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%I #52 Nov 05 2019 10:58:50
%S 1,2,4,28,304888344611713860501504000028
%N a(n) = a(n-1)! + a(n-1) with a(1)=1.
%C The next term is too large to include, since it has about 8.85695956*10^30 digits.
%C No term can be represented by a sum of three positive squares, because a(4) and the following terms can all be written as 4*(8*k+7) for k>=0.
%F a(n) = a(n-1)! + a(n-1)
%e a(3) = a(3-1)! + a(3-1) = a(2)! + a(2) = 4.
%t a[1]=1; a[n_] := a[n-1]! + a[n-1]; Array[a, 5] (* _Giovanni Resta_, Nov 03 2019 *)
%Y Cf. A000408 (sums of 3 squares).
%K nonn
%O 1,2
%A _Ananthakrishna Kaithalayil_, Nov 02 2019
%E Edited by _N. J. A. Sloane_, Nov 03 2019
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