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a(n)=floor(b(n)), where b(0)=1, b(n)=b(n-1)^2-n^b(n-1).
1

%I #17 Sep 06 2023 09:43:04

%S 1,0,-1,0,-3,4,-2092,4372582

%N a(n)=floor(b(n)), where b(0)=1, b(n)=b(n-1)^2-n^b(n-1).

%C The supplied Mathematica and PARI code corresponds to an initial value a(-1)=b(-1)=1 from which a(0)=b(0) is computed to be equal to 1.

%C The next term is approximately

%C -1.642155991293887705947531213655414816023738489781202992764677529*10^3948835

%C and is too large to display here. - _Robert G. Wilson v_, Nov 27 2010

%t a=1;lst={};Do[a=a^2-n^a;AppendTo[lst,Floor[a]],{n,0,7}];lst

%o (PARI) a=1;for(n=0,7, print1(floor(a=a^2-n^a)", ")) \\ - R. Gerbicz, Nov 27 2010

%Y Cf. A153059, A086851, A153060, A098152, A028300, A153061, A153062

%K sign

%O 0,5

%A _Vladimir Joseph Stephan Orlovsky_, Dec 17 2008

%E Definition clarified by _R. J. Mathar_, R. Gerbicz and _M. F. Hasler_, Nov 27 2010