A268137 a(n) = (1/n)*Sum_{k=0..n-1} A001850(k)*A245769(k).

-1, 1, 31, 417, 5919, 97217, 1828479, 38085249, 853450367, 20174707521, 496690317855, 12626836592289, 329476040177439, 8785359461936769, 238587766484265471, 6581966817521388033, 184067922884292651519, 5209333642085984431489, 148992465188631205367071, 4301514890878664802287777

Offset

1,3

Comments

Conjecture: (i) All the terms are odd integers. For any prime p, if p == 3 (mod 4) then a(p) == -5 (mod p^2), otherwise a(p) == -1 (mod p).

(ii) For n = 0,1,2,... let D_n(x) = Sum_{k=0..n} binomial(n,k)*binomial(n+k,k)*x^k and R_n(x) = Sum_{k=0..n} binomial(n,k)*binomial(n+k,k)*x^k/(2k-1). For any positive integer n, all the coefficients of the polynomial (1/n)*Sum_{k=0..n-1} D_k(x)*R_k(x) are integral and the polynomial is irreducible over the field of rational numbers.

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.

Zhi-Wei Sun, Arithmetic properties of Delannoy numbers and Schröder numbers, preprint, arXiv:1602.00574 [math.CO], 2016.

Example

a(3) = 31 since (A001850(0)*A245769(0) + A001850(1)*A245769(1) + A001850(2)*A245769(2))/3 = (1*(-1) + 3*1 + 13*7)/3 = 93/3 = 31.

Mathematica

d[n_]:=d[n]=Sum[Binomial[n, k]Binomial[n+k, k], {k, 0, n}]
R[n_]:=R[n]=Sum[Binomial[n, k]Binomial[n+k, k]/(2k-1), {k, 0, n}]
a[n_]:=a[n]=Sum[d[k]*R[k], {k, 0, n-1}]/n
Do[Print[n, " ", a[n]], {n, 1, 20}]

Crossrefs

Cf. A001850, A245769, A268136, A268138.

Sequence in context: A297019 A022691 A125443 * A089375 A069582 A175355

Adjacent sequences: A268134 A268135 A268136 * A268138 A268139 A268140

Keyword

sign

Author

Zhi-Wei Sun, Jan 26 2016

Status

approved