cp's OEIS Frontend

A269604 Number of length-n 0..6 arrays with no repeated value differing from the previous repeated value by one or less.

%I A269604 #12 Jan 24 2019 15:27:49
%S A269604 7,49,336,2298,15630,105892,714874,4811578,32300252,216337084,
%T A269604 1446056046,9648789758,64281141440,427655897226,2841661493142,
%U A269604 18861464959350,125070420653458,828618463551536,5485481885293294,36288577806336542
%N A269604 Number of length-n 0..6 arrays with no repeated value differing from the previous repeated value by one or less.
%H A269604 R. H. Hardin, <a href="/A269604/b269604.txt">Table of n, a(n) for n = 1..210</a>
%H A269604 Robert Israel, <a href="/A269604/a269604_1.pdf">Maple-assisted proof of empirical recursion</a>
%F A269604 Empirical: a(n) = 26*a(n-1) - 243*a(n-2) + 833*a(n-3) + 567*a(n-4) - 7567*a(n-5) - 1006*a(n-6) + 27361*a(n-7) + 31306*a(n-8) + 9984*a(n-9).
%F A269604 Empirical g.f.: x*(7 - 133*x + 763*x^2 - 362*x^3 - 7256*x^4 + 3224*x^5 + 32851*x^6 + 34133*x^7 + 10511*x^8) / ((1 - 6*x)*(1 - 20*x + 123*x^2 - 95*x^3 - 1137*x^4 + 745*x^5 + 5476*x^6 + 5495*x^7 + 1664*x^8)). - _Colin Barker_, Jan 24 2019
%F A269604 Empirical recursion verified: see link. - _Robert Israel_, Jan 24 2019
%e A269604 Some solutions for n=6:
%e A269604 ..3. .2. .6. .1. .5. .1. .2. .5. .2. .6. .4. .0. .1. .5. .5. .4
%e A269604 ..6. .0. .3. .2. .0. .1. .3. .2. .3. .1. .3. .5. .4. .4. .4. .0
%e A269604 ..6. .1. .6. .3. .1. .4. .6. .4. .1. .2. .1. .0. .5. .0. .3. .3
%e A269604 ..2. .4. .4. .6. .4. .6. .2. .6. .3. .4. .3. .0. .3. .1. .6. .5
%e A269604 ..2. .3. .2. .2. .1. .4. .3. .1. .2. .1. .1. .1. .0. .5. .3. .3
%e A269604 ..5. .1. .0. .4. .2. .0. .4. .1. .1. .1. .4. .6. .6. .6. .0. .2
%p A269604 with(LinearAlgebra):
%p A269604 T:= Matrix(56,56):
%p A269604 for x from 0 to 6 do
%p A269604   for v from 0 to 7 do
%p A269604     i:= 1 + x + 7*v;
%p A269604     for y in {$0..6} minus {x} do
%p A269604       T[i,1+y+7*v]:= 1;
%p A269604     od:
%p A269604     if abs(x-v) > 1 or v=7 then T[i,1+x+7*x]:= 1 fi
%p A269604 od od:
%p A269604 u:= Vector([0$49,1$7]): v:= Vector(56,1):
%p A269604 Tv[1]:= v:
%p A269604 for n from 2 to 50 do Tv[n]:= T . Tv[n-1] od:
%p A269604 seq(u^%T . Tv[n], n=1..50); # _Robert Israel_, Jan 24 2019
%Y A269604 Column 6 of A269606.
%K A269604 nonn
%O A269604 1,1
%A A269604 _R. H. Hardin_, Mar 01 2016