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A238603 A sixth-order linear divisibility sequence related to A000225: a(n) := (1/105)*(2^(3*n) - 1)*(2^(4*n) - 1)/(2^n - 1).

%I A238603 #30 Sep 08 2022 08:46:07
%S A238603 1,51,2847,170391,10555655,664857063,42215949223,2691226507047,
%T A238603 171901443816999,10990938133564455,703076406514657319,
%U A238603 44985901769992495143,2878746218051469266983,184228512166784552153127,11790264946382521291370535,754565442462197107544125479
%N A238603 A sixth-order linear divisibility sequence related to A000225: a(n) := (1/105)*(2^(3*n) - 1)*(2^(4*n) - 1)/(2^n - 1).
%C A238603 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 A238603 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 A238603 Here we take p = 3 and q = 4 with P = 3 and Q = 2, for which U(n) is the sequence A000225 (sometimes called the Mersenne numbers), and normalize the sequence {U(3*n)*U(4*n)/U(n)}n>=1 to have the initial term 1.
%C A238603 For other sequences of this type see A238600, A238601 and A238602. See also A238536.
%H A238603 G. C. Greubel, <a href="/A238603/b238603.txt">Table of n, a(n) for n = 1..500</a>
%H A238603 P. Bala, <a href="/A238600/a238600_1.pdf">Divisibility sequences from strong divisibility sequences</a>
%H A238603 Wikipedia, <a href="http://en.wikipedia.org/wiki/Divisibility_sequence">Divisibility sequence</a>
%H A238603 Wikipedia, <a href="http://en.wikipedia.org/wiki/Fibonacci_number">Fibonacci number</a>
%H A238603 Wikipedia, <a href="http://en.wikipedia.org/wiki/Lucas_sequence">Lucas Sequence</a>
%H A238603 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (119,-4382,59432,-280448,487424,-262144).
%F A238603 a(n) = (1/105)*(64^n + 32^n + 16^n - 4^n - 2^n - 1).
%F A238603 O.g.f.: x*(4096*x^4 - 4352*x^3 + 1160*x^2 - 68*x + 1 )/( (1-x)*(1-2*x)(1-4*x)*(1-16*x)*(1-32*x)*(1-64*x) ).
%F A238603 The formula for a(n) may be used to define it for all n in Z, and then we have a(n) = -(64)^n * a(-n). - _Michael Somos_, May 07 2017
%e A238603 G.f. = x + 51*x^2 + 2847*x^3 + 170391*x^4 + 10555655*x^5 + 664857063*x^6 + ... - _Michael Somos_, May 07 2017
%p A238603 seq(1/105*(2^(3*n)-1)*(2^(4*n)-1)/(2^n-1), n = 1..20);
%t A238603 Table[(1/105)*(64^n + 32^n + 16^n - 4^n - 2^n - 1), {n, 1, 50}] (* _G. C. Greubel_, Aug 07 2018 *)
%o A238603 (PARI) {a(n) = if( n, (8^n - 1) * (16^n - 1) / (105 * (2^n - 1)), 0)}; /* _Michael Somos_, May 07 2017 */
%o A238603 (Magma) [(1/105)*(64^n + 32^n + 16^n - 4^n - 2^n - 1): n in [1..50]]; // _G. C. Greubel_, Aug 07 2018
%Y A238603 Cf. A000225, A238536, A238600, A238601, A238602.
%K A238603 nonn,easy
%O A238603 1,2
%A A238603 _Peter Bala_, Mar 06 2014