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Numbers of n-length words on alphabet {0,1,...,7} with no subwords ii, where i is from {0,1,...,5}.
5

%I #30 Nov 28 2023 10:37:51

%S 1,8,58,422,3070,22334,162478,1182014,8599054,62557406,455099950,

%T 3310814462,24085901134,175222936862,1274732360302,9273572395838,

%U 67464471491470,490798445231966,3570518059606702,25975223307710846,188967599273189326,1374723641527746974

%N Numbers of n-length words on alphabet {0,1,...,7} with no subwords ii, where i is from {0,1,...,5}.

%C a(n) equals the number of octonary sequences of length n such that no two consecutive terms differ by 5. - _David Nacin_, May 31 2017

%H Colin Barker, <a href="/A254663/b254663.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (7,2).

%F G.f.: (1 + x)/(1 - 7*x - 2*x^2).

%F a(n) = 7*a(n-1) + 2*a(n-2) with n>1, a(0) = 1, a(1) = 8.

%F a(n) = 2^(-1-n)*((7-r)^n*(-9+r) + (7+r)^n*(9+r)) / r, where r=sqrt(57). - _Colin Barker_, Jan 22 2017

%t RecurrenceTable[{a[0] == 1, a[1] == 8, a[n] == 7 a[n - 1] + 2 a[n - 2]}, a[n], {n, 0, 20}]

%t LinearRecurrence[{7,2},{1,8},30] (* _Harvey P. Dale_, Nov 28 2023 *)

%o (Magma) [n le 1 select 8^n else 7*Self(n)+2*Self(n-1): n in [0..20]];

%o (PARI) Vec((1 + x)/(1 - 7*x - 2*x^2) + O(x^30)) \\ _Colin Barker_, Jan 22 2017

%Y Cf. A015555, A055099, A126473, A126501, A126528, A254598, A254602.

%K nonn,easy

%O 0,2

%A _Milan Janjic_, Feb 04 2015