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).

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

Offset: 1

Author

John W. Layman, Jul 25 2012

Keywords

• nonn
• easy

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

• G. C. Greubel, Table of n, a(n) for n = 1..10000
• 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).

Crossrefs

Cf. A010892, A049347, A078412, A078414, A132739.

Programs

Magma

I:=[1,1,2,3,1,4]; [n le 6 select I[n] else Self(n-6): n in [1..100]]; // G. C. Greubel, Mar 08 2024

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 *)

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

SageMath

def A214684_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x*(1+x+2*x^2+3*x^3+x^4+4*x^5)/(1-x^6) ).list()
a=A214684_list(100); a[1:] # G. C. Greubel, Mar 08 2024

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

a(n) = -4*[n=0] + (1/3)*(2 + 2*(-1)^n + A010892(n) - 2*A010892(n-1) + 3*A049347(n) + 3*A049347(n-1)). - G. C. Greubel, Mar 08 2024