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%I #28 Oct 13 2018 09:27:25
%S 0,0,0,0,0,0,0,0,120,120,1800,1800,16800,16800,126000,126000,834120,
%T 834120,5103000,5103000,29607600,29607600,165528000,165528000,
%U 901020120,901020120,4809004200,4809004200,25292030400
%N Number of palindromes of length n using exactly five different symbols.
%D 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.]
%H Vincenzo Librandi, <a href="/A056456/b056456.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (1,14,-14,-71,71,154,-154,-120,120).
%F a(n) = 5! * Stirling2( [(n+1)/2], 5).
%F 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
%F 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
%t k=5; Table[k! StirlingS2[Ceiling[n/2],k],{n,1,30}] (* _Robert A. Russell_, Sep 25 2018 *)
%t 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 *)
%o (PARI) a(n) = 5!*stirling((n+1)\2, 5, 2); \\ _Altug Alkan_, Sep 25 2018
%Y Cf. A056451, A001118.
%K nonn,easy
%O 1,9
%A _Marks R. Nester_
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