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%I #41 Dec 10 2023 15:45:21
%S 2,3,10,1001,1004006004002,
%T 1020191144865623440455270145683555422808365843606721760320033
%N a(0) = 2; thereafter a(n) = a(n - 1)^n + 1.
%C Note that this could be extended backwards to a(-1), and any nonzero value x for a(-1) would work, since x^0 + 1 = 2.
%H David Johnson-Davies, <a href="http://www.lispology.com/show?2X07">Recursive functions without a base case</a>
%p a:= proc(n) option remember; `if`(n<0, %,
%p 1 + a(n-1)^n)
%p end:
%p seq(a(n), n=0..5); # _Alois P. Heinz_, Dec 18 2019
%t a[0] = 2; a[n_] := a[n] = a[n - 1]^n + 1; Array[a, 6, 0] (* _Amiram Eldar_, Dec 19 2019 *)
%t nxt[{n_,a_}]:={n+1,a^(n+1)+1}; NestList[nxt,{0,2},5][[;;,2]] (* _Harvey P. Dale_, Dec 10 2023 *)
%o (Lisp) (defun a (n) (+ (if (zerop n) 1 (expt (a (- n 1)) n)) 1))
%o (PARI) a(n) = if(n==0, 2, a(n-1)^n+1) \\ _Felix Fröhlich_, Dec 18 2019
%Y Cf. A000522, A182242.
%K nonn
%O 0,1
%A _David Johnson-Davies_, Dec 18 2019
%E Edited by _N. J. A. Sloane_, Dec 27 2019
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