|
%I #22 May 30 2020 23:02:53
%S 1,3,20,169,1667,18084,208960,2527074,31630390,406680465,5342750699,
%T 71442850111,969548468960,13323571588607,185072895183632,
%U 2594890728951909,36681505784903758,522291180086851188,7484621370716999785,107876522368295972285,1562916545414144667559
%N Central elements in Parker's partition triangle.
%H Alois P. Heinz, <a href="/A051643/b051643.txt">Table of n, a(n) for n = 0..90</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.
%F 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]
%F 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
%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 a:= n-> b(2*n*(n+1), 2*n+1$2):
%p seq(a(n), n=0..20); # _Alois P. Heinz_, May 30 2020
%t a[n_] := SeriesCoefficient[QBinomial[2(2n+1), 2n+1, q], {q, 0, 2n(n+1)}];
%t Table[a[n], {n, 0, 17}] (* _Jean-François Alcover_, Aug 19 2019 *)
%Y Cf. A007042, A047812, A136621.
%K easy,nonn,nice
%O 0,2
%A _James A. Sellers_
%E a(18)-a(20) from _Alois P. Heinz_, May 30 2020
|
