A238601 A sixth-order linear divisibility sequence related to the Fibonacci numbers: a(n) := (1/10)*Fibonacci(3*n)*Fibonacci(5*n)/Fibonacci(n).

1, 44, 1037, 32472, 915305, 26874892, 776952553, 22595381424, 655633561309, 19040507781020, 552780012054689, 16050219184005336, 466002944275859873, 13530204273746536948, 392841165312292809085, 11405932444267712654688, 331164788382150547106857, 9615185834308570310716196

Offset: 1

Author

Peter Bala, Mar 06 2014

Keywords

• nonn
• easy

Comments

Let P and Q be relatively prime integers. The Lucas sequence U(n) (which depends on P and Q) is an integer sequence that satisfies the recurrence equation a(n) = P*a(n-1) - Q*a(n-2) with the initial conditions U(0) = 0, U(1) = 1. The sequence {U(n)}n>=1 is a strong divisibility sequence, i.e., gcd(U(n),U(m)) = |U(gcd(n,m))|. It follows that {U(n)} is a divisibility sequence, i.e., U(n) divides U(m) whenever n divides m and U(n) <> 0.

It can be shown that if p and q are a pair of relatively prime positive integers, and if U(n) never vanishes, then the sequence {U(p*n)*U(q*n)/U(n)}n>=1 is a linear divisibility sequence of order 2*min(p,q). For a proof and a generalization of this result see the Bala link.

Here we take p = 3 and q = 5 with P = 1 and Q = -1, for which U(n) is the sequence of Fibonacci numbers, A000045, and normalize the sequence {U(3*n)*U(5*n)/U(n)}n>=1 to have the initial term 1.

For other sequences of this type see A238600, A238602 and A238603. See also A238536.

Since Fibonacci(n) can be defined for all n, so can this sequence. - N. J. A. Sloane, May 07 2017

Examples

			G.f. = x + 44*x^2 + 1037*x^3 + 32472*x^4 + 915305*x^5 + 26874892*x^6 + ... - _Michael Somos_, May 07 2017
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..500
• P. Bala, Divisibility sequences from strong divisibility sequences
• Wikipedia, Divisibility sequence
• Wikipedia, Fibonacci number
• Wikipedia, Lucas Sequence
• Index entries for linear recurrences with constant coefficients, signature (22,250,-1320,-250,22,1).

Crossrefs

Cf. A000045, A215466, A238536, A238600, A238602, A238603.

Programs

Magma

[(Fibonacci(3*n) + (-1)^n*Fibonacci(5*n) + Fibonacci(7*n))/10: n in [1..30]]; // G. C. Greubel, Aug 07 2018

Maple

with(combinat):
seq(1/10*fibonacci(3*n)*fibonacci(5*n)/fibonacci(n), n = 1..20);

Mathematica

Table[(1/10)*(Fibonacci[3*n] + (-1)^n*Fibonacci[5*n] + Fibonacci[7*n]), {n, 0, 50}] (* G. C. Greubel, Aug 07 2018 *)

PARI

{a(n) = if(n, fibonacci(3*n) * fibonacci(5*n) / (10 * fibonacci(n)), 0)} /* Michael Somos, May 07 2017 */

Formula

a(n) = (1/10)*(Fibonacci(3*n) + (-1)^n*Fibonacci(5*n) + Fibonacci(7*n)).

The sequence can be extended to negative indices using a(-n) = (-1)^(n+1)*a(n).

O.g.f. x*(1 + 22*x - 181*x^2 - 22*x^3 + x^4)/( (1 - 4*x - x^2)*(1 + 11*x - x^2)*(1 - 29*x - x^2) ).

Recurrence equation: a(n) = 22*a(n-1) + 250*a(n-2) - 1320*a(n-3) - 250*a(n-4) + 22*a(n-5) + a(n-6).