A056490 Number of periodic palindromes using exactly four different symbols.

0, 0, 0, 0, 0, 12, 24, 132, 240, 900, 1560, 4980, 8400, 24612, 40824, 113652, 186480, 502500, 818520, 2158260, 3498000, 9087012, 14676024, 37728372, 60780720, 155091300, 249401880, 632972340, 1016542800, 2569858212, 4123173624, 10393634292, 16664094960

Offset: 1

Author

Marks R. Nester

Keywords

• nonn
• easy

Examples

			For example, aaabbb is not a (finite) palindrome but it is a periodic palindrome.
For n=6, the 12 arrangements are ABACDC, ABADCD, ACABDB, ACADBD, ADABCB, ADACBC, ABCDCB, ABDCDB, ACBDBC, ACDBDC, ADBCBD, and ADCBCD.
		

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

• Muniru A Asiru, Table of n, a(n) for n = 1..700
• Index entries for linear recurrences with constant coefficients, signature (1,9,-9,-26,26,24,-24).

Crossrefs

Cf. A056284, A056344, A056455.

Column 4 of A305540.

Programs

GAP

a:=[0,0,0,0,0,12,24];; for n in [8..35] do a[n]:=a[n-1]+9*a[n-2]-9*a[n-3]-26*a[n-4]+26*a[n-5]+24*a[n-6]-24*a[n-7]; od; a; # Muniru A Asiru, Sep 26 2018

Magma

m:=50; R:=PowerSeriesRing(Integers(), m); [0, 0, 0, 0, 0] cat Coefficients(R!(12*x^6*(1+x)/((1-x)*(1-2*x)*(1+2*x)*(1-2*x^2)*(1-3*x^2)))); // G. C. Greubel, Oct 13 2018

Maple

a:=n->(factorial(4)/2)*(Stirling2(floor((n+1)/2),4)+Stirling2(ceil((n+1)/2),4)): seq(a(n),n=1..35); # Muniru A Asiru, Sep 26 2018

Mathematica

k = 4; Table[(k!/2) (StirlingS2[Floor[(n + 1)/2], k] +
StirlingS2[Ceiling[(n + 1)/2], k]), {n, 1, 40}] (* Robert A. Russell, Jun 05 2018 *)
LinearRecurrence[{1,9,-9,-26,26,24,-24}, {0,0,0,0,0,12,24}, 40] (* Robert A. Russell, Sep 29 2018 *)

PARI

a(n) = my(k=4); (k!/2)*(stirling(floor((n+1)/2), k, 2) + stirling(ceil((n+1)/2), k, 2)); \\ Michel Marcus, Jun 05 2018

Formula

a(n) = 2*A056344(n) - A056284(n).

G.f.: 12*x^6*(1+x)/((1-x)*(1-2*x)*(1+2*x)*(1-2*x^2)*(1-3*x^2)). - Colin Barker, May 06 2012

a(n) = (k!/2)*(S2(floor((n+1)/2),k) + S2(ceiling((n+1)/2),k)), with k=4 different colors used and where S2(n,k) is the Stirling subset number A008277. - Robert A. Russell, Jun 05 2018