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a(n) = 4^n^2 + n!.
0

%I #31 Aug 24 2019 12:20:39

%S 2,5,258,262150,4294967320,1125899906842744,4722366482869645214416,

%T 316912650057057350374175806384,

%U 340282366920938463463374607431768251776,5846006549323611672814739330865132078623730534784,1606938044258990275541962092341162602522202993782792838930176

%N a(n) = 4^n^2 + n!.

%H Andrew M. Kamal, <a href="https://mentors4edu.github.io/New-Field/">New Field</a>

%e a(1) = 5 since 1^1=1, (4^1) + 1! = 5;

%e a(2) = 4^2^2 = 4^4 = 256, 256 + 2! = 256 + 2*1 = 258.

%Y Cf. A309669, A192366.

%K nonn

%O 0,1

%A _Andrew M. Kamal_, Aug 11 2019