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%I #31 Sep 08 2022 08:46:07
%S 1,44,1037,32472,915305,26874892,776952553,22595381424,655633561309,
%T 19040507781020,552780012054689,16050219184005336,466002944275859873,
%U 13530204273746536948,392841165312292809085,11405932444267712654688,331164788382150547106857,9615185834308570310716196
%N A sixth-order linear divisibility sequence related to the Fibonacci numbers: a(n) := (1/10)*Fibonacci(3*n)*Fibonacci(5*n)/Fibonacci(n).
%C 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.
%C 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.
%C 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.
%C For other sequences of this type see A238600, A238602 and A238603. See also A238536.
%C Since Fibonacci(n) can be defined for all n, so can this sequence. - _N. J. A. Sloane_, May 07 2017
%H G. C. Greubel, <a href="/A238601/b238601.txt">Table of n, a(n) for n = 1..500</a>
%H P. Bala, <a href="/A238600/a238600_1.pdf">Divisibility sequences from strong divisibility sequences</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Divisibility_sequence">Divisibility sequence</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Fibonacci_number">Fibonacci number</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Lucas_sequence">Lucas Sequence</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (22,250,-1320,-250,22,1).
%F a(n) = (1/10)*(Fibonacci(3*n) + (-1)^n*Fibonacci(5*n) + Fibonacci(7*n)).
%F The sequence can be extended to negative indices using a(-n) = (-1)^(n+1)*a(n).
%F 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) ).
%F 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).
%e G.f. = x + 44*x^2 + 1037*x^3 + 32472*x^4 + 915305*x^5 + 26874892*x^6 + ... - _Michael Somos_, May 07 2017
%p with(combinat):
%p seq(1/10*fibonacci(3*n)*fibonacci(5*n)/fibonacci(n), n = 1..20);
%t Table[(1/10)*(Fibonacci[3*n] + (-1)^n*Fibonacci[5*n] + Fibonacci[7*n]), {n, 0, 50}] (* _G. C. Greubel_, Aug 07 2018 *)
%o (PARI) {a(n) = if(n, fibonacci(3*n) * fibonacci(5*n) / (10 * fibonacci(n)), 0); /* _Michael Somos_, May 07 2017 */
%o (Magma) [(Fibonacci(3*n) + (-1)^n*Fibonacci(5*n) + Fibonacci(7*n))/10: n in [1..30]]; // _G. C. Greubel_, Aug 07 2018
%Y Cf. A000045, A215466, A238536, A238600, A238602, A238603.
%K nonn,easy
%O 1,2
%A _Peter Bala_, Mar 06 2014
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