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a(n) is the least value of k such that for the n-th prime p, p^2 divides Bell(p+k)-Bell(k+1)-Bell(k).
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%I #14 May 20 2017 22:46:17

%S 0,2,0,4,0,4,6,37,17,21,75,27,3,20,96,21,13,90,37,26,22,20,204,12,148,

%T 23,46,24,0,71,22,3,36,41,4,101,228,31,155,304,309,392,146,85,222,346,

%U 134,277,43,7,67,484,230,152,10,135,40,256,28,97,129,90,458

%N a(n) is the least value of k such that for the n-th prime p, p^2 divides Bell(p+k)-Bell(k+1)-Bell(k).

%C 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.

%D J. Touchard, "Propriétés arithmétiques de certains nombres récurrents", Ann. Soc. Sci. Bruxelles A 53 (1933), pp. 21-31.

%H Amiram Eldar, <a href="/A286663/b286663.txt">Table of n, a(n) for n = 1..100</a>

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

%e The 7th prime is p(7) = 17, and the least k such that Bell(k+17)-Bell(k)-Bell(k+1) is divisible by 17^2 is k = 6: Bell(23)-Bell(6)-Bell(7) = 44152005855083266 = 17^2*152775106764994, thus a(7) = 6.

%t a={};np = 100; p = Prime[Range[np]]; For[i = 0, i < np, i++; p1 = p[[i]];

%t n = 0; While[!Divisible[BellB[p1 + n] - BellB[n] - BellB[n + 1], p1^2], n++]; a=AppendTo[a,n]]; a

%o (PARI) bell(n) = polcoeff(sum(k=0, n, prod(i=1, k, x/(1-i*x)), x^n * O(x)), n);

%o a(n) = {my(k = 0, p = prime(n)); while ((bell(p+k)-bell(k+1)-bell(k)) % p^2, k++); k;} \\ _Michel Marcus_, May 20 2017

%Y Cf. A000110, A283300, A286664.

%K nonn

%O 1,2

%A _Amiram Eldar_, May 12 2017