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%I #18 Apr 25 2019 04:25:32
%S 1,3,8,18,39,75,141,251,433,724,1185,1892,2972,4588,6981,10480,15553,
%T 22821,33164,47746,68163,96542,135747,189550,262997,362691,497339,
%U 678300,920417,1242898,1670688,2235880,2979809,3955422,5230471,6891234
%N Column 2 of triangle A128545; a(n) is the coefficient of q^(2n+4) in the central q-binomial coefficient [2n+4,n+2].
%C Column 1 of triangle A128552 equals the partitions of n (A000041).
%C a(n) is the number of partitions of the integer 2n+4 into at most n+2 summands each of which is at most n+2. - _Geoffrey Critzer_, Sep 27 2013
%H Alois P. Heinz, <a href="/A128552/b128552.txt">Table of n, a(n) for n = 0..1000</a>
%H P. Flajolet and R. Sedgewick, <a href="http://algo.inria.fr/flajolet/Publications/AnaCombi/anacombi.html">Analytic Combinatorics</a>, 2009, page 45.
%H Shishuo Fu and James Sellers, <a href="https://www.emis.de/journals/INTEGERS/papers/m24/m24.Abstract.html">Enumeration of the degree sequences of line-Hamiltonian multigraphs</a>, INTEGERS 12 (2012), #A24. - From _N. J. A. Sloane_, Feb 04 2013
%F a(n) = A000041(2*n+4) - 2*Sum_{k=0..n+1} A000041(k), where A000041(n) = number of partitions of n, due to a formula given in the Fu and Sellers paper. - _Paul D. Hanna_, Feb 06 2013
%e a(2) = 8 because we have: 4+4 = 4+3+1 = 4+2+2 = 4+2+1+1 = 3+3+2 = 3+3+1+1 = 3+2+2+1 = 2+2+2+2. - _Geoffrey Critzer_, Sep 27 2013
%p with(combinat): p:= numbpart:
%p s:= proc(n) s(n):= p(n) +`if`(n>0, s(n-1), 0) end:
%p a:= n-> p(2*n+4) -2*s(n+1):
%p seq(a(n), n=0..40); # _Alois P. Heinz_, Sep 27 2013
%t Table[nn=2n;Coefficient[Series[Product[(1-x^(n+i))/(1-x^i),{i,1,n}],{x,0,nn}],x^(2n)],{n,1,37}] (* _Geoffrey Critzer_, Sep 27 2013 *)
%o (PARI) {a(n)=polcoeff(prod(j=n+3,2*n+4,1-q^j)/prod(j=1,n+2,1-q^j),2*n+4,q)}
%o (PARI) {a(n)=numbpart(2*n+4)-2*sum(k=0,n+1,numbpart(k))} \\ _Paul D. Hanna_, Feb 06 2013
%Y Cf. A128545; A128553, A128554.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Mar 10 2007
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