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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 6*p-5, where a(i,p) satisfies Sum_{i=1..n} C(i+5,6)^p = 7 * C(n+6,7) * Sum_{i=1..6*p-5} a(i,p) * C(n-1,i-1)/(i+6).
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%I #22 Mar 11 2018 13:26:51

%S 1,1,6,15,20,15,6,1,1,48,687,4850,20385,55908,104959,137886,127050,

%T 80640,33642,8316,924,1,342,21267,527876,7020525,58015362,324610399,

%U 1297791264,3839203452,8595153000,14760228672,19560928464,19987430694

%N 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 6*p-5, where a(i,p) satisfies Sum_{i=1..n} C(i+5,6)^p = 7 * C(n+6,7) * Sum_{i=1..6*p-5} a(i,p) * C(n-1,i-1)/(i+6).

%C From _Peter Bala_, Mar 11 2018: (Start)

%C 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)

%H G. C. Greubel, <a href="/A087110/b087110.txt">Table of n, a(n) for the first 40 rows, flattened</a>

%H 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 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) ]

%F From _Peter Bala_, Mar 11 2018: (Start)

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

%F 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.

%F 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.

%F 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.

%F R(n+1,x) = 1/6!*(1 + x)^6 * (d/dx)^6(x^6*R(n,x)).

%F R(n,x) = Sum_{i >= 0} binomial(i+6,6)^n*x^i/(1 + x)^(i+1).

%F (1 - x)^(6*n)*R(n,x/(1 - x)) appears to equal the n-th row polynomial of A237252. (End)

%e 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.

%p 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

%t 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 *)

%o (PARI) {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

%Y 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.

%Y Cf. A087127, A237252.

%K easy,nonn,tabf

%O 1,3

%A _André F. Labossière_, Aug 11 2003

%E Edited by _Dean Hickerson_, Aug 16 2003