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%I #24 Nov 09 2023 15:31:36
%S 1,0,2,0,0,3,0,0,1,4,0,0,0,2,5,0,0,0,2,3,6,0,0,0,0,4,4,7,0,0,0,0,3,6,
%T 5,8,0,0,0,0,1,6,8,6,9,0,0,0,0,0,6,9,10,7,10,0,0,0,0,0,2,11,12,12,8,
%U 11,0,0,0,0,0,2,9,16,15,14,9,12,0,0,0,0,0,0,7,16,21,18,16,10,13,0,0,0,0,0,0
%N Triangular array T(n,k) in which k-th column gives coefficients of sum of Gaussian polynomials [k,m] for m=0..k.
%C It appears that T(n-1,k-1) is the number of partitions of n with k objects in the first hook; i.e., with (largest part size) + (number of parts) - 1 = k. If this is correct, we have T(n-1,k-1) = sum_{j<=min(k,n-k-2)} (k-j) * T(k-1,j-1) with T(n-1,n-1) = n. Equivalently, T(n-1,k-1) = T(n-2,k-2) + sum(j<=min(k,n-k-2)} T(k-1,j-1) and thus T(n-1,k-1) = 2*T(n-2,k-2) - T(n-3,k-3) + T(k-1,n-k-3). - _Franklin T. Adams-Watters_, May 27 2008
%D G. E. Andrews, Theory of Partitions, 1976, pages 240-243
%F The g.f. for the n-th row as polynomial in q, sum(k=0..n, T(n,k)*q^k) is sum(k>=0, x^(k*(k+1))*q^(2*k)/(1-x^(k+1)*q)/prod(j=1..k, 1-x^j*q)^2). For example, the 5th row is the coefficient of x^6 of the g.f., 2*q^4 + 3*q^5 + 6*q^6. - _T. Amdeberhan_, Jul 31 2012
%e Table begins:
%e 1
%e 0 2
%e 0 0 3
%e 0 0 1 4
%e 0 0 0 2 5
%e 0 0 0 2 3 6
%e 0 0 0 0 4 4 7
%e 0 0 0 0 3 6 5 8
%e For k=4 the 5 polynomials have coefficients 1; 1 1 1 1; 1 1 2 1 1; 1 1 1 1; 1; which sum to 5 3 4 3 1, giving column 4.
%p a := n->sort(simplify(sum(product((1-q^i),i=n-r+1..n)/product((1-q^j),j=1..r), r=0..n))):T := proc(n,k) if k=n then n+1 elif k>n then 0 else coeff(a(k),q^(n-k)) fi end:seq(seq(T(n,k),k=0..n),n=0..21);
%t a [n_] := Sum[Product[1-q^i, {i, n-r+1, n}]/Product[1-q^j, {j, 1, r}], {r, 0, n}] // Simplify; T [n_, k_] := Which[k == n, n+1, k>n, 0, True, Coefficient[a[k], q^(n - k)]]; Table[Table[T[n, k], {k, 0, n}], {n, 0, 21}] // Flatten (* _Jean-François Alcover_, Feb 19 2015, after Maple *)
%Y The nonzero entries of the columns are the rows of A083906.
%K nonn,tabl
%O 0,3
%A _Alford Arnold_
%E More terms from _Emeric Deutsch_, Feb 23 2004
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