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A206451 Number of 0..4 arrays of length n avoiding the consecutive pattern 0..4 2

%I

%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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Last modified October 22 05:36 EDT 2021. Contains 348160 sequences. (Running on oeis4.)