A238602 - OEIS

%I #23 Sep 08 2022 08:46:07

%S 1,238,45507,9063516,1792708805,355009117386,70287911575687,

%T 13916722851826872,2755438412296182921,545562971271797876390,

%U 108018710075587599558731,21387159127038457710621972,4234549485214861760195346253,838419411023095574089504928386

%N A sixth-order linear divisibility sequence related to the Pell numbers: a(n) := (1/60)*Pell(3*n)*Pell(4*n)/Pell(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 = 2 and Q = -1, for which U(n) is the sequence of Pell numbers, A000129, and normalize the sequence {U(3*n)*U(4*n)/U(n)}n>=1 to have the initial term 1.

%C For other sequences of this type see A238600, A238601 and A238603. See also A238536.

%H G. C. Greubel, <a href="/A238602/b238602.txt">Table of n, a(n) for n = 1..435</a>

%H P. Bala, <a href="/A238600/a238600_1.pdf">Divisibility sequences from strong divisibility sequences</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Divisibility_sequence">Divisibility sequence</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Lucas_sequence">Lucas Sequence</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Pell_number">Pell number</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (170,5745,-40052,5745,170,-1).

%F a(n) = (1/60)*( Pell(2*n) + (-1)^n*Pell(4*n) + Pell(6*n) ).

%F The sequence can be extended to negative indices using a(-n) = -a(n).

%F O.g.f. x*(1 + 68*x - 698*x^2 + 68*x^3 + x^4)/( (1 - 6*x + x^2)*(1 + 34*x + x^2)*(1 - 198*x + x^2) ).

%F Recurrence equation: a(n) = 170*a(n-1) + 5745*a(n-2) - 40052*a(n-3) + 5745*a(n-4) + 170*a(n-5) - a(n-6).

%t Table[(1/60)*(Fibonacci[2*n, 2] + (-1)^n*Fibonacci[4*n, 2] + Fibonacci[6*n, 2]), {n, 1, 50}] (* _G. C. Greubel_, Aug 07 2018 *)

%o (PARI) x='x+O('x^30); Vec(x*(1+68*x-698*x^2+68*x^3+x^4)/((1-6*x+x^2)*(1 + 34*x+x^2)*(1-198*x+x^2))) \\ _G. C. Greubel_, Aug 07 2018

%o (Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(1 +68*x-698*x^2+68*x^3+x^4)/((1-6*x+x^2)*(1+34*x+x^2)*(1-198*x+x^2)))); // _G. C. Greubel_, Aug 07 2018

%Y Cf. A000129, A238536, A238600, A238601, A238603.

%K nonn,easy

%O 1,2

%A _Peter Bala_, Mar 06 2014