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A268136
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a(n) = (3/n)*Sum_{k=0..n-1} A245769(k)^2.
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3
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3, 3, 51, 507, 4947, 58243, 841443, 14240763, 269512483, 5524472451, 120183938835, 2738420763131, 64760819179635, 1579226738429187, 39515677808716739, 1010750709382934523, 26349289260686093379, 698387854199468231427, 18783213754115549685747, 511772677524431483886075
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OFFSET
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1,1
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COMMENTS
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Conjecture: (i) All the terms are odd integers.
(ii) For n = 0,1,2,... let R_n(x) denote the polynomial sum_{k=0..n} binomial(n,k)*binomial(n+k,k)*x^k/(2k-1). Then, for each n = 1,2,3,.., all the coefficients of the polynomial (3/n)*Sum_{k=0..n-1} R_k(x)^2 are integral and the polynomial is irreducible over the field of rational numbers.
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LINKS
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EXAMPLE
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MATHEMATICA
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R[n_]:=Sum[Binomial[n, k]Binomial[n+k, k]/(2k-1), {k, 0, n}]
a[n_]:=Sum[R[k]^2, {k, 0, n-1}]*3/n
Do[Print[n, " ", a[n]], {n, 1, 20}]
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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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