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a(0)=-1; a(n)=n^a(n-1)-a(n-1)^n.
5

%I #9 Jun 06 2022 15:21:50

%S -1,2,0,1,3,-118

%N a(0)=-1; a(n)=n^a(n-1)-a(n-1)^n.

%C Sequence is finite because followup terms are fractions. - _R. J. Mathar_, Jun 19 2021

%t lst={};a=1;Do[a=n^a-a^n;AppendTo[lst,a],{n,0,5}];lst

%t nxt[{n_,a_}]:={n+1,(n+1)^a-a^(n+1)}; NestList[nxt,{0,-1},5][[All,2]] (* _Harvey P. Dale_, Jun 06 2022 *)

%Y Cf. A084964, A152832, A152833, A152835

%K sign,fini,less

%O 0,2

%A _Vladimir Joseph Stephan Orlovsky_, Dec 14 2008

%E Definition corrected by _N. J. A. Sloane_, Jan 11 2009

%E Offset corrected. _R. J. Mathar_, Jun 19 2021