A268136 a(n) = (3/n)*Sum_{k=0..n-1} A245769(k)^2.

3, 3, 51, 507, 4947, 58243, 841443, 14240763, 269512483, 5524472451, 120183938835, 2738420763131, 64760819179635, 1579226738429187, 39515677808716739, 1010750709382934523, 26349289260686093379, 698387854199468231427, 18783213754115549685747, 511772677524431483886075

Offset: 1

Author

Zhi-Wei Sun, Jan 26 2016

Keywords

• nonn

Comments

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.

Examples

			a(3) = 51 since (3/3)*(A245769(0)^2 + A245769(1)^2 + A245769(2)^2) = (-1)^2 + 1^2 + 7^2 = 51.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..100
• Zhi-Wei Sun, Two new kinds of numbers and related divisibility results, preprint, arXiv:1408.5381 [math.NT], 2014.

Crossrefs

Cf. A000290, A245769, A268137, A268138.

Programs

Mathematica

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}]