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%I #26 Oct 02 2023 09:25:19
%S 1,1,2,4,10,27,81,262,910,3363,13150,54135,233671,1053911,4951997,
%T 24177536,122381035,640937746,3466900453,19337255086,111057640382,
%U 655892813805,3978591077096,24760700544301,157941950878839
%N G.f.: A(x) = Sum_{k>=0} x^k * (1+x)^(k*(k+1)/2).
%C a(n) is the number of length n permutations that simultaneously avoid the bivincular patterns (123,{2},{}) and (132,{},{2}). - _Christian Bean_, Jun 03 2015
%H Seiichi Manyama, <a href="/A121690/b121690.txt">Table of n, a(n) for n = 0..685</a>
%H Christian Bean, A Claesson, H Ulfarsson, <a href="http://arxiv.org/abs/1512.03226">Simultaneous Avoidance of a Vincular and a Covincular Pattern of Length 3</a>, arXiv preprint arXiv:1512.03226, 2015
%F a(n) = Sum_{k=0..n} C(k*(k+1)/2,n-k).
%F a(n) = A131338(n+1, n*(n+1)/2 + 1) for n>=0, where triangle A131338 starts with a '1' in row 0 and then for n>0 row n consists of n '1's followed by the partial sums of the prior row. - _Paul D. Hanna_, Aug 30 2007
%F From _Paul D. Hanna_, Apr 24 2010: (Start)
%F Let q = (1+x), then g.f. A(x) equals the continued fraction:
%F A(x) = 1/(1 - q*x/(1 - (q^2-q)*x/(1 - q^3*x/(1 - (q^4-q^2)*x/(1 - q^5*x/(1- (q^6-q^3)*x/(1 - q^7*x/(1 - (q^8-q^4)*x/(1 - ...)))))))))
%F due to an identity of a partial elliptic theta function.
%F (End)
%F G.f.: Sum_{n>=0} x^n * Product_{k=1..n} (1 - x*(1+x)^(2*k-2))/(1 - x*(1+x)^(2*k-1)). - _Paul D. Hanna_, Mar 21 2018
%t Table[Sum[Binomial[k*(k+1)/2,n-k],{k,0,n}],{n,0,30}] (* _Vaclav Kotesovec_, Jun 03 2015 *)
%o (PARI) a(n) = sum(k=0,n, binomial(k*(k+1)/2, n-k))
%o for(n=0, 30, print1(a(n), ", "))
%o (PARI) {a(n)=polcoeff(sum(m=0, n, x^m*prod(k=1, m, (1 - x*(1+x)^(2*k-2))/(1 - x*(1+x)^(2*k-1) + x*O(x^n)))), n)}
%o for(n=0, 30, print1(a(n), ", ")) \\ _Paul D. Hanna_, Mar 21 2018
%Y Cf. A121689, A131338.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Aug 15 2006
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