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a(n) = (n!)^n + n^n.
0

%I #16 Dec 30 2024 06:20:48

%S 2,2,8,243,332032,24883203125,139314069504046656,

%T 82606411253903523840823543,6984964247141514123629140377616777216,

%U 109110688415571316480344899355894085582848387420489,395940866122425193243875570782668457763038822400000000010000000000,409933016554924328182440935903164918932547530146724293451448320000000000285311670611

%N a(n) = (n!)^n + n^n.

%C The equation (k!)^n + k^n = (n!)^k + n^k holds if and only if n = k. See the Proof of the Theorem 2.2 (p.182-183) in the Alzer and Luca article in Links section.

%H Horst Alzer and Florian Luca, <a href="http://eudml.org/doc/288107">Diophantine equations involving factorials</a>, Mathematica Bohemica 142.2 (2017), 181-184.

%F a(n) = A036740(n) + A000312(n).

%e n = 3: a(3) = (3!)^3 + 3^3 = 243.

%p seq((n!)^n + n^n, n=0..12); # _Georg Fischer_, Dec 30 2024

%t a[n_]:=(n!)^n + n^n; Array[a,12,0] (* _Stefano Spezia_, Dec 26 2024 *)

%Y Cf. A000142, A000312, A036740.

%K nonn,easy

%O 0,1

%A _Ctibor O. Zizka_, Dec 26 2024

%E a(9)-a(11) corrected by _Georg Fischer_, Dec 30 2024