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
From Wolfdieter Lang, Jan 02 2024: (Start)
This sequence {a(n-1)}, with a(-1) = 0, appears in the formula for powers of phi21 := (1 + sqrt(21))/2 = A222134 = 2.791287..., together with A(n) = A365824(n), as phi21^n = A(n) + a(n-1)*phi21(n), for n >= 0.
Limit_{n->oo} a(n)/a(n-1) = phi21. (End)
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Joerg Arndt, Matters Computational (The Fxtbook), section 14.8, pp. 317-318
Milan 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 and 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.
Paul Thomas Young, p-adic congruences for generalized Fibonacci sequences, The Fibonacci Quarterly, Vol. 32, No. 1, 1994.
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
With an initial 0 prepended, the sequence [0, 1, 1, 6, 11, 41, 96, ...] satisfies the congruences a(n*p^k) == (3|p)*(7|p)*a(n*p^(k-1)) (mod p^k) for positive integers k and n and all primes p, where (n|p) denotes the Legendre symbol. See Young, Theorem 1, Corollary 1(i). - Peter Bala, Dec 28 2022
a(n) = sqrt(-5)^(n-1)*S(n-1,1/sqrt(-5)), for n >= 0, with the Chebyshev polynomial S(n, x) (see A049310). - Wolfdieter Lang, Nov 17 2023
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
KEYWORD
nonn,easy
AUTHOR
STATUS
approved