A015451 a(n) = 6*a(n-1) + a(n-2) for n > 1, with a(0) = a(1) = 1.

1, 1, 7, 43, 265, 1633, 10063, 62011, 382129, 2354785, 14510839, 89419819, 551029753, 3395598337, 20924619775, 128943316987, 794584521697, 4896450447169, 30173287204711, 185936173675435, 1145790329257321

Offset: 0

Author

Olivier Gérard

Keywords

• nonn
• easy

Comments

Row m=6 of A135597.

a(n) = term (1,1) in the 2 X 2 matrix [1,2; 3,5]^n. - Gary W. Adamson, May 30 2008

a(n)/a(n-1) tends to sqrt(10) + 3 = 6.16227766... - Gary W. Adamson, May 30 2008

For n >= 1, row sums of triangle for numbers 6^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 13 2012

Z[sqrt(10)] is not a unique factorization domain, since, for example, 6 = 2 * 3 = (-1)(2 - sqrt(10))(2 + sqrt(10)) = (4 - sqrt(10))(4 + sqrt(10)). However, the latter two factorizations are not distinct, because 3 + sqrt(10) is a unit in Z[sqrt(10)], and (2 - sqrt(10))(-3 - sqrt(10)) = 4 + sqrt(10). In fact, (2 - sqrt(10))(-3 - sqrt(10))^n gives an algebraic integer b + a(n) * sqrt(10) which, when multiplied by its associate (and by -1 when n is even) is equal to 6. - Alonso del Arte, Mar 15 2014

For n >= 1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,2,3,5,6} containing no subwords 00, 11, 22, 33, 44, 55. - Milan Janjic, Jan 31 2015

a(n+1) equals the number of sequences over the alphabet {0,1,2,3,4,5,6} of length n such that no two consecutive terms differ by 4. - David Nacin, May 31 2017

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• M. Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
• Tanya Khovanova, Recursive Sequences
• Index entries for linear recurrences with constant coefficients, signature (6,1).

Crossrefs

Cf. A049310, A055830.

Programs

Magma

[n le 2 select 1 else 6*Self(n-1) + Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 08 2012

Maple

a[0]:=1: a[1]:=1: for n from 2 to 26 do a[n]:=6*a[n-1]+a[n-2] od: seq(a[n], n=0..20); # Zerinvary Lajos, Jul 26 2006

Mathematica

LinearRecurrence[{6, 1}, {1, 1}, 30] (* Vincenzo Librandi, Nov 08 2012 *)
CoefficientList[Series[(1-5*x)/(1-6*x-x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)

PARI

x='x+O('x^30); Vec((1-5*x)/(1-6*x-x^2)) \\ G. C. Greubel, Dec 19 2017

Formula

a(n) = Sum_{k=0..n} 5^k * A055830(n, k). - Philippe Deléham, Oct 18 2006

G.f.: (1-5*x)/(1-6*x-x^2). - Philippe Deléham, Nov 20 2008

For n >= 2, a(n) = F_n(6) + F_(n+1)(6), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i = 0..floor((n-1)/2)} C(n-i-1,i) * x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012

a(n) = Sum_{k=0..n} A046854(n-1,k)*6^k. - R. J. Mathar, Feb 14 2024