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a(1) = 1; a(n) = (Sum_{i=1..n-1} a(i))^n for n >= 2.
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%I #11 Oct 02 2023 20:17:05

%S 1,1,8,10000,100501001000500100000

%N a(1) = 1; a(n) = (Sum_{i=1..n-1} a(i))^n for n >= 2.

%C a(6) has 120 digits (1006016...1000000).

%H G. C. Greubel, <a href="/A165429/b165429.txt">Table of n, a(n) for n = 1..7</a>

%e a(4) = (a(1) + a(2) + a(3))^4 = (1 + 1 + 8)^4 = 10^4 = 10000.

%t a[1]:= 1; a[n_]:= (Sum[a[j], {j, 1, n-1}])^n; Table[a[n], {n,1,7}] (* _G. C. Greubel_, Oct 19 2018 *)

%o (PARI) {a(n) = if(n==1, 1, (sum(j=1,n-1, a(j)))^n)};

%o for(n=1,7, print1(a(n), ", ")) \\ _G. C. Greubel_, Oct 19 2018

%K nonn

%O 1,3

%A _Jaroslav Krizek_, Sep 17 2009

%E Edited by _N. J. A. Sloane_, Oct 15 2009