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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
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
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
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
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
Sequence in context: A003464 A022036 A277670 * A194779 A126502 A286911
KEYWORD
nonn,easy
STATUS
approved