A294362 E.g.f.: exp(Sum_{n>=1} sigma_2(n) * x^n).

1, 1, 11, 91, 1105, 13841, 230731, 3955771, 80483201, 1738065025, 41800101931, 1070731623611, 29804263624081, 878224530964561, 27672361220570795, 919409968480087771, 32304618825218432641, 1191168445737728717441, 46119903359374012564171

Offset

0,3

Comments

From Peter Bala, Nov 14 2017: (Start)

It appears that the sequence taken modulo 10 is periodic with period (1, 1, 1, 1, 5) of length 5.

More generally, we conjecture that for k = 2,3,4,... the sequence a(n+k) - a(n) is divisible by k: if true, then for each k the sequence a(n) taken modulo k would be periodic with the exact period dividing k. (End)

From Peter Bala, Mar 28 2022: (Start)

The above conjectures are true. See the Bala link.

a(5*n+4) = 0 (mod 5); a(7*n+3) == 0 (mod 7); a(11*n+2) == 0 (mod 11); a(13*n+3) == 0 (mod 13); a(17*n+4) == 0 (mod 17); a(19*n+12) == 0 (mod 19). (End)

Links

Seiichi Manyama, Table of n, a(n) for n = 0..417

Peter Bala, Integer sequences that become periodic on reduction modulo k for all k

Formula

a(0) = 1 and a(n) = (n-1)! * Sum_{k=1..n} k*A001157(k)*a(n-k)/(n-k)! for n > 0.

E.g.f.: Product_{k>=1} exp(k^2*x^k/(1 - x^k)). - Ilya Gutkovskiy, Nov 27 2017

a(n) ~ (3*Zeta(3))^(1/8) * exp(2^(9/4) * Zeta(3)^(1/4) * n^(3/4) / 3^(3/4) - n^(1/4) / (2^(9/4) * 3^(5/4) * Zeta(3)^(1/4)) - n) * n^(n - 1/8) / 2^(7/8). - Vaclav Kotesovec, Sep 04 2018

Mathematica

nmax = 20; CoefficientList[Series[Exp[Sum[DivisorSigma[2, k]*x^k, {k, 1, nmax}]], {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Sep 04 2018 *)

Prog

(PARI) N=66; x='x+O('x^N); Vec(serlaplace(exp(sum(k=1, N, sigma(k, 2)*x^k))))

Crossrefs

E.g.f.: exp(Sum_{n>=1} sigma_k(n) * x^n): A294363 (k=0), A294361 (k=1), this sequence (k=2).

Cf. A001157.

Sequence in context: A199678 A201549 A117611 * A363669 A044262 A044643

Adjacent sequences: A294359 A294360 A294361 * A294363 A294364 A294365

Keyword

nonn,easy

Author

Seiichi Manyama, Oct 29 2017

Status

approved