A015440 a(n) = a(n-1) + 5*a(n-2), with a(0) = a(1) = 1.
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, 662854323131, 1850229480761, 5164501096416
Offset: 0
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).
Programs
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Magma
[n le 2 select 1 else Self(n-1)+5*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 06 2012
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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
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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 *) -
PARI
a(n)=abs([1,3;1,-2]^n*[1;1])[2,1] \\ Charles R Greathouse IV, Feb 03 2014
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Sage
[lucas_number1(n,1,-5) for n in range(1, 28)] # Zerinvary Lajos, Apr 22 2009
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
From Peter Bala, Jun 27 2025: (Start)
The following products telescope:
Product_{k >= 0} (1 + 5^k/a(2*k+1)) = 1 + sqrt(21).
Product_{k >= 1} (1 - 5^k/a(2*k+1)) = 1/22 * (1 + sqrt(21)).
Product_{k >= 0} (1 + (-5)^k/a(2*k+1)) = (1/21) * (21 + sqrt(21)).
Product_{k >= 1} (1 - (-5)^k/a(2*k+1)) = (1/22) * (21 + sqrt(21)). (End)
E.g.f.: exp(x/2)*(sqrt(21)*cosh(sqrt(21)*x/2) + sinh(sqrt(21)*x/2))/sqrt(21). - Stefano Spezia, Jul 04 2025
Comments