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
KEYWORD
nonn
AUTHOR
Enrique Pérez Herrero, Sep 10 2013
STATUS
approved