A056456 Number of palindromes of length n using exactly five different symbols.

0, 0, 0, 0, 0, 0, 0, 0, 120, 120, 1800, 1800, 16800, 16800, 126000, 126000, 834120, 834120, 5103000, 5103000, 29607600, 29607600, 165528000, 165528000, 901020120, 901020120, 4809004200, 4809004200, 25292030400

Offset: 1

Author

Marks R. Nester

Keywords

• nonn
• easy

References

• M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2.]

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (1,14,-14,-71,71,154,-154,-120,120).

Crossrefs

Cf. A056451, A001118.

Programs

Mathematica

k=5; Table[k! StirlingS2[Ceiling[n/2],k],{n,1,30}] (* Robert A. Russell, Sep 25 2018 *)
LinearRecurrence[{1, 14, -14, -71, 71, 154, -154, -120, 120}, {0, 0, 0, 0, 0, 0, 0, 0, 120}, 30] (* Vincenzo Librandi, Sep 29 2018 *)

PARI

a(n) = 5!*stirling((n+1)\2, 5, 2); \\ Altug Alkan, Sep 25 2018

Formula

a(n) = 5! * Stirling2( [(n+1)/2], 5).

G.f.: -120*x^9/((x-1)*(2*x-1)*(2*x+1)*(2*x^2-1)*(3*x^2-1)*(5*x^2-1)). - Colin Barker, Sep 03 2012

G.f.: k!(x^(2k-1)+x^(2k))/Product_{i=1..k}(1-ix^2), where k=5 is the number of symbols. - Robert A. Russell, Sep 25 2018