cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A238538 A fourth-order linear divisibility sequence: a(n) = (2^n + 1)*(2^(3*n) - 1)/ ( (2 + 1)*(2^3 - 1) ).

Original entry on oeis.org

1, 15, 219, 3315, 51491, 811395, 12882499, 205321155, 3278747331, 52408827075, 838132189379, 13406842675395, 214483303960771, 3431523432591555, 54902699475185859, 878429788032676035, 14054769379960303811, 224875452250864496835, 3598000373385828511939
Offset: 1

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Author

Peter Bala, Feb 28 2014

Keywords

Comments

This is a fourth-order linear divisibility sequence, that is, the sequence satisfies a linear recurrence of order 4 and if n | m then a(n) | a(m). This is a consequence of the following more general result: The polynomials P(n,x,y) := (x^n + y^n)*(x^(3*n) - y^(3*n)) form a fourth-order linear divisibility sequence in the polynomial ring Z[x,y]. See the Bala link.
Hence, for a fixed integers M and N, the normalized sequence (M^n + N^n)*(M^(3*n) - N^(3*n))/ ( (M + N)*(M^3 - N^3) ) for n = 1,2,3,... is a linear divisibility sequence of order 4. It has the rational o.g.f. x*(1 - 2*M*N*(M^2 - M*N + N^2)*x + (M*N)^4*x^2)/( (1 - M^4*x)*(1 - M^3*N*x)*(1 - M*N^3*x)*(1 - N^4*x) ). This is the case M = 2, N = 1. For other cases see A238539(M = 2, N = -1), A238540(M = 3, N = 1) and A238541(M = 3, N = 2). See also A238536, A238537 and A215466.
Note, these sequences do not belong to the family of linear divisibility sequences of the fourth order studied by Williams and Guy, which have o.g.f.s of the form x*(1 - q*x^2)/Q(x), Q(x) a quartic polynomial and q an integer parameter.

Crossrefs

Programs

  • Maple
    seq(1/21*(2^n + 1)*(2^(3*n) - 1), n = 1..20);
  • Mathematica
    LinearRecurrence[{27,-202,432,-256},{1,15,219,3315},20] (* Harvey P. Dale, Jul 04 2019 *)

Formula

a(n) = (1/21)*(2^n + 1)*(2^(3*n) - 1) = A000051(n)*A024088(n)/21.
a(n) = (1/21)*(4^n - 1)*(8^n - 1)/(2^n - 1).
O.g.f.: x*(1 - 12*x + 16*x^2)/((1 - x)*(1 - 2*x)*(1 - 8*x)*(1 - 16*x)).
Recurrence equation: a(n) = 27*a(n-1) - 202*a(n-2) + 432*a(n-4) - 256*a(n-4).