OFFSET
0,4
COMMENTS
Motivated by Peter Bala's identity described in A158690:
Sum_{n>=0} Product_{k=1..n} (q^k - 1) =
Sum_{n>=0} q^(-n^2) * Product_{k=1..n} (q^(2*k-1) - 1),
here q = (1+x)/(1+x^2). See cross-references for other examples.
At present Bala's identity is conjectural and needs formal proof.
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 0..240
Hsien-Kuei Hwang, Emma Yu Jin, Asymptotics and statistics on Fishburn matrices and their generalizations, arXiv:1911.06690 [math.CO], 2019.
FORMULA
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]
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
EXAMPLE
G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 11*x^4 + 74*x^5 + 557*x^6 + 4799*x^7 +...
Let q = (1+x)/(1+x^2), then
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) +...
which also is proposed to equal:
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 +...
PROG
(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)}
(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)}
for(n=0, 25, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Feb 17 2012
STATUS
approved