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A015451 a(n) = 6*a(n-1) + a(n-2) for n > 1, with a(0) = a(1) = 1. 12

%I

%S 1,1,7,43,265,1633,10063,62011,382129,2354785,14510839,89419819,

%T 551029753,3395598337,20924619775,128943316987,794584521697,

%U 4896450447169,30173287204711,185936173675435,1145790329257321

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

%C Row m=6 of A135597.

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

%C a(n)/a(n-1) tends to sqrt(10) + 3 = 6.16227766... - _Gary W. Adamson_, May 30 2008

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

%C 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

%C 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

%C 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

%H Vincenzo Librandi, <a href="/A015451/b015451.txt">Table of n, a(n) for n = 0..1000</a>

%H M. Janjic, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Janjic/janjic63.html">On Linear Recurrence Equations Arising from Compositions of Positive Integers</a>, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6,1).

%F a(n) = Sum_{k=0..n} 5^k * A055830(n, k). - _Philippe Deléham_, Oct 18 2006

%F 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

%F G.f.: (1-5*x)/(1-6*x-x^2). - _Philippe Deléham_, Nov 20 2008

%F 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

%p 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

%t LinearRecurrence[{6, 1}, {1, 1}, 30] (* _Vincenzo Librandi_, Nov 08 2012 *)

%t CoefficientList[Series[(1-5*x)/(1-6*x-x^2), {x, 0, 50}], x] (* _G. C. Greubel_, Dec 19 2017 *)

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

%o (PARI) x='x+O('x^30); Vec((1-5*x)/(1-6*x-x^2)) \\ _G. C. Greubel_, Dec 19 2017

%Y Cf. A049310, A055830.

%K nonn,easy

%O 0,3

%A _Olivier Gérard_

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Last modified December 12 03:23 EST 2018. Contains 318052 sequences. (Running on oeis4.)