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A207386
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G.f.: Sum_{n>=0} Product_{k=1..n} (q^k - 1) where q = (1+x)/(1+x^3).
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6
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1, 1, 2, 6, 28, 172, 1269, 10879, 106343, 1167970, 14241792, 190919195, 2790920003, 44184957237, 753152722642, 13752229833566, 267809474619299, 5540559819166056, 121355678158129804, 2805498395990301867, 68265999939081386947, 1744058001878302097109
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OFFSET
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0,3
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COMMENTS
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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^3). See cross-references for other examples.
At present Bala's identity is conjectural and needs formal proof.
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LINKS
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FORMULA
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G.f.: Sum_{n>=0} q^(-n^2) * Product_{k=1..n} (q^(2*k-1) - 1) where q = (1+x)/(1+x^3). [Based on Peter Bala's conjecture in A158690]
a(n) ~ n! * 2^(2*n+3/2) * 3^(n+1) / (exp(Pi^2/24) * Pi^(2*n+2)). - Vaclav Kotesovec, Aug 22 2017
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EXAMPLE
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G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 28*x^4 + 172*x^5 + 1269*x^6 +...
Let q = (1+x)/(1+x^3) = 1/(1-x+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 +...
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PROG
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(PARI) {a(n)=local(A=1+x, q=(1+x)/(1+x^3 +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^3 +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, 21, print1(a(n), ", "))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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