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Compare g.f. to the Lambert series identity: Sum_{n>=1} lamdalambda(n)*x^n/(1-x^n) = Sum_{n>=1} x^(n^2).
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_Paul D. Hanna (pauldhanna(AT)juno.com), _, Jan 12 2012
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G.f.: Sum_{n>=1} fibonacci(n^2)*x^(n^2).
1, 0, 0, 3, 0, 0, 0, 0, 34, 0, 0, 0, 0, 0, 0, 987, 0, 0, 0, 0, 0, 0, 0, 0, 75025, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 14930352, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7778742049, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10610209857723, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 37889062373143906, 0, 0, 0, 0, 0
Compare g.f. to the Lambert series identity: Sum_{n>=1} lamda(n)*x^n/(1-x^n) = Sum_{n>=1} x^(n^2).
G.f.: A(x) = x + 3*x^4 + 34*x^9 + 987*x^16 + 75025*x^25 + 14930352*x^36 +...
(PARI) {a(n)=issquare(n)*fibonacci(n)}
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