cp's OEIS Frontend

A269603 Number of length-n 0..5 arrays with no repeated value differing from the previous repeated value by one or less.

%I A269603 #12 Jan 24 2019 15:43:11
%S A269603 6,36,210,1220,7030,40288,229754,1304934,7385898,41679780,234601902,
%T A269603 1317558578,7385249086,41325945826,230904832646,1288466651340,
%U A269603 7181415415962,39985405920156,222432351559566,1236355637246456
%N A269603 Number of length-n 0..5 arrays with no repeated value differing from the previous repeated value by one or less.
%H A269603 R. H. Hardin, <a href="/A269603/b269603.txt">Table of n, a(n) for n = 1..210</a>
%H A269603 Robert Israel, <a href="/A269603/a269603_1.pdf">Maple-assisted proof for empirical recursion</a>
%F A269603 Empirical: a(n) = 17*a(n-1) - 91*a(n-2) + 83*a(n-3) + 542*a(n-4) - 550*a(n-5) - 1651*a(n-6) - 745*a(n-7).
%F A269603 Empirical g.f.: 2*x*(3 - 33*x + 72*x^2 + 214*x^3 - 420*x^4 - 922*x^5 - 393*x^6) / ((1 - 5*x)*(1 - 12*x + 31*x^2 + 72*x^3 - 182*x^4 - 360*x^5 - 149*x^6)). - _Colin Barker_, Jan 24 2019
%F A269603 Empirical recursion verified - see link. - _Robert Israel_, Jan 24 2019
%e A269603 Some solutions for n=6:
%e A269603 ..4. .4. .1. .3. .0. .3. .5. .1. .1. .5. .5. .1. .3. .0. .2. .0
%e A269603 ..3. .5. .5. .2. .3. .1. .1. .3. .1. .5. .3. .2. .2. .0. .5. .3
%e A269603 ..4. .3. .3. .1. .3. .5. .2. .2. .3. .2. .4. .1. .5. .2. .4. .5
%e A269603 ..2. .2. .1. .3. .2. .1. .1. .0. .0. .5. .3. .4. .2. .3. .1. .0
%e A269603 ..3. .5. .2. .0. .1. .3. .4. .4. .3. .2. .1. .0. .2. .1. .4. .4
%e A269603 ..0. .1. .5. .3. .2. .3. .1. .4. .5. .2. .2. .2. .4. .3. .1. .4
%p A269603 T:= Matrix(42,42):
%p A269603 for x from 0 to 5 do
%p A269603   for v from 0 to 6 do
%p A269603     i:= 1 + x + 6*v;
%p A269603     for y in {$0..5} minus {x} do
%p A269603       T[i,1+y+6*v]:= 1;
%p A269603     od:
%p A269603     if abs(x-v) > 1 or v=6 then T[i,1+x+6*x]:= 1 fi
%p A269603 od od:
%p A269603 u:= Vector([0$36,1$6]): v:= Vector(42,1):
%p A269603 Tv[1]:= v:
%p A269603 for n from 2 to 50 do Tv[n]:= T . Tv[n-1] od:
%p A269603 seq(u^%T . Tv[n], n=1..50); # _Robert Israel_, Jan 24 2019
%Y A269603 Column 5 of A269606.
%K A269603 nonn
%O A269603 1,1
%A A269603 _R. H. Hardin_, Mar 01 2016