A206451 Number of 0..4 arrays of length n avoiding the consecutive pattern 0..4.
5, 25, 125, 625, 3124, 15615, 78050, 390125, 1950000, 9746876, 48718765, 243515775, 1217188750, 6083993750, 30410221874, 152002390605, 759768437250, 3797624997500, 18982040993750, 94879794746876, 474246971343775
Offset: 1
Keywords
Links
- R. H. Hardin, Table of n, a(n) for n = 1..210
- Index entries for linear recurrences with constant coefficients, signature (5,0,0,0,-1).
Programs
-
Maple
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>>: seq(<1|1|1|1|1> . M^n . <1,0,0,0,0>, n=1..30); # Robert Israel, Jan 08 2016
Formula
a(n) = 5*a(n-1) -a(n-5)
Empirical: a(n) = sum{i in 0..floor(n/5)} ((-1)^i*5^(n-5*i)*binomial(n-4*i,i))
From Robert Israel, Jan 08 2016: (Start) The recursion can be proved using the matrix representation
a(n) = [ 1 1 1 1 1] M^n [ 1 0 0 0 0 ]^T, where
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 ]
which satisfies M^5 = 5 M^4 - I.
G.f.: -x*(-5+x^4) / ( 1-5*x+x^5 ).. (End)
Comments