A300293 A sequence based on the period 6 sequence A151899.

0, 0, 1, 1, 1, 2, 3, 3, 4, 4, 4, 5, 6, 6, 7, 7, 7, 8, 9, 9, 10, 10, 10, 11, 12, 12, 13, 13, 13, 14, 15, 15, 16, 16, 16, 17, 18, 18, 19, 19, 19, 20, 21, 21, 22, 22, 22, 23, 24, 24, 25, 25, 25, 26, 27, 27, 28, 28, 28, 29, 30, 30, 31, 31, 31, 32, 33, 33, 34, 34, 34, 35, 36, 36, 37, 37, 37, 38, 39, 39, 40, 40, 40, 41, 42, 42, 43, 43, 43, 44

Offset: 0

Author

Wolfdieter Lang, Mar 05 2018

Keywords

• nonn
• easy

Comments

a(k-1) + 2 =: v2(k), k >= 1, is used to obtain for 2^(v2(k))*V_{-k}(2) as well as 2^(v2(k))*V_{-k}(5) integer coordinates in the quadratic number field Q(sqrt(3)), where V_{-k}(j), j = 0..5, are the vertices of a k-family of regular hexagons H_{-k} whose centers O_{-k} form part of a discrete spiral. See the linked paper, Lemma 6, eqs. (47) and (48), and the Table 19. - Wolfdieter Lang, Mar 30 2018

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..1000
• Wolfdieter Lang, On a Conformal Mapping of Regular Hexagons and the Spiral of its Centers
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,1,-1).

Crossrefs

Cf. A174257, A300068, A300076, A151899.

Programs

Mathematica

LinearRecurrence[{1,0,0,0,0,1,-1},{0,0,1,1,1,2,3},100] (* Harvey P. Dale, Dec 29 2024 *)

PARI

a151899(n) = [0, 0, 1, 1, 1, 2][n%6+1]
a(n) = a151899(n) + 3*floor(n/6) \\ Felix Fröhlich, Mar 30 2018

Formula

a(n) = A151899(n) + 3*floor(n/6), n >= 0.

a(n) = A300076(n+1) - 1.

G.f.: x^2*(1 + x^3 + x^4)/((1 - x^6)*(1 - x)) = G(x) + 3*x^6/((1-x)*(1-x^6)), with the g.f. G(x) of A151899.

a(n) = a(n-1) + a(n-6) - a(n-7). - Wesley Ivan Hurt, Jun 19 2025