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a(n) = phi( 2^(2*n) + 2^n + 1 ) / (6*n), where phi() is Euler's totient function (A000010).
1

%I #10 Sep 25 2024 10:27:59

%S 1,1,4,6,30,72,336,720,4864,9900,54498,139968,728208,1820448,11748240,

%T 23224320,142888536,424189440,2066689584,4704480000,34426570752,

%U 75016279008,437072997306,1171108896768,6391042560000,16287748233120,111155542830144,225354102607872,1419514150386528,3983428743840000,21252009311404938,49614674429214720

%N a(n) = phi( 2^(2*n) + 2^n + 1 ) / (6*n), where phi() is Euler's totient function (A000010).

%C In fact, phi(m^(2*n) + m^n + 1) is a multiple of 6*n for all m > 1 and n >= 1. This sequence gives the corresponding quotients for m = 2.

%H M. Alekseyev et al., <a href="https://dxdy.ru/topic149967.html">Divisibility of φ(n^(2k) + n^k + 1) by 6k</a> (in Russian), dxdy.ru, 2022.

%t a[n_]:= EulerPhi[2^(2*n) + 2^n + 1 ]/(6*n); Array[a,32] (* _Stefano Spezia_, Sep 25 2024 *)

%Y Cf. A000010, A001576, A066791.

%K nonn

%O 1,3

%A _Max Alekseyev_, Sep 24 2024