%I #45 Sep 08 2022 08:46:04
%S -1,-3,-1,45,359,2037,10079,46365,204119,873477,3666959,15191085,
%T 62342279,254119317,1030760639,4165958205,16792710839,67557739557,
%U 271392171119,1089053371725,4366669645799,17498051254197,70086331418399,280627721655645
%N Poly-Cauchy numbers c_3^(-n).
%C Definition of poly-Cauchy numbers in A222627.
%H Vincenzo Librandi, <a href="/A223173/b223173.txt">Table of n, a(n) for n = 1..300</a>
%H Takao Komatsu, <a href="https://doi.org/10.1007/s11139-012-9452-0">Poly-Cauchy numbers with a q parameter</a>, Ramanujan J. 31 (2013), 353-371.
%H Takao Komatsu, <a href="http://doi.org/10.2206/kyushujm.67.143">Poly-Cauchy numbers</a>, Kyushu J. Math. 67 (2013), 143-153.
%H M. Z. Spivey, <a href="http://dx.doi.org/10.1016/j.disc.2007.03.052">Combinatorial sums and finite differences</a>, Discr. Math. 307 (24) (2007) 3130-3146
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Stirling_transform">Stirling transform</a>
%F a(n) = Sum_{k=0..3} Stirling1(3,k)*(k+1)^n.
%F From _Colin Barker_, Mar 31 2013: (Start)
%F Conjecture:
%F a(n) = 2^(1+n) - 3^(1+n) + 4^n;
%F g.f.: -x*(6*x-1) / ((2*x-1)*(3*x-1)*(4*x-1)). (End)
%F Conjecture verified by _Robert Israel_, Jun 21 2018
%p seq(2^(n+1)-3^(n+1)+4^n, n=0..30); # _Robert Israel_, Jun 21 2018
%t Table[Sum[StirlingS1[3, k] (k + 1)^n, {k, 0, 3}], {n, 25}]
%o (Magma) [&+[StirlingFirst(3,k)*(k+1)^n: k in [0..3]]: n in [1..25]]; // _Bruno Berselli_, Mar 28 2013
%o (PARI) a(n) = sum(k=0, 3, stirling(3, k, 1)*(k+1)^n); \\ _Michel Marcus_, Nov 14 2015
%K sign
%O 1,2
%A _Takao Komatsu_, Mar 28 2013