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a(n) is the unique integer such that Sum_{k=0..p-1} b(k)/(-n)^k == a(n) (mod p) for any prime p not dividing n, where b(0), b(1), b(2), ... are Bell numbers given by A000110.
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%I #29 Jul 28 2021 23:07:51

%S 2,1,2,-1,10,-43,266,-1853,14834,-133495,1334962,-14684569,176214842,

%T -2290792931,32071101050,-481066515733,7697064251746,-130850092279663,

%U 2355301661033954,-44750731559645105,895014631192902122,-18795307255050944539

%N a(n) is the unique integer such that Sum_{k=0..p-1} b(k)/(-n)^k == a(n) (mod p) for any prime p not dividing n, where b(0), b(1), b(2), ... are Bell numbers given by A000110.

%C On July 17, 2010 Zhi-Wei Sun conjectured that a(n) exists for every n=1,2,3,... He noted that a(1)=2 since Sum_{k=0..p-1} (-1)^k * b(k) == b(p) (mod p), and conjectured that a(2)=1, a(3)=2, a(4)=-1, a(5)=10, a(6)=-43, a(7)=266, a(8)=-1853, a(9)=14834, a(10)=-133495. It seems that (-1)^(n-1)*a(n) > 0 for all n=3,4,5,...

%C I guess that a(2n) == (-1)^(n-1) (mod 4) and a(2n-1) == 2 (mod 4) for all n=1,2,3,... Perhaps a(2n-1) == 2 (mod 8) for every positive integer n. - _Zhi-Wei Sun_, Jul 18 2010

%C On August 5, 2010 Zhi-Wei Sun and Don Zagier proved that a(n) actually equals (-1)^(n-1)*D(n-1)+1, where D(0), D(1), D(2), ... are derangement numbers given by A000166. - _Zhi-Wei Sun_, Aug 07 2010

%H Seiichi Manyama, <a href="/A179508/b179508.txt">Table of n, a(n) for n = 1..451</a>

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/0911.5665">Open Conjectures on Congruences</a>, preprint, arXiv:0911.5665 [math.NT], 2009-2010.

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1006.2776">On Apery numbers and generalized central trinomial coefficients</a>, preprint, arXiv:1006.2776 [math.NT], 2010-2011.

%H Zhi-Wei Sun, <a href="https://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;7bbd51e4.1007">A conjecture on Bell numbers</a> (a message to Number Theory List on July 17, 2010) [From _Zhi-Wei Sun_, Jul 18 2010]

%H Zhi-Wei Sun and Don Zagier, <a href="https://doi.org/10.1017/S0004972711002218">On a curious property of Bell numbers</a>, Bulletin of the Australian Mathematical Society, Volume 84, Issue 1, August 2011. [_Zhi-Wei Sun_, Aug 07 2010]

%F a(n) = a(n-1)*(1-n)+n+1. - _Jon Maiga_, Jul 10 2021

%p A179508:= n-> (-1)^n*(n!*add((-1)^(k)/k!, k=0..n))+1 : seq(A179508(n), n=0..21);

%p # second program:

%p G(x):=(2-x)*exp(-x)/(1-x): f[0]:=G(x): for n from 1 to 21 do f[n]:=diff(f[n-1], x) od: x:=0; seq((-1)^n*f[n], n=0..21); # _Mélika Tebni_, Jul 10 2021

%t a[1] = 2;

%t a[n_]:=a[n]=a[n-1]*(1-n)+n+1;

%t Array[a, 30] (* _Jon Maiga_, Jul 10 2021 *)

%Y Cf. A000110, A000166.

%K sign

%O 1,1

%A _Zhi-Wei Sun_, Jul 17 2010