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A214684 a(1)=1, a(2)=1, and, for n>2, a(n)=(a(n-1)+a(n-2))/5^k, where 5^k is the highest power of 5 dividing a(n-1)+a(n-2). 5
1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1, 2, 3, 1, 4, 1, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

COMMENTS

This sequence is periodic with period 1,1,2,3,1,4 of length 6.

It appears that for most choices of a(1), a(2), and divisor b^k (replacing 5^k), the resulting sequence is not periodic.

LINKS

Table of n, a(n) for n=1..80.

B. Avila and T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614, 2014 and J. Int. Seq. 17 (2014) # 14.8.5

Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 1).

FORMULA

a(n) = A132739(a(n-1)+a(n-2)), for n>2, and a(1)=1, a(2)=1. - Michel Marcus, Jul 08 2014

G.f.: -x*(4*x^5+x^4+3*x^3+2*x^2+x+1) / ((x-1)*(x+1)*(x^2-x+1)*(x^2+x+1)). - Colin Barker, Jul 08 2014

MATHEMATICA

CoefficientList[Series[(4*x^5 + x^4 + 3*x^3 + 2*x^2 + x + 1)/((1 - x)*(x + 1)*(x^2 - x + 1)*(x^2 + x + 1)), {x, 0, 100}], x] (* Wesley Ivan Hurt, Jul 08 2014 *)

LinearRecurrence[{0, 0, 0, 0, 0, 1}, {1, 1, 2, 3, 1, 4}, 80] (* Ray Chandler, Aug 25 2015 *)

PROG

(PARI) lista(nn) = {va = vector(nn); va[1] = 1; va[2] = 1; for (n=3, nn, sump = va[n-1] + va[n-2]; va[n] = sump/5^(valuation(sump, 5)); ); va; } \\ Michel Marcus, Jul 08 2014

(PARI) Vec(-x*(4*x^5+x^4+3*x^3+2*x^2+x+1)/((x-1)*(x+1)*(x^2-x+1)*(x^2+x+1)) + O(x^100)) \\ Colin Barker, Jul 08 2014

CROSSREFS

Cf. A078412, A078414, A132739.

Sequence in context: A211343 A039661 A293668 * A268727 A325542 A081877

Adjacent sequences:  A214681 A214682 A214683 * A214685 A214686 A214687

KEYWORD

nonn,easy

AUTHOR

John W. Layman, Jul 25 2012

STATUS

approved

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Last modified April 6 15:23 EDT 2020. Contains 333276 sequences. (Running on oeis4.)