%I #19 Feb 05 2020 23:52:22
%S 1,1,1,2,11,74,557,4799,47004,516717,6302993,84502346,1235198136,
%T 19552296646,333212892221,6083009119262,118433569748072,
%U 2449663066933397,53643715882853914,1239875630317731463,30163779836127304106,770476745704778418686
%N G.f.: Sum_{n>=0} Product_{k=1..n} (q^k - 1) where q = (1+x)/(1+x^2).
%C Motivated by Peter Bala's identity described in A158690:
%C Sum_{n>=0} Product_{k=1..n} (q^k - 1) =
%C Sum_{n>=0} q^(-n^2) * Product_{k=1..n} (q^(2*k-1) - 1),
%C here q = (1+x)/(1+x^2). See cross-references for other examples.
%C At present Bala's identity is conjectural and needs formal proof.
%H Vaclav Kotesovec, <a href="/A207397/b207397.txt">Table of n, a(n) for n = 0..240</a>
%H Hsien-Kuei Hwang, Emma Yu Jin, <a href="https://arxiv.org/abs/1911.06690">Asymptotics and statistics on Fishburn matrices and their generalizations</a>, arXiv:1911.06690 [math.CO], 2019.
%F G.f.: Sum_{n>=0} q^(-n^2) * Product_{k=1..n} (q^(2*k-1) - 1) where q = (1+x)/(1+x^2). [Based on Peter Bala's conjecture in A158690]
%F a(n) ~ c * 12^n * n! / Pi^(2*n), where c = 6*sqrt(2) / (Pi^2 * exp(Pi^2/8)) = 0.250367043877216848533826021231826... . - _Vaclav Kotesovec_, May 06 2014, updated Aug 22 2017
%e G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 11*x^4 + 74*x^5 + 557*x^6 + 4799*x^7 +...
%e Let q = (1+x)/(1+x^2), then
%e A(x) = 1 + (q-1) + (q-1)*(q^2-1) + (q-1)*(q^2-1)*(q^3-1) + (q-1)*(q^2-1)*(q^3-1)*(q^4-1) + (q-1)*(q^2-1)*(q^3-1)*(q^4-1)*(q^5-1) +...
%e which also is proposed to equal:
%e A(x) = 1 + (q-1)/q + (q-1)*(q^3-1)/q^4 + (q-1)*(q^3-1)*(q^5-1)/q^9 + (q-1)*(q^3-1)*(q^5-1)*(q^7-1)/q^16 + (q-1)*(q^3-1)*(q^5-1)*(q^7-1)*(q^9-1)/q^25 +...
%o (PARI) {a(n)=local(A=1+x,q=(1+x)/(1+x^2 +x*O(x^n))); A=sum(m=0,n,prod(k=1,m,(q^k-1)));polcoeff(A,n)}
%o (PARI) {a(n)=local(A=1+x,q=(1+x)/(1+x^2 +x*O(x^n))); A=sum(m=0,n,q^(-m^2)*prod(k=1,m,(q^(2*k-1)-1)));polcoeff(A,n)}
%o for(n=0,25,print1(a(n),", "))
%Y Cf. A158690, A158691, A179525, A207386, A207433.
%K nonn
%O 0,4
%A _Paul D. Hanna_, Feb 17 2012
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