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A015451 a(n) = 6*a(n-1) + a(n-2) for n > 1, with a(0) = a(1) = 1. 12
1, 1, 7, 43, 265, 1633, 10063, 62011, 382129, 2354785, 14510839, 89419819, 551029753, 3395598337, 20924619775, 128943316987, 794584521697, 4896450447169, 30173287204711, 185936173675435, 1145790329257321 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

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).

FORMULA

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

a(n) = (1/10)*[3 - sqrt(10)]^n*sqrt(10) - (1/10)*[3 + sqrt(10)]^n*sqrt(10) + (1/2)*[3 + sqrt(10)]^n + (1/2) *[3 - sqrt(10)]^n, with n >= 0. - Paolo P. Lava, Jun 25 2008

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

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 *)

PROG

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

CROSSREFS

Row m=6 of A135597.

Cf. A049310, A055830.

Sequence in context: A003464 A022036 A277670 * A194779 A126502 A286911

Adjacent sequences:  A015448 A015449 A015450 * A015452 A015453 A015454

KEYWORD

nonn,easy

AUTHOR

Olivier Gérard

STATUS

approved

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Last modified August 21 23:43 EDT 2017. Contains 290940 sequences.