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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 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). 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 *) CoefficientList[Series[(1-5*x)/(1-6*x-x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *) PROG (MAGMA) [n le 2 select 1 else 6*Self(n-1) + Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 08 2012 (PARI) x='x+O('x^30); Vec((1-5*x)/(1-6*x-x^2)) \\ G. C. Greubel, Dec 19 2017 CROSSREFS Cf. A049310, A055830. Sequence in context: A003464 A022036 A277670 * A194779 A126502 A286911 Adjacent sequences:  A015448 A015449 A015450 * A015452 A015453 A015454 KEYWORD nonn,easy AUTHOR STATUS approved

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Last modified July 22 20:36 EDT 2018. Contains 312918 sequences. (Running on oeis4.)