A087127 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of triangular numbers. The p-th row (p>=1) contains a(i,p) for i=1 to 2*p-1, where a(i,p) satisfies Sum_{i=1..n} C(i+1,2)^p = 3 * C(n+2,3) * Sum_{i=1..2*p-1} a(i,p) * C(n-1,i-1)/(i+2).

1, 1, 2, 1, 1, 8, 19, 18, 6, 1, 26, 163, 432, 564, 360, 90, 1, 80, 1135, 6354, 18078, 28800, 26100, 12600, 2520, 1, 242, 7291, 77400, 405060, 1210680, 2211570, 2520000, 1751400, 680400, 113400, 1, 728, 45199, 862218, 7667646, 38350080, 118848420

Offset: 1

Author

André F. Labossière, Aug 11 2003

Keywords

• easy
• nonn
• tabf

Comments

From Peter Bala, Mar 08 2018: (Start)

The table entries T(n,k) are the coefficients when expressing the polynomial C(x+2,2)^p of degree 2*p in terms of falling factorials: C(x+2,2)^p = Sum_{k = 0..2*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+2,2)^p = Sum_{k = 0..2*p} T(p,k)*C(n,k+1).

The sum of the p-th powers of the triangular numbers is also given by Sum_{i = 0..n-1} C(i+2,2)^p = Sum_{k = 2..2*p} A122193(p,k)*C(n+2,k+1) for p >= 1. (End)

Examples

			Row 3 contains 1,8,19,18,6, so Sum_{i=1..n} C(i+1,2)^3 = (n+2) * C(n+1,2) * [ a(1,3)/3 + a(2,3)*C(n-1,1)/4 + a(3,3)*C(n-1,2)/5 + a(4,3)*C(n-1,3)/6 + a(5,3)*C(n-1,4)/7 ] = [ (n+2)*(n+1)*n/2 ] * [ 1/3 + (8/4)*C(n-1,1) + (19/5)*C(n-1,2) + (18/6)*C(n-1,3) + (6/7)*C(n-1,4). Cf. A085438 for more details.
From _Peter Bala_, Mar 08 2018: (Start)
Table begins
n=0 |1
n=1 |1   2     1
n=2 |1   8    19    18      6
n=3 |1  26   163   432    564    360     90
n=4 |1  80  1135  6354  18078  28800  26100  12600  2520
...
Row 2: C(i+2,2)^2 = C(i,0) + 8*C(i,1) + 19*C(i,2) + 18*C(i,3) + 6*C(i,4). Hence, Sum_{i = 0..n-1} C(i+2,2)^2 =  C(n,1) + 8*C(n,2) + 19*C(n,3) + 18*C(n,4) + 6*C(n,5). (End)
		

Links

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

Crossrefs

Cf. A000292, A024166, A024166, A085438, A085439, A085440, A085441, A085442, A087107, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A000217, A122193.

Programs

GAP

Flat(List([0..6],n->List([0..2*n],k->Sum([0..k],i->(-1)^(k-i)*Binomial(k,i)*Binomial(i+2,2)^n)))); # Muniru A Asiru, Mar 22 2018

Maple

seq(seq(add( (-1)^(k-i)*binomial(k,i)*binomial(i+2,2)^n, i = 0..k), k = 0..2*n), n = 0..8); # Peter Bala, Mar 08 2018

Mathematica

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 3, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 2, 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, 2*p - 1}]//Flatten (* G. C. Greubel, Nov 23 2017 *)
a[i_,p_]:=(-1)^i HypergeometricPFQ[ConstantArray[3,p]~Join~{2-i},ConstantArray[1,p],1];Table[a[i,p],{p,0,10},{i,2,2 p+2}]//Flatten (* Jonathan Burns, Mar 20 2018 *)

PARI

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

Formula

a(1, p) = 1, a(2, p) = 3^(p-1)-1, a(3, p) = 3^(p-1)*[2^(p-1)-2]+1, ..., a(2*p-3, p) = [ (6*p^4-20*p^3+21*p^2-7*p)*(2*p-4)! ]/[3*2^(p-1)], a(2*p-2, p) = [ (p^2-p)*(2*p-3)! ]/2^(p-2), a(2*p-1, p) = [ (p-1)*(2*p-3)! ]/2^(p-2).

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+3, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+2, i-2*k)^(p-1) ]

From Peter Bala, Mar 08 2018: (Start)

The following remarks assume row and column indices start at 0.

T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*C(k,i)*C(i+2,2)^n. Equivalently, let v_n denote the sequence (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. 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. Cf. A122193.

T(n+1,k) = C(k+2,2)*T(n,k) + 2*C(k+1,2)*T(n,k-1) + C(k,2)*T(n,k-2), with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 2*n.

Let R(n,x) denote the n-th row polynomial.

R(n+1,x) = 1/2!*(1 + x)^2*(d/dx)^2 (x^2*R(n,x)).

R(n,x) = Sum_{i >= 0} binomial(i+2,2)^n*x^i/(1 + x)^(i+1).

R(n,x) = (1 + x)^2 o (1 + x)^2 o ... o (1 + x)^2 (n factors), where o denotes the black diamond product of power series defined in Dukes and White. Note the polynomial x^2 o ... o x^2 (n factors) is the n-th row polynomial of A122193.

x^2*R(n,x) = (1 + x)^2 * the n-th row polynomial of A122193 (End)

Extensions

Edited by Dean Hickerson, Aug 16 2003