A224520 Numbers a(n) with property a(n) + a(n+4) = 2^(n+4) - 1 = A000225(n+4).

0, 1, 3, 7, 15, 30, 60, 120, 240, 481, 963, 1927, 3855, 7710, 15420, 30840, 61680, 123361, 246723, 493447, 986895, 1973790, 3947580, 7895160, 15790320, 31580641, 63161283, 126322567, 252645135, 505290270, 1010580540

Offset

0,3

Comments

This is the case k=4 of a(n) + a(n+k) = 2^(n+k) - 1 = A000225(n+k). The sequences A000975, A077854 and A153234 correspond to cases k=1,2 and 3, respectively.

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (3,-2,0,-1,3,-2).

Formula

a(n) + a(n+4) = 2^(n+4) - 1.

From Joerg Arndt, Apr 09 2013: (Start)

G.f.: x/((1-x)*(1-2*x)*(1+x^4)).

a(n) = +3*a(n-1) -2*a(n-2) -1*a(n-4) +3*a(n-5) -2*a(n-6). (End)

a(n) = floor(2^(n+4)/17). - Karl V. Keller, Jr., Jun 30 2021

Mathematica

CoefficientList[Series[x/((1 - x)*(1 - 2*x)*(1 + x^4)), {x, 0, 50}], x] (* G. C. Greubel, Oct 11 2017 *)
LinearRecurrence[{3, -2, 0, -1, 3, -2}, {0, 1, 3, 7, 15, 30}, 40] (* Harvey P. Dale, Aug 23 2021 *)

Prog

(PARI) x='x+O('x^50); concat([0], Vec(x/((1-x)*(1-2*x)*(1+x^4)))) \\ G. C. Greubel, Oct 11 2017
(Python) print([2**(n+4)//17 for n in range(31)]) # Karl V. Keller, Jr., Jun 30 2021

Crossrefs

Cf. A000975, A077854, A153234.

Sequence in context: A062544 A120411 A350606 * A291754 A336685 A069112

Adjacent sequences: A224517 A224518 A224519 * A224521 A224522 A224523

Keyword

nonn,easy

Author

Arie Bos, Apr 09 2013

Status

approved