%I #24 Dec 03 2021 15:43:04

%S 1,1,0,0,1,-1,0,6,-19,29,48,-524,2057,-3901,-9632,129034,-664363,

%T 1837905,2388688,-67004696,478198545,-1994889945,1669470784,

%U 56929813934,-615188040195,3794477505573,-12028579019536,-50780206473220

%N a(n) = Sum_{k=0..n} (k-n)^k.

%H Seiichi Manyama, <a href="/A038125/b038125.txt">Table of n, a(n) for n = 0..500</a>

%F G.f.: 1+ sum(k>=0, x^(k+1)/(1+x^(k+1)) ) = 1/Q(0), where Q(k) = 1 - x + x^2*(k+1)/(1 + (k+1)*x/Q(k+1) ); (continued fraction). - _Sergei N. Gladkovskii_, Jan 10 2014

%e 0^0 = 1,

%e 1^0 - 0^1 = 1,

%e 2^0 - 1^1 + 0^2 = 0,

%e 3^0 - 2^1 + 1^2 - 0^3 = 0,

%e ...

%t Prepend[ Table[ Sum[ (k-n)^k, {k, 0, n} ], {n, 30} ], 1 ]

%o (PARI) my(N=40, x='x+O('x^N)); Vec(sum(k=0, N, x^k/(1+k*x))) \\ _Seiichi Manyama_, Dec 02 2021

%o (PARI) a(n) = sum(k=0, n, (k-n)^k); \\ _Michel Marcus_, Dec 03 2021

%Y Cf. A003101, A026898, A120485.

%K sign,easy

%O 0,8

%A Jim Ferry (jferry(AT)alum.mit.edu)