A113497 Ascending descending base exponent transform of sequence A000034(n) = 1 + n mod 2.

1, 3, 6, 6, 11, 9, 16, 12, 21, 15, 26, 18, 31, 21, 36, 24, 41, 27, 46, 30, 51, 33, 56, 36, 61, 39, 66, 42, 71, 45, 76, 48, 81, 51, 86, 54, 91, 57, 96, 60, 101, 63, 106, 66, 111, 69, 116, 72, 121, 75, 126, 78, 131, 81, 136, 84, 141, 87, 146, 90, 151, 93, 156, 96, 161, 99, 166, 102, 171

Offset

1,2

Comments

A000034 = 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, ... = continued fraction for (sqrt(3)+1)/2 (cf. A040001) = base 3 digital root of n+1. In general, the ascending descending base exponent transform of any simple periodic sequence can be written as a periodic set of interleaved sequences.

Links

G. C. Greubel, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Formula

a(n) = Sum_{i=1..n} A000034(i)^A000034(n-i+1).

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

From Colin Barker, Jun 16 2012: (Start)

a(n) = (-3+3*(-1)^n+8*n-2*(-1)^n*n)/4.

a(n) = 2*a(n-2)-a(n-4).

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

E.g.f.: (1/2)*(3*(x-1)*sinh(x) + 5*x*cosh(x)). - G. C. Greubel, Mar 12 2017

Example

a(1) = 1^1 = 1.
a(2) = 1^2 + 2^1 = 3.
a(3) = 1^1 + 2^2 + 1^1 = 6.
a(4) = 1^2 + 2^1 + 1^2 + 2^1 = 6.
a(5) = 1^1 + 2^2 + 1^1 + 2^2 + 1^1 = 11.
a(6) = 1^2 + 2^1 + 1^2 + 2^1 + 1^2 + 2^1 = 9.

Mathematica

Table[(-3 + 3*(-1)^n + 8*n - 2*(-1)^n*n)/4, {n, 1, 50}] (* G. C. Greubel, Mar 12 2017 *)

Prog

(PARI) x='x +O('x^50); Vec(x*(1+3*x+4*x^2)/((1-x)^2*(1+x)^2)) \\ G. C. Greubel, Mar 12 2017

Crossrefs

Cf. A000034, A113320, A005408, A113122, A113153, A113154, A113336, A113271, A113258, A113257, A113231, A087316, A113208.

Sequence in context: A157018 A203330 A197442 * A210692 A158662 A119980

Adjacent sequences: A113494 A113495 A113496 * A113498 A113499 A113500

Keyword

nonn,easy

Author

Jonathan Vos Post, Jan 10 2006

Extensions

Definition improved by M. F. Hasler, Jan 13 2012

Status

approved