A268368 An eventually quasi-quadratic sequence satisfying a Hofstadter-like recurrence.

0, 1, 0, 4, 4, 4, 3, 12, 8, 4, 3, 24, 12, 4, 3, 40, 16, 4, 3, 60, 20, 4, 3, 84, 24, 4, 3, 112, 28, 4, 3, 144, 32, 4, 3, 180, 36, 4, 3, 220, 40, 4, 3, 264, 44, 4, 3, 312, 48, 4, 3, 364, 52, 4, 3, 420, 56, 4, 3, 480, 60, 4, 3, 544, 64, 4, 3, 612, 68, 4, 3, 684

Offset: 1

Author

Nathan Fox, Feb 23 2016

Keywords

• nonn
• easy

Comments

a(n) is the solution to the recurrence relation a(n) = a(n-a(n-1)) + a(n-a(n-2)) + a(n-a(n-3)), with the initial conditions: a(n) = 0 if n <= 0; a(1) = 0, a(2) = 1, a(3) = 0, a(4) = 4, a(5) = 4, a(6) = 4, a(7) = 3.

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Crossrefs

Cf. A005185, A188670, A244477, A264757, A267501.

Programs

PARI

concat(0, Vec(x^2*(1 + 4*x^2 + 4*x^3 + x^4 + 3*x^5 - 4*x^7 - 5*x^8 - 6*x^9 + 3*x^12 + 3*x^13) / ((1 - x)^3*(1 + x)^3*(1 + x^2)^3) + O(x^100))) \\ Colin Barker, Jun 22 2017

Formula

a(2) = 1, a(3) = 0; otherwise a(4n) = 2n^2+2n, a(4n+1) = 4n, a(4n+2) = 4, a(4n+3) = 3.

From Colin Barker, Jun 22 2017: (Start)

G.f.: x^2*(1 + 4*x^2 + 4*x^3 + x^4 + 3*x^5 - 4*x^7 - 5*x^8 - 6*x^9 + 3*x^12 + 3*x^13) / ((1 - x)^3*(1 + x)^3*(1 + x^2)^3).

a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12) for n>12.

(End)