A136621 -id:A136621 - cp's OEIS frontend

A047812 Parker's partition triangle T(n,k) read by rows (n >= 1 and 0 <= k <= n-1).

1, 1, 1, 1, 3, 1, 1, 5, 7, 1, 1, 9, 20, 11, 1, 1, 13, 48, 51, 18, 1, 1, 20, 100, 169, 112, 26, 1, 1, 28, 194, 461, 486, 221, 38, 1, 1, 40, 352, 1128, 1667, 1210, 411, 52, 1, 1, 54, 615, 2517, 4959, 5095, 2761, 720, 73, 1, 1, 75, 1034, 5288, 13241, 18084, 13894, 5850, 1221, 97, 1

Offset: 1

N. J. A. Sloane

The entries in row n are the coefficients of q^(k*(n+1)) in the q-binomial coefficient [2n, n], where k runs from 0 to n-1. - James Sellers
T(n,k) is the number of partitions of k*(n+1) into at most n parts each no bigger than n (see the links). - Petros Hadjicostas, May 30 2020
Named after the American mathematician Ernest Tilden Parker (1926-1991). - Amiram Eldar, Jun 20 2021

			Triangle T(n,k) (with rows n >= 1 and columns k = 0..n-1) starts:
  1;
  1,  1;
  1,  3,  1;
  1,  5,  7   1;
  1,  9, 20, 11,  1;
  1, 13, 48, 51, 18, 1;
  ...

Alois P. Heinz, Rows n = 1..141, flattened
Richard K. Guy, Letter to N. J. A. Sloane, Aug. 1992.
Richard K. Guy, Parker's permutation problem involves the Catalan numbers, Preprint, 1992. (Annotated scanned copy)
Richard K. Guy, Parker's permutation problem involves the Catalan numbers, Amer. Math. Monthly, Vol. 100, No. 3 (1993), pp. 287-289.
Wikipedia, E. T. Parker.

Cf. A000108 (row sums), A136621 (mirror image).

b:= proc(n, i, t) option remember; `if`(n=0, 1, `if`(t*i
       b(k*(n+1), n$2):
seq(seq(T(n, k), k=0..n-1), n=1..12);  # Alois P. Heinz, May 30 2020

s[n_] := s[n] = Series[Product[(1-q^(2n-k)) / (1-q^(k+1)), {k, 0, n-1}], {q, 0, n^2}];
t[n_, k_] := SeriesCoefficient[s[n], k(n+1)];
Flatten[Table[t[n, k], {n, 1, 12}, {k, 0, n-1}]] (* Jean-François Alcover, Jan 27 2012 *)
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[n_, k_] := b[k(n+1), n, n];
Table[T[n, k], {n, 1, 12}, {k, 0, n - 1}] // Flatten (* Jean-François Alcover, Nov 27 2020, after Alois P. Heinz *)

T(n,k) = #partitions(k*(n+1), n,n);
for (n=1, 10, for (k=0, n-1, print1(T(n,k), ", "); ); print(); ); \\ Petros Hadjicostas, May 30 2020
/* Second program, courtesy of G. C. Greubel */
T(n,k) = polcoeff(prod(j=0, n-1, (1-q^(2*n-j))/(1-q^(j+1)) ), k*(n+1) );
vector(12, n, vector(n, k, T(n,k-1))); \\ Petros Hadjicostas, May 31 2020

More terms from James Sellers

Offset corrected by Alois P. Heinz, May 30 2020

A051643 Central elements in Parker's partition triangle.

Original entry on oeis.org

1, 3, 20, 169, 1667, 18084, 208960, 2527074, 31630390, 406680465, 5342750699, 71442850111, 969548468960, 13323571588607, 185072895183632, 2594890728951909, 36681505784903758, 522291180086851188, 7484621370716999785, 107876522368295972285, 1562916545414144667559

Offset: 0

James Sellers

Alois P. Heinz, Table of n, a(n) for n = 0..90
R. K. Guy, Parker's permutation problem involves the Catalan numbers, Amer. Math. Monthly 100 (1993), 287-289.

Cf. A007042, A047812, A136621.

b:= proc(n, i, t) option remember; `if`(n=0, 1, `if`(t*i
       b(2*n*(n+1), 2*n+1$2):
seq(a(n), n=0..20);  # Alois P. Heinz, May 30 2020

a[n_] := SeriesCoefficient[QBinomial[2(2n+1), 2n+1, q], {q, 0, 2n(n+1)}];
Table[a[n], {n, 0, 17}] (* Jean-François Alcover, Aug 19 2019 *)

a(n) = coefficient of q^((m^2-1)/2) = q(2*n*(n+1)) in the q-binomial coefficient [2*m, m] = [2*(2*n+1), 2*n+1], where m = 2*n+1. [Corrected by Petros Hadjicostas, May 30 2020]

a(n) is the number of partitions of 2*n*(n+1) into at most 2*n+1 parts each no bigger than 2*n+1. - Petros Hadjicostas, May 30 2020

a(18)-a(20) from Alois P. Heinz, May 30 2020

A136622 Partial sums of the irregular table A136624.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 3, 5, 2, 1, 1, 3, 8, 4, 2, 1, 1, 3, 9, 10, 4, 2, 1, 1, 3, 9, 17, 8, 4, 2, 1, 1, 3, 9, 23, 16, 8, 4, 2, 1, 1, 3, 9, 27, 28, 14, 8, 4, 2, 1, 1, 3, 9, 28, 43, 26, 14, 8, 4, 2, 1, 1, 3, 9, 28, 60, 41, 24, 14, 8, 4, 2

Offset: 1

Alford Arnold, Jan 29 2008

A129176 can also be viewed as partial sums, but are perpendicular to the sequences of A136622.

			A136624 begins
  1
  ...1
  ......2...1
  ..........2...3...3...1
  ..............2...2...6...7
  ..................2...2...4
  ......................2...2
  ..........................2
therefore this sequence begins
  1...1...1...1...1...1...1...1
  ....1...1...1...1...1...1...1
  ........2...3...3...3...3...3
  ............2...5...8...9...9
  ................2...4..10..17
  ....................2...4...8
  ........................2...4
  ............................2

Cf. A129176, A136621, A136624, A136625.

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A047812 Parker's partition triangle T(n,k) read by rows (n >= 1 and 0 <= k <= n-1).

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A051643 Central elements in Parker's partition triangle.

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A136622 Partial sums of the irregular table A136624.

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