A242602 Integers repeated thrice in a canonical order.

0, 0, 0, 1, 1, 1, -1, -1, -1, 2, 2, 2, -2, -2, -2, 3, 3, 3, -3, -3, -3, 4, 4, 4, -4, -4, -4, 5, 5, 5, -5, -5, -5, 6, 6, 6, -6, -6, -6, 7, 7, 7, -7, -7, -7, 8, 8, 8, -8, -8, -8, 9, 9, 9, -9, -9, -9, 10, 10, 10, -10, -10, -10, 11, 11, 11, -11, -11, -11, 12, 12, 12, -12, -12, -12, 13, 13, 13, -13, -13, -13, 14, 14, 14, -14, -14, -14

Offset

0,10

Comments

See the comments under A242601 for the k-family of sequences s(k,n), k = 1, 2, ..., and n >= 0. The present sequence is s(3,k). See the Myerson-van der Poorten link, p. 4.

Links

Table of n, a(n) for n=0..86.

G. Myerson and A. J. van der Poorten, Some problems concerning recurrence sequences, Amer. Math. Monthly 102 (1995), no. 8, 698-705.

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

Formula

O.g.f.: x^3/((1 + x^3)^2*(1 - x)) = x^3/(1 - x + 2*x^3 - 2*x^4 + x^6 - x^7).

a(n) = a(n-1) - 2*a(n-3) + 2*a(n-4) - a(n-6) + a(n-7), n >= 7, with a(0) = a(1) = a(2) = 0, a(3) = a(4) = a(5) = 1 and a(6) = -1.

a(n) = floor((n+3)/6)*(-1)^floor((n+3)/3), n >= 0.

Crossrefs

Cf. A242601, A152467 (unsigned version with three additional leading zeros).

Sequence in context: A321018 A340764 A152467 * A097992 A195177 A147583

Adjacent sequences: A242599 A242600 A242601 * A242603 A242604 A242605

Keyword

sign,easy

Author

Wolfdieter Lang, Jun 17 2014

Status

approved