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a(0)=1; a(n)=Floor[n^(n+a(n-1))-a(n-1)^(n+a(n-1))].
0

%I #4 Mar 31 2012 12:38:17

%S 1,0,-4,-1,-115,-1

%N a(0)=1; a(n)=Floor[n^(n+a(n-1))-a(n-1)^(n+a(n-1))].

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

%Y Cf. A084964, A152832, A152833, A152835, A152836, A152837, A152838, A152839, A152840

%K sign

%O 0,3

%A _Vladimir Joseph Stephan Orlovsky_, Dec 14 2008

%E Indices added to definition, offset corrected - _R. J. Mathar_, Jan 08 2009