A289060 - OEIS

A289060

a(n) = 3*a(n-1) - 3*a(n-2) + *a(n-3) for n >= 8, where a(0) = 2, a(1) = 4, a(2) = 7, a(3) = 11, a(4) = 17, a(5) = 25, a(6) = 36, a(7) = 51.

2, 4, 7, 11, 17, 25, 36, 51, 70, 93, 120, 151, 186, 225, 268, 315, 366, 421, 480, 543, 610, 681, 756, 835, 918, 1005, 1096, 1191, 1290, 1393, 1500, 1611, 1726, 1845, 1968, 2095, 2226, 2361, 2500, 2643, 2790, 2941, 3096, 3255, 3418, 3585, 3756, 3931, 4110

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OFFSET

0,1

COMMENTS

Conjecture: a(n) is the number of letters (0's and 1's) in the n-th iterate of the mapping 00->0010, 01->010, 10->011, starting with 00; see A289057.

LINKS

Clark Kimberling, Table of n, a(n) for n = 0..10000

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

FORMULA

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 8, where a(0) = 2, a(1) = 4, a(2) = 7, a(3) = 11, a(4) = 17, a(5) = 25, a(6) = 36, a(7) = 51.

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

a(n) = 30 - 11*n + 2*n^2 for n>4. - Colin Barker, Jul 02 2017