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A253368
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a(n) = F(12*n)/(12^2) with the Fibonacci numbers F = A000045.
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7
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1, 322, 103683, 33385604, 10750060805, 3461486193606, 1114587804280327, 358893811492071688, 115562692712642803209, 37210828159659490561610, 11981771104717643318035211, 3858093084890921488916776332, 1242293991563772001787883943693
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OFFSET
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1,2
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COMMENTS
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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, ... .
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).
For the digital root of this sequence see A253298.
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
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LINKS
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FORMULA
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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 ).
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 ).
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 ).
(End)
a(n) = (((161+72*sqrt(5))^(-n)*(-1+(161+72*sqrt(5))^(2*n))))/(144*sqrt(5)). - Colin Barker, Jun 02 2016
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MATHEMATICA
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LinearRecurrence[{322, -1}, {1, 322}, 11] (* Ray Chandler, Aug 12 2015 *)
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PROG
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(PARI) Vec(x/(x^2-322*x+1) + O(x^20)) \\ Colin Barker, Dec 31 2014
(PARI) vector(20, n, fibonacci(12*n)/(9*2^4)) \\ Altug Alkan, Nov 11 2015
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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Errors in name, data and formula corrected by Colin Barker, Dec 31 2014
Edited: numbers and name changed, formula and programs adjusted by Wolfdieter Lang, Jan 20 2015
Name simplified using "12" as the common number.Peter M. Chema, Mar 31 2016
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STATUS
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approved
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