This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A253368 #52 Jun 02 2016 11:42:27
%S A253368 1,322,103683,33385604,10750060805,3461486193606,1114587804280327,
%T A253368 358893811492071688,115562692712642803209,37210828159659490561610,
%U A253368 11981771104717643318035211,3858093084890921488916776332,1242293991563772001787883943693
%N A253368 a(n) = F(12*n)/(12^2) with the Fibonacci numbers F = A000045.
%C A253368 The Fibonacci sequence with seeds 10/9, 10/9 for n = 0 and 1 has integer values precisely for every 12th entry, namely (10/9)*F(12*n), n >= 1: 160, 51520, 16589280, 5341696640, 1720009728800, ... .
%C A253368 Because F(12*n)/(9*2^4) = (F(6*n)/2^3)*(L(6*n)/2) = A049660(n)*A023039(n), this is indeed an integer sequence. Here L = A000032 (Lucas).
%C A253368 For the digital root of this sequence see A253298.
%C A253368 The final digits cycle a sequence of period 10 [1,2,3,4,5,6,7,8,9,0...], see A010879. - _Peter M. Chema_, Nov 11 2015
%H A253368 Michael De Vlieger, <a href="/A253368/b253368.txt">Table of n, a(n) for n = 1..399</a>
%H A253368 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (322, -1).
%F A253368 a(n) = F(12*n)/(9*2^4) = A049660(n)*A023039(n), n >= 1.
%F A253368 G.f.: x / (x^2 - 322*x + 1). - _Colin Barker_, Dec 31 2014
%F A253368 From _Peter Bala_, Apr 03 2015: (Start)
%F A253368 For integer k, 1 + k*(36 - k)*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + k/8*Sum_{n >= 1} Fibonacci(6*n)*x^n )*( 1 + k/8*Sum_{n >= 1} Fibonacci(6*n)*(-x)^n ).
%F A253368 1 + 64*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + Sum_{n >= 1} Fibonacci(6*n+3)*x^n )*( 1 + Sum_{n >= 1} Fibonacci(6*n+3)*(-x)^n ).
%F A253368 1 + 320*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + Sum_{n >= 1} Lucas(6*n)*x^n )*( 1 + Sum_{n >= 1} Lucas(6*n)*(-x)^n ).
%F A253368 (End)
%F A253368 a(n) = (((161+72*sqrt(5))^(-n)*(-1+(161+72*sqrt(5))^(2*n))))/(144*sqrt(5)). - _Colin Barker_, Jun 02 2016
%t A253368 Table[Fibonacci[12 n]/144, {n, 11}] (* _Michael De Vlieger_, Apr 03 2015 *)
%t A253368 LinearRecurrence[{322, -1},{1, 322},11] (* _Ray Chandler_, Aug 12 2015 *)
%o A253368 (PARI) Vec(x/(x^2-322*x+1) + O(x^20)) \\ _Colin Barker_, Dec 31 2014
%o A253368 (PARI) vector(20, n, fibonacci(12*n)/(9*2^4)) \\ _Altug Alkan_, Nov 11 2015
%Y A253368 Cf. A000045, A010879, A023039, A049660, A253298.
%K A253368 nonn,easy
%O A253368 1,2
%A A253368 _Peter M. Chema_, Dec 30 2014
%E A253368 Errors in name, data and formula corrected by _Colin Barker_, Dec 31 2014
%E A253368 Edited: numbers and name changed, formula and programs adjusted by _Wolfdieter Lang_, Jan 20 2015
%E A253368 Name simplified using "12" as the common number._Peter M. Chema_, Mar 31 2016