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a(n) = (6*n + 5)^12.
1

%I #16 Apr 01 2022 09:11:35

%S 244140625,3138428376721,582622237229761,21914624432020321,

%T 353814783205469041,3379220508056640625,22563490300366186081,

%U 116191483108948578241,491258904256726154641,1779197418239532716881,5688009063105712890625,16409682740640811134241,43439888521963583647921

%N a(n) = (6*n + 5)^12.

%H Vincenzo Librandi, <a href="/A016980/b016980.txt">Table of n, a(n) for n = 0..1000</a>

%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 01 2022: (Start)

%F a(n) = A016969(n)^12 = A016970(n)^6 = A016971(n)^4 = A016972(n)^3 = A016974(n)^2.

%F Sum_{n>=0} 1/a(n) = PolyGamma(11, 5/6)/86890185149644800. (End)

%o (Magma) [(6*n+5)^12: n in [0..40]]; // _Vincenzo Librandi_, May 19 2011

%Y Cf. A016969 (6*n + 5), A016970, A016971, A016972, A016973, A016974, A016975, A016976, A016977, A016978, A016979.

%Y Subsequence of A008456 (n^12).

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_