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a(n) = (9*n + 3)^11.
1

%I #18 Oct 03 2024 03:48:50

%S 177147,743008370688,350277500542221,17714700000000000,

%T 317475837322472439,3116402981210161152,20635899893042801193,

%U 103510234140112521216,422351360321044921875,1469170321634239709184,4501035456767426597157,12433743083946522728448,31517572945366073781711

%N a(n) = (9*n + 3)^11.

%H Vincenzo Librandi, <a href="/A017207/b017207.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_, Oct 03 2024: (Start)

%F a(n) = A017197(n)^11 = 3^11 * A016787(n).

%F Sum_{n>=0} 1/a(n) = 7388*Pi^11/(444826519957575*sqrt(3)) + 88573*zeta(11)/31381059609. (End)

%t Table[(9*n + 3)^11, {n, 0, 15}] (* _Amiram Eldar_, Oct 03 2024 *)

%o (Magma) [(9*n+3)^11: n in [0..15]]; // _Vincenzo Librandi_, Jul 23 2011

%Y Cf. A013669, A016787, A017197.

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_