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a(n) = (4n+2)^12.
1

%I #12 Apr 21 2023 05:47:07

%S 4096,2176782336,1000000000000,56693912375296,1156831381426176,

%T 12855002631049216,95428956661682176,531441000000000000,

%U 2386420683693101056,9065737908494995456,30129469486639681536,89762301673555234816,244140625000000000000,614787626176508399616

%N a(n) = (4n+2)^12.

%H <a href="/index/Rec#order_13">Index entries for linear recurrences with constant coefficients</a>, signature (13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1).

%F From _Amiram Eldar_, Apr 21 2023: (Start)

%F a(n) = A016825(n)^12.

%F a(n) = 2^12*A016764(n).

%F Sum_{n>=0} 1/a(n) = 691*Pi^12/2615987404800. (End)

%t Table[(4*n+2)^12, {n, 0, 20}] (* _Amiram Eldar_, Apr 21 2023 *)

%Y Cf. A016764, A016825.

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_