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%I #10 Dec 06 2013 08:32:38
%S 1,1,1,1,1,1,1,1,2,2,3,2,2,1,1,1,1,2,3,4,4,5,4,4,3,2,1,1,1,1,2,3,5,5,
%T 7,7,8,7,7,5,5,3,2,1,1,1,1,2,3,5,6,8,9,11,11,12,11,11,9,8,6,5,3,2,1,1,
%U 1,1,2,3,5,6,9,10,13,14,16,16,18,16,16,14,13,10,9,6,5,3,2,1,1
%N Coefficient of Gauss polynomials [n+4,4]_q (q-binomials).
%C The length of row n of this table is 4*n + 1 = A016813(n).
%C The sum of row n is binomial(n+4,4) = A000332(n+4), n>= 0.
%C The Gauss polynomial [n+4,4]_q := [n+4]_q/([n]_q*[4]_q, with [n]_q = product(1-q^j,j=1..n) = (q;q)_n (in q-shifted factorials notation), n>=0. [n+4,4]_q = product(1-q^j,j=(n+1)..(n+4))/product(1-q^j,j=1..4). This is a polynomial in q (of degree 4*n) because it is the o.g.f. of the numbers p(n,4,k), the number of partitions of k into at most 4 parts, each <= n (see Andrews, p. 33 and 35). p(n,4,k) is also the number of partitions of k into at most n parts, each <= 4, due to the symmetry property [n+4,4]_q = [n+4,n]_q (Andrews, (3,3,2), p.35). With the latter interpretation p(n,4,k) is the number of solutions of the two Diophantine equations sum(j*m(j),j=1..4) = k and sum(m(j),j=0..m) = n, i.e. sum(m(j),j=1..m) = n - m(0), with 0 <= m(j) <= n. Therefore p(n,4,k) = [q^k] [x^n] G(4;x,q) with o.g.f. G(4;x,q) = 1/product(1-x*q^j,j=0..4). Here we will call p(n,4,k) = a(n,k), n >= 0, 0 <= k <= 4*n.
%C See the comments in A008967 concerning a counting problem of Cayley (there m = 4, Theta = n and q = k), described also in the Hawkins reference (N(p->n,4,w->k) = a(n,k)) given there.
%D G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976, p. 240, 242-3.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/q-BinomialCoefficient.html">q-Binomial Coefficient</a>.
%F a(n,k) = [q^k] [x^n](1/product(1-x*q^j,j=0..4)), n >= 0, 0 <= k <= 4*n.
%F a(n,k) = [q^k]([n+4,4]_q), n >= 0, 0 <= k <= 4*n.
%F See the comments above.
%e The table a(n,k) begins:
%e n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
%e 0: 1
%e 1: 1 1 1 1 1
%e 2: 1 1 2 2 3 2 2 1 1
%e 3: 1 1 2 3 4 4 5 4 4 3 2 1 1
%e 4: 1 1 2 3 5 5 7 7 8 7 7 5 5 3 2 1 1
%e n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
%e ...
%e Row n = 5: [1, 1, 2, 3, 5, 6, 8, 9, 11, 11, 12, 11, 11, 9, 8, 6, 5, 3, 2, 1, 1],
%e Row n = 6: [1, 1, 2, 3, 5, 6, 9, 10, 13, 14, 16, 16, 18, 16, 16, 14, 13, 10, 9, 6, 5, 3, 2, 1, 1].
%e Partition interpretation: a(3,5) = 4 because there are 4 partitions of 5 into at most 4 parts, each <= 3, namely 23, 113, 122 and 1112. here are also 4 partitions of 5 into at most 3 parts, each <= 4, namely 14, 23, 113 and 122. Note the conjugacy of the partitions 1112 and 14.
%e The 4 solutions of the two Diophantine equations given in a comment, with k=5 and n=3, are for (m(0), m(1), m(2), m(3), m(4)): (1,1,0,0,1), (1,0,1,1,0), (0,2,0,1,0) and (0,1,2,0,0).
%t a[0, 0] = 1; a[n_, k_] := SeriesCoefficient[ QBinomial[n+4, 4, q], {q, 0, k}]; Table[a[n, k], {n, 0, 6}, {k, 0, 4*n}] // Flatten (* _Jean-François Alcover_, Dec 04 2013 *)
%Y Cf. A000012 (as triangle for m=1), A008967 (m=2), A047971 (m=3).
%K nonn,easy,tabf
%O 0,9
%A _Wolfdieter Lang_, Dec 04 2012
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