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%I #24 May 25 2023 14:06:22
%S 1,4,25,150,900,5400,32400,194400,1166400,6998400,41990400,251942400,
%T 1511654400,9069926400,54419558400,326517350400,1959104102400,
%U 11754624614400,70527747686400,423166486118400,2538998916710400
%N a(n) = 25*6^(n-2), with a(0)=1 and a(1)=4.
%C For n>=2, a(n) is equal to the number of functions f:{1,2,...,n}->{1,2,3,4,5,6} such that for fixed, different x_1, x_2 in {1,2,...,n} and fixed y_1, y_2 in {1,2,3,4,5,6} we have f(x_1)<>y_1 and f(x_2)<> y_2. - _Milan Janjic_, Apr 19 2007
%C a(n) is the number of generalized compositions of n when there are 5*i-1 different types of i, (i=1,2,...). - _Milan Janjic_, Aug 26 2010
%D A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
%H Milan Janjic, <a href="http://www.pmfbl.org/janjic/">Enumerative Formulas for Some Functions on Finite Sets</a>
%H <a href="/index/Rec#order_01">Index entries for linear recurrences with constant coefficients</a>, signature (6)
%F a(n) = 25*6^(n-2), a(0)=1, a(1)=4. a(n) = 6a(n-1) + ((-1)^n)*binomial(2, 2-n); g.f.(x)=(1-x)^2/(1-6x).
%F a(n) = Sum_{k, 0<=k<=n} A201780(n,k)*4^k. - _Philippe Deléham_, Dec 05 2011
%t LinearRecurrence[{6},{1,4,25},30] (* _Harvey P. Dale_, May 25 2023 *)
%Y First differences of A052934. Cf. A000400.
%K easy,nonn
%O 0,2
%A _Barry E. Williams_, Jun 03 2000
%E More terms from _James A. Sellers_, Jun 05 2000