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A207397
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G.f.: Sum_{n>=0} Product_{k=1..n} (q^k - 1) where q = (1+x)/(1+x^2).
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4
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1, 1, 1, 2, 11, 74, 557, 4799, 47004, 516717, 6302993, 84502346, 1235198136, 19552296646, 333212892221, 6083009119262, 118433569748072, 2449663066933397, 53643715882853914, 1239875630317731463, 30163779836127304106, 770476745704778418686
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OFFSET
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0,4
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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^2). 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^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
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EXAMPLE
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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 +...
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PROG
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(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), ", "))
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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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