A032088 Number of reversible strings with n beads of 5 colors. If more than 1 bead, not palindromic.

5, 10, 50, 300, 1500, 7750, 38750, 195000, 975000, 4881250, 24406250, 122062500, 610312500, 3051718750, 15258593750, 76293750000, 381468750000, 1907347656250, 9536738281250, 47683710937500

Offset

1,1

Links

Andrew Howroyd, Table of n, a(n) for n = 1..200

C. G. Bower, Transforms (2)

Index entries for linear recurrences with constant coefficients, signature (5, 5, -25).

Formula

"BHK" (reversible, identity, unlabeled) transform of 5, 0, 0, 0, ...

Conjectures from Colin Barker, Jul 07 2012: (Start)

a(n) = 5*a(n-1) + 5*a(n-2) - 25*a(n-3) for n > 4.

G.f.: 5*x*(1 - 3*x - 5*x^2 + 25*x^3)/((1 - 5*x)*(1 - 5*x^2)).

(End)

Conjectures from Colin Barker, Mar 09 2017: (Start)

a(n) = 5^(n/2)*(5^(n/2) - 1) / 2 for n > 1 and even.

a(n) = -5*(5^(n/2-1/2) - 5^(n-1)) / 2 for n > 1 and odd.

(End)

The above conjectures are true: The second set follows from the definition and the first set can be derived from that. - Andrew Howroyd, Oct 10 2017

a(n) = (5^n - 5^(ceiling(n/2)) / 2 = (A000351(n) - A056451(n)) / 2 for n>1. - Robert A. Russell and Danny Rorabaugh, Jun 22 2018

Mathematica

Join[{5}, LinearRecurrence[{5, 5, -25}, {10, 50, 300}, 19]] (* Jean-François Alcover, Oct 11 2017 *)

Prog

(PARI) a(n) = if(n<2, [5][n], (5^n - 5^(ceil(n/2)))/2); \\ Andrew Howroyd, Oct 10 2017

Crossrefs

Column 5 of A293500 for n>1.

Cf. A032122.

Equals (A000351 - A056451) / 2 for n>1.

Sequence in context: A003587 A268100 A205884 * A081076 A174462 A277247

Adjacent sequences: A032085 A032086 A032087 * A032089 A032090 A032091

Keyword

nonn

Author

Christian G. Bower

Status

approved