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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).

%I A236463 #42 Jan 28 2025 10:47:34
%S A236463 1,1,16,36,16,1,1,112,1828,8464,13840,8464,1828,112,1,1,608,40136,
%T A236463 724320,4961755,15018688,21571984,15018688,4961755,724320,40136,608,1,
%U A236463 1,3104,693960,37229920,733059110,6501577152,29066972368,69830127680,93200908410,69830127680
%N 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).
%C A236463 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).
%C A236463 Using these coefficients we can obtain formulas for binomial(n,e)^p and for Sum_{i=1..n} binomial(e-1+i,e)^p.
%C A236463 In particular:
%C A236463   binomial(n, e)^p = Sum_{k=0..e*(p-1)} b(k,e,p) * binomial(n+k, e*p).
%C A236463   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).
%C A236463 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
%H A236463 Vincenzo Librandi, <a href="/A236463/b236463.txt">Table of n, a(n) for n = 1..780</a>
%H A236463 Feihu Liu, Guoce Xin, and Chen Zhang, <a href="https://arxiv.org/abs/2412.18744">Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS</a>, arXiv:2412.18744 [math.CO], 2024. See p. 12.
%F A236463 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).
%F A236463 binomial(n,4)^p = Sum_{k=0..4*(p-1)} T(p,k) * binomial(n+k, 4*p).
%e A236463 T(n,0) = 1;
%e A236463 T(n,1) = 5^n - (4*n+1);
%e A236463 T(n,2) = 15^n - (4*n+1)*5^n + C(4*n+1,2);
%e A236463 T(n,3) = 35^n - (4*n+1)*15^n + C(4*n+1,2)*5^n - C(4*n+1,3);
%e A236463 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).
%e A236463 Triangle T(n,k) begins:
%e A236463 1,
%e A236463 1, 16, 36, 16, 1;
%e A236463 1, 112, 1828, 8464, 13840, 8464, 1828, 112, 1;
%e A236463 1, 608, 40136, 724320, 4961755, 15018688, 21571984, 15018688, 4961755, 724320, 40136, 608, 1;
%e A236463 1, 3104, 693960, 37229920, 733059110, 6501577152, 29066972368, 69830127680, 93200908410, 69830127680, 29066972368, 6501577152, 733059110, 37229920, 693960, 3104, 1;
%e A236463 1, 15600, 11000300, 1558185200, 75073622025, 1585757994496, 16938467955200, 99825129369600, 342907451401150, 710228619472800, 903546399077256, 710228619472800, 342907451401150, 99825129369600, 16938467955200, 1585757994496, 75073622025, 1558185200, 11000300, 15600, 1;
%e A236463   ...
%e A236463 Example:
%e A236463 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).
%e A236463 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).
%t A236463 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 *)
%o A236463 (PARI) 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
%Y A236463 Columns k=2..8 are A151640, A151641, A151642, A151643, A151644, A151645, A151646.
%Y A236463 Row sums are A014608.
%Y A236463 Similar triangles for e=1..6: A173018 (or A008292), A154283, A174266, this sequence, A237202, A237252.
%Y A236463 Sum_{i=1..n} binomial(3+i,4)^p for p=2..3 gives: A086023, A086024.
%Y A236463 Cf. A087127, A087107, A087108, A087109, A087110, A087111, A181544.
%K A236463 nonn,tabf
%O A236463 1,3
%A A236463 _Yahia Kahloune_, Feb 01 2014
%E A236463 a(36) corrected by _Vincenzo Librandi_, Feb 14 2014
%E A236463 Edited by _Andrew Howroyd_, May 08 2020