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a(n) = (12*n + 4)^11.
1

%I #20 Dec 21 2025 17:20:58

%S 4194304,17592186044416,8293509467471872,419430400000000000,

%T 7516865509350965248,73786976294838206464,488595558857835544576,

%U 2450808588882738675712,10000000000000000000000,34785499933455142617088,106570876280498368282624,294393347626373780340736

%N a(n) = (12*n + 4)^11.

%H Paolo Xausa, <a href="/A017579/b017579.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).

%F From _Amiram Eldar_, Jul 14 2024: (Start)

%F a(n) = A008455(A017569(n)) = A017569(n)^11.

%F a(n) = 4194304 * A016787(n).

%F Sum_{n>=0} 1/a(n) = 1847*Pi^11/(2633035913625600*sqrt(3)) + 88573*zeta(11)/743008370688. (End)

%t (12*Range[0,20]+4)^11 (* _Harvey P. Dale_, Jan 07 2015 *)

%Y Cf. A008455, A013669, A016787, A017569.

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_

%E More terms from _Amiram Eldar_, Jul 14 2024