 (n factors), where o is the black diamond product of power series defined in Dukes and White.
R(n,x) = coefficient of (z_1)^3*...*(z_n)^3 in the expansion of the rational function 1/(1 + x - x*(1 + z_1)*...*(1 + z_n)).
The polynomials Sum_{k = 3..3*n} T(n,k)*x^(k-3)*(1 - x)^(3*n-k) are the row polynomials of A174266.
Sum_{i = 3..n-1} binomial(i,3)^m = Sum_{k = 3..3*m} T(m,k)*binomial(n,k+1) for m >= 1. See Examples below.
x^3*R(n,-1 - x) = (-1)^n*(1 + x)^3*R(n,x).
R(n+1,x) = 1/3!*x^3*(d/dx)^3 ((1 + x)^3*R(n,x)) for n >= 1.
The zeros of R(n,x) belong to the interval [-1, 0].
Row sums R(n,1) = A062208(n); alternating row sums R(n,-1) = (-1)^n.
For k a nonzero integer, the power series A(k,x) := exp( Sum_{n >= 1} 1/k^3*R(n,k)*x^n/n ) appear to have integer coefficients. See the Example section.
Sum_{k = 3..3*n} T(n,k)*binomial(x,k) = ( binomial(x,3) )^n. Equivalently, Sum_{k = 3..3*n} (-1)^(n+k)*T(n,k)*binomial(x+k,k) = ( binomial(x+3,3) )^n. Cf. the Worpitzky-type identity Sum_{k = 1..n} A019538(n,k)* binomial(x,k) = x^n.
Sum_{k = 3..3*n} T(n,k)*binomial(x,k-3) = -binomial(x,3)^n + 3*binomial(x+1,3)^n - 3*binomial(x+2,3)^n + binomial(x+3,3)^n. These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane.

A151632 Number of permutations of 3 indistinguishable copies of 1..n with exactly 2 adjacent element pairs in decreasing order.

Original entry on oeis.org

0, 9, 405, 6750, 83736, 922347, 9639783, 98361900, 992660346, 9967494609, 99857394225, 999379243674, 9997315646220, 99988457276295, 999950607877131, 9999789546603672, 99999106646803758, 999996220428781005, 9999984057081398901, 99999932929790707494

Offset: 1

R. H. Hardin, May 29 2009

Andrew Howroyd, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (21,-153,503,-786,576,-160).

Column k=2 of A174266.

T[n_, k_]:= T[n, k]= Sum[(-1)^(k-j)*Binomial[3*n+1, k-j+2]*(Binomial[j+1,3])^n, {j, 0, k+2}];
Table[T[n, 2], {n, 30}] (* G. C. Greubel, Mar 26 2022 *)

a(n) = {10^n - (3*n + 1)*4^n + 3*n*(3*n + 1)/2} \\ Andrew Howroyd, May 06 2020

concat(0, Vec(9*x^2*(1 + 24*x - 42*x^2 - 64*x^3) / ((1 - x)^3*(1 - 4*x)^2*(1 - 10*x)) + O(x^40))) \\ Colin Barker, Jul 17 2020

@CachedFunction
def T(n, k): return sum( (-1)^(k-j)*binomial(3*n+1, k-j+2)*(binomial(j+1,3))^n for j in (0..k+2) )
[T(n, 2) for n in (1..30)] # G. C. Greubel, Mar 26 2022

a(n) = 10^n - (3*n + 1)*4^n + 3*n*(3*n + 1)/2. - Andrew Howroyd, May 06 2020
From Colin Barker, Jul 17 2020: (Start)
G.f.: 9*x^2*(1 + 24*x - 42*x^2 - 64*x^3) / ((1 - x)^3*(1 - 4*x)^2*(1 - 10*x)).
a(n) = 21*a(n-1) - 153*a(n-2) + 503*a(n-3) - 786*a(n-4) + 576*a(n-5) - 160*a(n-6) for n>6.
(End)

Terms a(10) and beyond from Andrew Howroyd, May 06 2020

A151633 Number of permutations of 3 indistinguishable copies of 1..n with exactly 3 adjacent element pairs in decreasing order.

Original entry on oeis.org

0, 1, 760, 49682, 1722320, 45699447, 1063783164, 23119658500, 484099087156, 9930487583345, 201402352998560, 4059011173618086, 81520052344904040, 1634100242397204427, 32722001111322772660, 654870005050881521672, 13102000022780506515884

Offset: 1

R. H. Hardin, May 29 2009

nonn

Andrew Howroyd, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (56, -1242, 14412, -96873, 394308, -984324, 1492224, -1330560, 640000, -128000).

Column k=3 of A174266.

T[n_, k_]:= T[n, k]= Sum[(-1)^(k-j)*Binomial[3*n+1, k-j+2]*(Binomial[j+1,3])^n, {j, 0, k+2}];
Table[T[n, 3], {n, 30}] (* G. C. Greubel, Mar 26 2022 *)

a(n) = {20^n - (3*n + 1)*10^n + binomial(3*n+1, 2)*4^n - binomial(3*n+1, 3)} \\ Andrew Howroyd, May 07 2020

@CachedFunction
def T(n, k): return sum( (-1)^(k-j)*binomial(3*n+1, k-j+2)*(binomial(j+1,3))^n for j in (0..k+2) )
[T(n, 3) for n in (1..30)] # G. C. Greubel, Mar 26 2022

a(n) = 20^n - (3*n + 1)*10^n + binomial(3*n+1, 2)*4^n - binomial(3*n+1, 3). - Andrew Howroyd, May 07 2020

a(n) = Sum_{j=0..5} (-1)^(j+1)*binomial(3*n+1, 5-j)*(binomial(j+1, 3))^n. - G. C. Greubel, Mar 26 2022

Terms a(10) and beyond from Andrew Howroyd, May 07 2020

A151634 Number of permutations of 3 indistinguishable copies of 1..n with exactly 4 adjacent element pairs in decreasing order.

Original entry on oeis.org

0, 0, 405, 128124, 12750255, 789300477, 38464072830, 1641724670475, 64856779908606, 2445752640197970, 89642032274378115, 3228334377697738350, 115003717118946936945, 4069184219056622926539, 143377786266629066071740, 5038841894823365860640997, 176801555321207696717476200

Offset: 1

R. H. Hardin, May 29 2009

nonn

Andrew Howroyd, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (126, -6741, 203286, -3863391, 48979386, -427502471, 2613017466, -11265590916, 34232982136, -72719412480, 106245417600, -103853184000, 64584960000, -23040000000, 3584000000).

Column k=4 of A174266.

T[n_, k_]:= T[n, k]= Sum[(-1)^(k-j)*Binomial[3*n+1, k-j+2]*(Binomial[j+1,3])^n, {j, 0, k+2}];
Table[T[n, 4], {n, 30}] (* G. C. Greubel, Mar 26 2022 *)

a(n) = {35^n - (3*n + 1)*20^n + binomial(3*n+1, 2)*10^n - binomial(3*n+1, 3)*4^n + binomial(3*n+1, 4)} \\ Andrew Howroyd, May 07 2020

@CachedFunction
def T(n, k): return sum( (-1)^(k-j)*binomial(3*n+1, k-j+2)*(binomial(j+1, 3))^n for j in (0..k+2) )
[T(n, 4) for n in (1..30)] # G. C. Greubel, Mar 26 2022

a(n) = 35^n - (3*n + 1)*20^n + binomial(3*n+1, 2)*10^n - binomial(3*n+1, 3)*4^n + binomial(3*n+1, 4). - Andrew Howroyd, May 07 2020

a(n) = Sum_{j=0..6} (-1)^j*binomial(3*n+1, 6-j)*(binomial(j+1, 3))^n. - G. C. Greubel, Mar 26 2022

Terms a(9) and beyond from Andrew Howroyd, May 07 2020

A151635 Number of permutations of 3 indistinguishable copies of 1..n with exactly 5 adjacent element pairs in decreasing order.

Original entry on oeis.org

0, 0, 54, 128124, 40241088, 5904797049, 592030140912, 47871255785661, 3399596932632516, 222507204130403730, 13816730633213564154, 828855022115369147634, 48598186867956968680368, 2806334420165022553155783, 160409202733612103932779012, 9106532681255976991378628043

Offset: 1

R. H. Hardin, May 29 2009

nonn

Andrew Howroyd, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (252, -28116, 1847460, -80186430, 2443408020, -54222394300, 897042522780, -11233051883145, 107495660310160, -790365294823704, 4473663278780448, -19473246213545104, 64926170063690880, -164639495047219200, 314180023114240000, -444424489989120000, 455945899622400000, -328038555648000000, 156378808320000000, -44255232000000000, 5619712000000000).

Column k=5 of A174266.

T[n_, k_]:= T[n, k]= Sum[(-1)^(k-j)*Binomial[3*n+1, k-j+2]*(Binomial[j+1,3])^n, {j, 0, k+2}];
Table[T[n, 5], {n, 30}] (* G. C. Greubel, Mar 26 2022 *)

@CachedFunction
def T(n, k): return sum( (-1)^(k-j)*binomial(3*n+1, k-j+2)*(binomial(j+1,3))^n for j in (0..k+2) )
[T(n, 5) for n in (1..30)] # G. C. Greubel, Mar 26 2022

a(n) = Sum_{j=0..7} (-1)^(j+1)*binomial(3*n+1, 7-j)*(binomial(j+1, 3))^n. - G. C. Greubel, Mar 26 2022

Terms a(9) and beyond from Andrew Howroyd, May 06 2020

cp's OEIS Frontend

A154283 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(2*n+1,i) * binomial(k+2-i,2)^n, 0 <= k <= 2*(n-1).

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

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A299041 Irregular table: T(n,k) equals the number of alignments of length k of n strings each of length 3.

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A154283 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(2n+1,i) binomial(k+2-i,2)^n, 0 <= k <= 2*(n-1).

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).