%I #37 Nov 27 2020 23:43:13

%S 1,1,1,1,3,1,1,7,5,1,1,11,20,9,1,1,18,51,48,13,1,1,26,112,169,100,20,

%T 1,1,38,221,486,461,194,28,1,1,52,411,1210,1667,1128,352,40,1,1,73,

%U 720,2761,5095,4959,2517,615,54,1,1,97,1221,5850,13894,18084,13241,5288,1034,75,1

%N Transpose T(n,k) of Parker's partition triangle A047812 (n >= 1 and 0 <= k <= n-1).

%C Parker's triangle is closely associated with q-binomial coefficients and Gaussian polynomials; cf. A063746. For example, row 4 of A063746 is 1, 1, 2, 3, 5, 5, 7, 7, 8, 7, 7, 5, 5, 3, 2, 1, 1, the coefficients of [8, 4], while the entries in row 4 of A047812 are the coefficients of q^(k*(4+1)) = q^(5*k) in [8, 4] where k runs from 0 to n-1 = 3. Likewise, by symmetry, "1 7 5 1" is embedded also because they are the coefficients of q^(5*(3-k)), where k runs from 0 to n-1 = 3. [Edited by _Petros Hadjicostas_, May 30 2020]

%H Alois P. Heinz, <a href="/A136621/b136621.txt">Rows n = 1..141, flattened</a>

%H R. K. Guy, <a href="http://www.jstor.org/stable/2324467">Parker's permutation problem involves the Catalan numbers</a>, Amer. Math. Monthly 100 (1993), 287-289.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/E._T._Parker">E. T. Parker</a>.

%e Row four of A047812 is 1 5 7 1, so row four of the present entry is 1 7 5 1.

%e From _Petros Hadjicostas_, May 30 2020: (Start)

%e Triangle T(n,k) (with rows n >= 1 and columns k = 0..n-1) begins:

%e 1;

%e 1, 1;

%e 1, 3, 1;

%e 1, 7, 5, 1;

%e 1, 11, 20, 9, 1;

%e 1, 18, 51, 48, 13, 1;

%e 1, 26, 112, 169, 100, 20, 1;

%e 1, 38, 221, 486, 461, 194, 28, 1;

%e 1, 52, 411, 1210, 1667, 1128, 352, 40, 1;

%e ... (End)

%p b:= proc(n, i, t) option remember; `if`(n=0, 1, `if`(t*i

%p <n, 0, b(n, i-1, t)+b(n-i, min(i, n-i), t-1)))

%p end:

%p T:= (n, k)-> b((n-k-1)*(n+1), n$2):

%p seq(seq(T(n, k), k=0..n-1), n=1..12); # _Alois P. Heinz_, May 30 2020

%t T[n_, k_]:= SeriesCoefficient[QBinomial[2*n, n, q], {q, 0, k*(n+1)}];

%t Table[T[n, n-k], {n,12}, {k,n}]//Flatten (* _G. C. Greubel_, May 31 2020 *)

%t b[n_, i_, t_] := b[n, i, t] = If[n == 0, 1, If[n < 0 || t i < n, 0, b[n, i - 1, t] + b[n - i, Min[i, n - i], t - 1]]];

%t T[n_, k_] := b[(n-k-1)(n+1), n, n];

%t Table[T[n, k], {n, 1, 12}, {k, 0, n-1}] // Flatten (* _Jean-François Alcover_, Nov 27 2020, after _Alois P. Heinz_ *)

%o (PARI) T(n, k) = #partitions(k*(n+1), n, n);

%o for (n=1, 10, for (k=0, n-1, print1(T(n, n-1-k), ", "); ); print(); ); \\ _Petros Hadjicostas_, May 30 2020

%o /* Second program, courtesy of _G. C. Greubel_ */

%o T(n,k) = polcoeff(prod(j=0, n-1, (1-q^(2*n-j))/(1-q^(j+1)) ), k*(n+1) );

%o vector(12, n, vector(n, k, T(n,n-k))) \\ _Petros Hadjicostas_, May 31 2020

%o (Sage)

%o def T(n,k):

%o P.<x> = PowerSeriesRing(ZZ, k*(n+1)+1)

%o return P( q_binomial(2*n, n, x) ).list()[k*(n+1)]

%o [[ T(n,n-k) for k in (1..n)] for n in (1..12)] # _G. C. Greubel_, May 31 2020

%Y Cf. A000108 (Catalan row sums), A047812, A063746.

%K nonn,tabl

%O 1,5

%A _Alford Arnold_, Jan 26 2008

%E Name edited by _Petros Hadjicostas_, May 30 2020