A126501 Number of n-tuples of numbers [0..5] (leading zeros allowed) in which adjacent digits differ by 4 or less.

1, 6, 34, 194, 1106, 6306, 35954, 204994, 1168786, 6663906, 37994674, 216628994, 1235123666, 7042134306, 40151166194, 228924368194, 1305226505746, 7441830001506, 42430056030514, 241917600158594, 1379308224915026

Offset: 0

Author

R. H. Hardin, Dec 28 2006

Keywords

• nonn
• easy

Comments

For n>=1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,...,5} containing no subwords 00 and 11. - Milan Janjic, Jan 31 2015

See PARI script for proof of g.f. - Andrew Howroyd, Apr 15 2017

Links

• Index entries for linear recurrences with constant coefficients, signature (5, 4).

Crossrefs

Cf. Base 6 differing by three or less A126474, two or less A126393, one or less A126360.

Programs

Mathematica

LinearRecurrence[{5, 4}, {1, 6}, 21] (* Jean-François Alcover, Oct 07 2017 *)

PARI

\\ Proof of generating function
TransferGf(m,u,t,v,z)=vector(m,i,u(i))*matsolve(matid(m)-z*matrix(m,m,i,j,t(i,j)),vectorv(m,i,v(i)));
RowGf(d,m,z)=1+z*TransferGf(m, i->1, (i,j)->abs(i-j)<=d, j->1, z);
print(RowGf(4,6,x)); \\ Andrew Howroyd, Apr 15 2017

Formula

[Empirical] a(base,n) = a(base-1,n)+9^(n-1) for base>=4n-3; a(base,n) = a(base-1,n)+9^(n-1)-2 when base=4n-4.

From Philippe Deléham, Mar 24 2012: (Start)

G.f.: (1+x)/(1-5*x-4*x^2).

a(n) = 5*a(n-1) + 4*a(n-2), a(0) = 1, a(1) = 6.

a(n) = Sum_{k, 0<=k<=n} A054458(n,k)*3^k. (End)

Conjecture: a(n) = (2^(-1-n)*((5-sqrt(41))^n*(-7+sqrt(41)) + (5+sqrt(41))^n*(7+sqrt(41)))) / sqrt(41). - Colin Barker, Jan 20 2017