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%I #30 Sep 08 2022 08:46:07
%S 1,28,408,7896,137555,2496144,44599477,801617712,14375440584,
%T 258018516140,4629531440711,83076469908768,1490726895438793,
%U 26750144944686436,480010941060482040,8613453244178393184,154562103244937408987,2773504708179098411952
%N A sixth-order linear divisibility sequence related to the Fibonacci numbers: a(n) := (1/6)*Fibonacci(3*n)*Fibonacci(4*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 = 4 with P = 1 and Q = -1, for which U(n) is the sequence of Fibonacci numbers, A000045, and normalize the sequence to have the initial term 1.
%C For other sequences of this type see A238601, A238602 and A238603. See also A238536.
%H G. C. Greubel, <a href="/A238600/b238600.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 (14,90,-350,90,14,-1).
%F a(n) = (1/6)*Fibonacci(3*n)*Fibonacci(4*n)/Fibonacci(n).
%F a(n) = (1/6)*( Fibonacci(2*n) + (-1)^n*Fibonacci(4*n) + Fibonacci(6*n) ).
%F The sequence can be extended to negative indices when a(-n) = -a(n).
%F O.g.f. x*(1 + 14*x - 74*x^2 + 14*x^3 + x^4)/( (1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2) ).
%F Recurrence equation: a(n) = 14*a(n-1) + 90*a(n-2) - 350*a(n-3) + 90*a(n-4) + 14*a(n-5) - a(n-6).
%p with(combinat):
%p seq(1/6*fibonacci(3*n)*fibonacci(4*n)/fibonacci(n), n = 1..20);
%t Table[(1/6)*(Fibonacci[2*n] + (-1)^n*Fibonacci[4*n] + Fibonacci[6*n]), {n, 1, 500}] (* _G. C. Greubel_, Aug 07 2018 *)
%t LinearRecurrence[{14,90,-350,90,14,-1},{1,28,408,7896,137555,2496144},20] (* _Harvey P. Dale_, Aug 26 2020 *)
%o (PARI) vector(30, n, (fibonacci(2*n) + (-1)^n*fibonacci(4*n) + fibonacci(6*n))/6) \\ _G. C. Greubel_, Aug 07 2018
%o (Magma) [(Fibonacci(2*n) + (-1)^n*Fibonacci(4*n) + Fibonacci(6*n))/6: n in [1..30]]; // _G. C. Greubel_, Aug 07 2018
%Y Cf. A000045, A215466, A238536, A238601, A238602, A238603.
%K nonn,easy
%O 1,2
%A _Peter Bala_, Mar 01 2014
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