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 A015440 Generalized Fibonacci numbers. 22
 1, 1, 6, 11, 41, 96, 301, 781, 2286, 6191, 17621, 48576, 136681, 379561, 1062966, 2960771, 8275601, 23079456, 64457461, 179854741, 502142046, 1401415751, 3912125981, 10919204736, 30479834641, 85075858321, 237475031526 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 2, 6*a(n-2) equals the number of 6-colored compositions of n with all parts >= 2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011 Pisano period lengths: 1, 3, 6, 6, 1, 6, 21, 12, 18, 3, 40, 6, 56, 21, 6, 24, 16, 18, 360, 6, .... - R. J. Mathar, Aug 10 2012 LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..1000 Joerg Arndt, Matters Computational (The Fxtbook), section 14.8, pp. 317-318 M. Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7. A. G. Shannon, J. V. Leyendekkers, The Golden Ratio family and the Binet equation, Notes on Number Theory and Discrete Mathematics, Vol. 21, No. 2, (2015), 35-42. Index entries for linear recurrences with constant coefficients, signature (1,5). FORMULA a(n) = a(n-1) + 5 a(n-2). a(n) = (( (1+sqrt(21))/2 )^(n+1) - ( (1-sqrt(21))/2 )^(n+1))/sqrt(21). a(n) = Sum_{k=0..ceiling(n/2)} 5^k*binomial(n-k, k). - Benoit Cloitre, Mar 06 2004 G.f.: 1/(1 - x - 5x^2). - R. J. Mathar, Sep 03 2008 a(n) = Sum_{k=0..n} A109466(n,k)*(-5)^(n-k). - Philippe Deléham, Oct 26 2008 From Jeffrey R. Goodwin, May 28 2011: (Start) A special case of a more general class of Lucas sequences given by U(n) = U(n-1) + (4^(m-1)-1)/3 U(n-2). U(n) = (( (1+sqrt((4^(m)-1)/3))/2 )^(n+1) - ( (1-sqrt((4^(m)-1)/3))/2 )^(n+1))/sqrt((4^(m)-1)/3). Fix m = 2 to get the formula for the Fibonacci sequence, fix m = 3 to get the formula for a(n). (End) G.f.: G(0)/(2-x), where G(k)= 1 + 1/(1 - x*(21*k-1)/(x*(21*k+20) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 20 2013 G.f.: Q(0)/x -1/x, where Q(k) = 1 + 5*x^2 + (k+2)*x - x*(k+1 + 5*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 06 2013 a(n) = (Sum_{k=1..n+1, k odd} binomial(n+1,k)*21^((k-1)/2))/2^n. - Vladimir Shevelev, Feb 05 2014 MAPLE A015440 := proc(n)     if n <= 1 then         1;     else         procname(n-1)+5*procname(n-2) ;     end if; end proc: # R. J. Mathar, May 15 2016 MATHEMATICA a[n_]:=(MatrixPower[{{1, 3}, {1, -2}}, n].{{1}, {1}})[[2, 1]]; Table[Abs[a[n]], {n, -1, 40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *) LinearRecurrence[{1, 5}, {1, 1}, 100] (* Vincenzo Librandi, Nov 06 2012 *) PROG (Sage) [lucas_number1(n, 1, -5) for n in range(1, 28)] # Zerinvary Lajos, Apr 22 2009 (MAGMA) [n le 2 select 1 else Self(n-1)+5*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 06 2012 (PARI) a(n)=abs([1, 3; 1, -2]^n*[1; 1])[2, 1] \\ Charles R Greathouse IV, Feb 03 2014 CROSSREFS Cf. A006130, A006131, A015441. Sequence in context: A038265 A243139 A288822 * A253209 A193488 A270223 Adjacent sequences:  A015437 A015438 A015439 * A015441 A015442 A015443 KEYWORD nonn,easy AUTHOR STATUS approved

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Last modified February 17 18:37 EST 2020. Contains 332005 sequences. (Running on oeis4.)