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

Author

Enrique Pérez Herrero, Sep 10 2013

Keywords

• nonn

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

Crossrefs

Cf. A000110, A016789, A155730, A054767.

Programs

Mathematica

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

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)