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a(n) = (6*n + 4)^11.
2

%I #16 Apr 01 2022 09:10:35

%S 4194304,100000000000,17592186044416,584318301411328,8293509467471872,

%T 70188843638032384,419430400000000000,1951354384207722496,

%U 7516865509350965248,24986644000165537792,73786976294838206464,197732674300000000000,488595558857835544576,1127073856954876807168

%N a(n) = (6*n + 4)^11.

%H Vincenzo Librandi, <a href="/A016967/b016967.txt">Table of n, a(n) for n = 0..1000</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_, Apr 01 2022: (Start)

%F a(n) = A016957(n)^11.

%F a(n) = 2^11*A016799(n).

%F Sum_{n>=0} 1/a(n) = 88573*zeta(11)/362797056 - 1847*Pi^11/(1285662067200*sqrt(3)). (End)

%t (6*Range[0,20]+4)^11 (* _Harvey P. Dale_, Sep 06 2012 *)

%o (Magma) [(6*n+4)^11: n in [0..20]]; // _Vincenzo Librandi_, May 07 2011

%Y Cf. A016799, A016957, A016958, A016959, A016960, A016961, A016962, A016963, A016964, A016965, A016966.

%Y Subsequence of A008455.

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_