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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). 4
1, 51, 2847, 170391, 10555655, 664857063, 42215949223, 2691226507047, 171901443816999, 10990938133564455, 703076406514657319, 44985901769992495143, 2878746218051469266983, 184228512166784552153127, 11790264946382521291370535, 754565442462197107544125479 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
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 = 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.
For other sequences of this type see A238600, A238601 and A238602. See also A238536.
LINKS
Wikipedia, Fibonacci number
Wikipedia, Lucas Sequence
Index entries for linear recurrences with constant coefficients, signature (119,-4382,59432,-280448,487424,-262144).
FORMULA
a(n) = (1/105)*(64^n + 32^n + 16^n - 4^n - 2^n - 1).
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) ).
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
EXAMPLE
G.f. = x + 51*x^2 + 2847*x^3 + 170391*x^4 + 10555655*x^5 + 664857063*x^6 + ... - Michael Somos, May 07 2017
MAPLE
seq(1/105*(2^(3*n)-1)*(2^(4*n)-1)/(2^n-1), n = 1..20);
MATHEMATICA
Table[(1/105)*(64^n + 32^n + 16^n - 4^n - 2^n - 1), {n, 1, 50}] (* G. C. Greubel, Aug 07 2018 *)
PROG
(PARI) {a(n) = if( n, (8^n - 1) * (16^n - 1) / (105 * (2^n - 1)), 0)}; /* Michael Somos, May 07 2017 */
(Magma) [(1/105)*(64^n + 32^n + 16^n - 4^n - 2^n - 1): n in [1..50]]; // G. C. Greubel, Aug 07 2018
CROSSREFS
Sequence in context: A097836 A267786 A267733 * A128917 A172742 A172821
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Mar 06 2014
STATUS
approved

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Last modified April 25 05:18 EDT 2024. Contains 371964 sequences. (Running on oeis4.)