# Greetings from The On-Line Encyclopedia of Integer Sequences! http://oeis.org/

Search: id:a286664
Showing 1-1 of 1

%I A286664 #16 Sep 01 2018 03:30:24
%S A286664 2,5,2,2,2,20663,2,229,2,2,2,11,2,5,2,2,2,23,2,3,2,2,2,101,2,3,2,2,2
%N A286664 a(n) is the smallest prime p such that p^2 divides Bell(p+n) - Bell(n+1) - Bell(n).
%C A286664 Jacques Touchard proved in 1933 that for the Bell numbers (A000110), Bell(p+k) == Bell(k+1) + Bell(k) (mod p) for all primes p and k >= 0.
%C A286664 a(29) > 242000 and a(89) > 90000, if they exist. The terms from a(30) to a(89) are 2, 163, 2, 2, 2, 7, 2, 19, 2, 2, 2, 3, 2, 3, 2, 2, 2, 3, 2, 7, 2, 2, 2, 359, 2, 3, 2, 2, 2, 7, 2, 43, 2, 2, 2, 3, 2, 5, 2, 2, 2, 5, 2, 547, 2, 2, 2, 3, 2, 7, 2, 2, 2, 59, 2, 5, 2, 2, 2. - _Giovanni Resta_, Aug 26 2018
%C A286664 a(29) > 10^7. - _Hiroaki Yamanouchi_, Sep 01 2018
%D A286664 J. Touchard, "Propriétés arithmétiques de certains nombres récurrents", Ann. Soc. Sci. Bruxelles A 53 (1933), pp. 21-31.
%H A286664 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/TouchardsCongruence.html">Touchard's Congruence</a>
%e A286664 The smallest prime p such that Bell(p+1) == Bell(2)+Bell(1)(mod p^2) is 5, since Bell(6) - Bell(2) - Bell(1) = 203 - 2 - 1 = 200 = 5^2 * 8, thus a(1) = 5.
%t A286664 a = {}; n = 0; While[n < 101, p = 2; While[!Divisible[BellB[p + n] - BellB[n] - BellB[n + 1], p^2], p = NextPrime@p]; a = AppendTo[a, p]; n++]; a
%Y A286664 Cf. A000110, A283300, A286663.
%K A286664 nonn,more
%O A286664 0,1
%A A286664 _Amiram Eldar_, May 12 2017

# Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE