A276508 a(n) = (2*5^n + 3*(-1)^(floor((n-1)/3)) + (-1)^n)/6.

0, 2, 9, 42, 208, 1041, 5208, 26042, 130209, 651042, 3255208, 16276041, 81380208, 406901042, 2034505209, 10172526042, 50862630208, 254313151041, 1271565755208, 6357828776042, 31789143880209, 158945719401042, 794728597005208, 3973642985026041, 19868214925130208, 99341074625651042

Offset

0,2

Comments

Number of 1’s in substitution system {1 -> 12321, 2 -> 23132, 3 -> 31213} at step n from initial string "3" (see example). Number of 2’s: A000351(n) - A010892(n+1) - 2*a(n). Number of 3’s: A010892(n+1) + a(n).

Excluding zero, convolution of A000351 and A174737.

Links

Table of n, a(n) for n=0..25.

Ilya Gutkovskiy, Illustration (substitution system {1 -> 12321, 2 -> 23132, 3 -> 31213})

Eric Weisstein's World of Mathematics, Substitution System

Index entries for linear recurrences with constant coefficients, signature (6,-6,5)

Formula

O.g.f.: x*(2 - 3*x)/((1 - 5 x)*(1 - x + x^2)).

E.g.f.: (exp(9*x/2) - 2*sin(Pi/6-sqrt(3)*x/2))*exp(x/2)/3.

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

a(n) = (5^n + sqrt(3)*sin(Pi*n/3) - cos(Pi*n/3))/3.

a(n) = (A020729(n) + A057079(n-1))/3.

Example

Evolution from initial string "3": 3 -> 31213 -> 3121312321231321232131213 -> ...
Therefore, number of 1’s at step n:
a(0) = 0;
a(1) = 2;
a(2) = 9, etc.

Maple

A276508:=n->(2*5^n + 3*(-1)^(floor((n-1)/3)) + (-1)^n)/6: seq(A276508(n), n=0..30); # Wesley Ivan Hurt, Sep 07 2016

Mathematica

Table[(2 5^n + 3 (-1)^Floor[(n - 1)/3] + (-1)^n)/6, {n, 0, 25}]
LinearRecurrence[{6, -6, 5}, {0, 2, 9}, 26]

Prog

(PARI) concat(0, Vec(x*(2-3*x)/((1-5*x)*(1-x+x^2)) + O(x^99))) \\ Altug Alkan, Sep 06 2016

Crossrefs

Cf. A000351, A010892, A020729, A057079, A174737.

Sequence in context: A289684 A368764 A280955 * A381309 A347996 A092239

Adjacent sequences: A276505 A276506 A276507 * A276509 A276510 A276511

Keyword

nonn,easy

Author

Ilya Gutkovskiy, Sep 06 2016

Status

approved