A092900 A Jacobsthal sequence (A078008) to base 4.

1, 0, 2, 2, 12, 22, 112, 222, 1112, 2222, 11112, 22222, 111112, 222222, 1111112, 2222222, 11111112, 22222222, 111111112, 222222222, 1111111112, 2222222222, 11111111112, 22222222222, 111111111112, 222222222222, 1111111111112

Offset: 0

Author

Paul Barry, Mar 12 2004

Keywords

• easy
• base
• nonn

Examples

			a(8)= 1112 because A078008(8) = 86 (in base 10) = 64 + 16 + 4 + 2 = 1*(4^3) + 1*(4^2) + 1*(4^1) + 2.
		

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (0,11,0,-10).

Crossrefs

Cf. A081857.

Programs

PARI

Vec((1-9*x^2+2*x^3)/((1-x)*(1+x)*(1-10*x^2)) + O(x^30)) \\ Colin Barker, Apr 01 2016

Formula

For n > 0, a(2*n+1) is represented as a string of n 2's and a(2*n) as a string of (n-1) 1's followed by a 2.

From Colin Barker, Apr 01 2016: (Start)

a(n) = (6+10*(-1)^n+10^(1/2*(-1+n))*(2-2*(-1)^n+sqrt(10)+(-1)^n*sqrt(10)))/18.

a(n) = (10^(n/2)+8)/9 for n even.

a(n) = (2^((n+1)/2)*5^((n-1)/2)-2)/9 for n odd.

a(n) = 11*a(n-2)-10*a(n-4) for n>3.

G.f.: (1-9*x^2+2*x^3) / ((1-x)*(1+x)*(1-10*x^2)).

(End)