A229004 Indices of Bell numbers divisible by 3.

4, 8, 9, 11, 17, 21, 22, 24, 30, 34, 35, 37, 43, 47, 48, 50, 56, 60, 61, 63, 69, 73, 74, 76, 82, 86, 87, 89, 95, 99, 100, 102, 108, 112, 113, 115, 121, 125, 126, 128, 134, 138, 139, 141, 147, 151, 152, 154, 160, 164, 165, 167, 173, 177, 178, 180, 186, 190, 191

Offset

1,1

Comments

a(n) appears to be congruent 4, 8, 9, 11 mod 13. - Ralf Stephan, Sep 12 2013

Wagstaff shows that N(p) = (p^p-1)/(p-1) is the period for all primes p < 102, for p=3 then N(3) = A054767(3) = 13, Bell numbers with indices less than or equal to 13 that are divisible by 3 are those with indices: 4, 8, 9, 11, so the conjecture holds. - Enrique Pérez Herrero, Sep 12 2013

Links

Vaclav Kotesovec, Table of n, a(n) for n = 1..15384 (terms 1..1200 from Enrique Pérez Herrero)

J. Levine and R. E. Dalton, Minimum Periods, Modulo p, of First Order Bell Exponential Integrals, Mathematics of Computation, 16 (1962), 416-423.

Samuel S. Wagstaff Jr., Aurifeuillian factorizations and the period of the Bell numbers modulo a prime, Math. Comp. 65 (1996), 383-391.

Eric Weisstein's World of Mathematics, Bell Number

Formula

Conjectures from Colin Barker, Jul 16 2014: (Start)

a(n) = a(n-1) + a(n-4) - a(n-5).

G.f.: x*(2*x^4+2*x^3+x^2+4*x+4) / ((x-1)^2*(x+1)*(x^2+1)). (End)

Mathematica

Select[Range[1000], Mod[BellB[#], 3] == 0&]

Crossrefs

Cf. A000110, A016789, A155730, A054767.

Sequence in context: A158758 A317253 A368812 * A306976 A266142 A297252

Adjacent sequences: A229001 A229002 A229003 * A229005 A229006 A229007

Keyword

nonn

Author

Enrique Pérez Herrero, Sep 10 2013

Status

approved