A229004 - OEIS

%I #28 Jul 23 2021 03:36:08

%S 4,8,9,11,17,21,22,24,30,34,35,37,43,47,48,50,56,60,61,63,69,73,74,76,

%T 82,86,87,89,95,99,100,102,108,112,113,115,121,125,126,128,134,138,

%U 139,141,147,151,152,154,160,164,165,167,173,177,178,180,186,190,191

%N Indices of Bell numbers divisible by 3.

%C a(n) appears to be congruent 4, 8, 9, 11 mod 13. - _Ralf Stephan_, Sep 12 2013

%C 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

%H Vaclav Kotesovec, <a href="/A229004/b229004.txt">Table of n, a(n) for n = 1..15384</a> (terms 1..1200 from Enrique Pérez Herrero)

%H J. Levine and R. E. Dalton, <a href="http://dx.doi.org/10.1090/S0025-5718-1962-0148604-2">Minimum Periods, Modulo p, of First Order Bell Exponential Integrals</a>, Mathematics of Computation, 16 (1962), 416-423.

%H Samuel S. Wagstaff Jr., <a href="http://www.ams.org/mcom/1996-65-213/S0025-5718-96-00683-7/home.html">Aurifeuillian factorizations and the period of the Bell numbers modulo a prime</a>, Math. Comp. 65 (1996), 383-391.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/BellNumber.html">Bell Number</a>

%F Conjectures from _Colin Barker_, Jul 16 2014: (Start)

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

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

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

%Y Cf. A000110, A016789, A155730, A054767.

%K nonn

%O 1,1

%A _Enrique Pérez Herrero_, Sep 10 2013