OFFSET
0,3
COMMENTS
Every fifth term of A000129 is divisible by 29. Dividing every fifth term by 29 gives this sequence.
LINKS
Colin Barker, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (82,1).
FORMULA
a(n) = A000129(5*n)/29.
From Colin Barker, Sep 20 2017: (Start)
G.f.: x / (1 - 82*x - x^2).
a(n) = (((-41-29*sqrt(2))^(-n)*(-1 + (-3363-2378*sqrt(2))^n))) / (58*sqrt(2)).
(End)
MAPLE
a:= n-> (<<0|1>, <1|82>>^n)[1, 2]:
seq(a(n), n=0..20); # Alois P. Heinz, Sep 18 2017
MATHEMATICA
CoefficientList[Series[x/(1-82*x-x^2), {x, 0, 20}], x] (* G. C. Greubel, Feb 02 2019 *)
PROG
(PARI) a(n) = ([82, 1; 1, 0]^n)[2, 1]; \\ Altug Alkan, Sep 18 2017
(PARI) concat(0, Vec(x / (1 - 82*x - x^2) + O(x^20))) \\ Colin Barker, Sep 20 2017
(Magma) m:=20; R<x>:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!( x/(1-82*x-x^2) )); // G. C. Greubel, Feb 02 2019
(Sage) (x/(1-82*x-x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Feb 02 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bobby Jacobs, Sep 18 2017
STATUS
approved