This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A087127 #28 Apr 20 2018 01:07:28
%S A087127 1,1,2,1,1,8,19,18,6,1,26,163,432,564,360,90,1,80,1135,6354,18078,
%T A087127 28800,26100,12600,2520,1,242,7291,77400,405060,1210680,2211570,
%U A087127 2520000,1751400,680400,113400,1,728,45199,862218,7667646,38350080,118848420
%N 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).
%C A087127 From _Peter Bala_, Mar 08 2018: (Start)
%C A087127 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).
%C A087127 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)
%H A087127 G. C. Greubel, <a href="/A087127/b087127.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%H A087127 M. Dukes, C. D. White, <a href="http://arxiv.org/abs/1603.01589">Web Matrices: Structural Properties and Generating Combinatorial Identities</a>, arXiv:1603.01589 [math.CO], 2016.
%F A087127 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).
%F A087127 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) ]
%F A087127 From _Peter Bala_, Mar 08 2018: (Start)
%F A087127 The following remarks assume row and column indices start at 0.
%F A087127 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.
%F A087127 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.
%F A087127 Let R(n,x) denote the n-th row polynomial.
%F A087127 R(n+1,x) = 1/2!*(1 + x)^2*(d/dx)^2 (x^2*R(n,x)).
%F A087127 R(n,x) = Sum_{i >= 0} binomial(i+2,2)^n*x^i/(1 + x)^(i+1).
%F A087127 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.
%F A087127 x^2*R(n,x) = (1 + x)^2 * the n-th row polynomial of A122193 (End)
%e A087127 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.
%e A087127 From _Peter Bala_, Mar 08 2018: (Start)
%e A087127 Table begins
%e A087127 n=0 |1
%e A087127 n=1 |1 2 1
%e A087127 n=2 |1 8 19 18 6
%e A087127 n=3 |1 26 163 432 564 360 90
%e A087127 n=4 |1 80 1135 6354 18078 28800 26100 12600 2520
%e A087127 ...
%e A087127 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)
%p A087127 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
%t A087127 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 *)
%t A087127 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 *)
%o A087127 (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
%o A087127 (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
%Y A087127 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.
%Y A087127 Cf. A000217, A122193.
%K A087127 easy,nonn,tabf
%O A087127 1,3
%A A087127 _André F. Labossière_, Aug 11 2003
%E A087127 Edited by _Dean Hickerson_, Aug 16 2003