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%I #12 Jun 30 2021 12:30:53
%S 5,25,125,625,3124,15615,78050,390125,1950000,9746876,48718765,
%T 243515775,1217188750,6083993750,30410221874,152002390605,
%U 759768437250,3797624997500,18982040993750,94879794746876,474246971343775
%N Number of 0..4 arrays of length n avoiding the consecutive pattern 0..4
%C Column 4 of A206455
%H R. H. Hardin, <a href="/A206451/b206451.txt">Table of n, a(n) for n = 1..210</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,0,0,0,-1).
%F a(n) = 5*a(n-1) -a(n-5)
%F Empirical: a(n) = sum{i in 0..floor(n/5)} ((-1)^i*5^(n-5*i)*binomial(n-4*i,i))
%F From _Robert Israel_, Jan 08 2016: (Start) The recursion can be proved using the matrix representation
%F a(n) = [ 1 1 1 1 1] M^n [ 1 0 0 0 0 ]^T, where
%F M = [ 4 3 3 3 3 ]
%F [ 1 1 1 1 1 ]
%F [ 0 1 0 0 0 ]
%F [ 0 0 1 0 0 ]
%F [ 0 0 0 1 0 ]
%F which satisfies M^5 = 5 M^4 - I.
%F G.f.: -x*(-5+x^4) / ( 1-5*x+x^5 ).. (End)
%p M:= <<4|3|3|3|3>,<1|1|1|1|1>,<0|1|0|0|0>,<0|0|1|0|0>,<0|0|0|1|0>>:
%p seq(<1|1|1|1|1> . M^n . <1,0,0,0,0>, n=1..30); # _Robert Israel_, Jan 08 2016
%K nonn
%O 1,1
%A _R. H. Hardin_, Feb 07 2012
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