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A247207 4^(2^n) + 3^(2^n) + 2^(2^n) + 1. 0

%I #13 Nov 29 2014 05:51:08

%S 10,30,354,72354,4338079554,18448597098193370754,

%T 340282370354622283774333836163326852354,

%U 115792089237316207213755562747271079374483128445080168181132550891186006948354

%N 4^(2^n) + 3^(2^n) + 2^(2^n) + 1.

%F a(n) = sum_{k = 1 .. 4} k^(2^n).

%F For all n > 1, a(n) = 54 mod 100.

%e a(2) = 4^4 + 3^4 + 2^4 + 1 = 256 + 81 + 16 + 1 = 354.

%e a(3) = 4^8 + 3^8 + 2^8 + 1 = 65536 + 6561 + 256 + 1 = 72354.

%t Table[1 + Sum[k^(2^n), {k, 2, 4}], {n, 0, 7}]

%Y Cf. A091775, A000855.

%K nonn,easy

%O 0,1

%A _Alonso del Arte_, Nov 25 2014

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