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 A087107 #28 Mar 11 2018 13:26:12
%S A087107 1,1,3,3,1,1,15,69,147,162,90,20,1,63,873,5191,16620,31560,36750,
%T A087107 25830,10080,1680,1,255,9489,130767,919602,3832650,10238000,18244380,
%U A087107 21990360,17745000,9198000,2772000,369600,1,1023,97953,2903071,40317780
%N 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 3*p-2, where a(i,p) satisfies Sum_{i=1..n} C(i+2,3)^p = 4 * C(n+3,4) * Sum_{i=1..3*p-2} a(i,p) * C(n-1,i-1)/(i+3).
%C A087107 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
%C A087107 From _Peter Bala_, Mar 11 2018: (Start)
%C A087107 The observation above is correct.
%C A087107 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).
%C A087107 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)
%H A087107 G. C. Greubel, <a href="/A087107/b087107.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%H A087107 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 A087107 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) ].
%F A087107 From _Peter Bala_, Nov 26 2017: (Start)
%F A087107 Conjectural formula for table entries: T(n,k) = Sum_{j = 0..k} (-1)^(k+j)*binomial(k,j)*binomial(j+3,3)^n.
%F A087107 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)
%F A087107 From _Peter Bala_, Mar 11 2018: (Start)
%F A087107 The conjectures above are correct.
%F A087107 The following remarks assume the row and column indices start at 0.
%F A087107 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.
%F A087107 Sum_{k = 0..3*n} T(n,k)*binomial(x,k) = (binomial(x+3,3))^n.
%F A087107 x^3*R(n,x) = (1 + x)^3 * the n-th row polynomial of A299041.
%F A087107 R(n+1,x) = 1/3!*(1 + x)^3*(d/dx)^3 (x^3*R(n,x)).
%F A087107 (1 - x)^(3*n)*R(n,x/(1 - x)) gives the n-th row polynomial of A174266.
%F A087107 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)
%e A087107 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.
%e A087107 From _Peter Bala_, Mar 11 2018: (Start)
%e A087107 Table begins
%e A087107 n=0 | 1
%e A087107 n=1 | 1 3 3 1
%e A087107 n=2 | 1 15 69 147 162 90 20
%e A087107 n=3 | 1 63 873 5191 16620 31560 36750 25830 10080 1680
%e A087107 ...
%e A087107 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)
%p A087107 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
%t A087107 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 *)
%o A087107 (PARI) {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
%Y A087107 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.
%Y A087107 Cf. A087127, A087110, A174266, A299041.
%K A087107 easy,nonn,tabf
%O A087107 1,3
%A A087107 _André F. Labossière_, Aug 11 2003
%E A087107 Edited by _Dean Hickerson_, Aug 16 2003