OFFSET
1,2
COMMENTS
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
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..399
Index entries for linear recurrences with constant coefficients, signature (322, -1).
FORMULA
G.f.: x / (x^2 - 322*x + 1). - Colin Barker, Dec 31 2014
From Peter Bala, Apr 03 2015: (Start)
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
MATHEMATICA
Table[Fibonacci[12 n]/144, {n, 11}] (* Michael De Vlieger, Apr 03 2015 *)
LinearRecurrence[{322, -1}, {1, 322}, 11] (* Ray Chandler, Aug 12 2015 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter M. Chema, Dec 30 2014
EXTENSIONS
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
STATUS
approved