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a(1) = 1, a(2) = a(1) + 1, a(3) = a(2)^2 + a(2) + 1; a(n + 1) = a(n)^n + a(n)^(n-1) + ... + a(n)^2 + a(n) + 1 = {a(n)^(n + 1)-1}/{a(n) - 1}.
1

%I #6 Sep 03 2017 03:31:21

%S 1,2,7,400,25664160401,

%T 11133592066966265230223312802560009320400030085606006

%N a(1) = 1, a(2) = a(1) + 1, a(3) = a(2)^2 + a(2) + 1; a(n + 1) = a(n)^n + a(n)^(n-1) + ... + a(n)^2 + a(n) + 1 = {a(n)^(n + 1)-1}/{a(n) - 1}.

%C The next term is too large to include.

%e a(3) = 2^2 + 2 + 1 = 7.

%p a[1]:=1: for n from 1 to 7 do a[n+1]:=sum(a[n]^j,j=0..n) od: seq(a[n],n=1..7); # _Emeric Deutsch_, Jul 31 2005

%K easy,nonn

%O 1,2

%A _Amarnath Murthy_, Jul 26 2005

%E More terms from _Emeric Deutsch_, Jul 31 2005