OFFSET
0,2
COMMENTS
In general, sequences with recurrence a(n) = k*a(n-1) + a(n-2) and a(0)=1 (and a(-1)=0) have the generating function 1/(1-k*x-x^2). If k is odd (k>=3) they satisfy a(3n) = b(5n), a(3n+1) = b(5n+3), a(3n+2) = 2*b(5n+4) where b(n) is the sequence of denominators of continued fraction convergents to sqrt(k^2+4). [If k is even then a(n) is the sequence of denominators of continued fraction convergents to sqrt(k^2/4+1).]
a(p) == 53^((p-1)/2)) (mod p), for odd primes p. - Gary W. Adamson, Feb 22 2009
From Johannes W. Meijer, Jun 12 2010: (Start)
For the sequence given above k=7 which implies that it is associated with A041091.
For a similar statement about sequences with recurrence a(n) = k*a(n-1) + a(n-2) but with a(0) = 2, and a(-1) = 0, see A086902; a sequence that is associated with A041090.
(End)
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 7's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
a(n) equals the number of words of length n on alphabet {0,1,...,7} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Feb 21 2023: (Start)
Also called the 7-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 7 kinds of squares available. (End)
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Sergio Falcón and Ángel Plaza, On the Fibonacci k-numbers, Chaos, Solitons & Fractals 2007; 32(5): 1615-24.
Sergio Falcón and Ángel Plaza, The k-Fibonacci sequence and the Pascal 2-triangle Chaos, Solitons & Fractals 2007; 33(1): 38-49.
Milan 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
Index entries for linear recurrences with constant coefficients, signature (7,1).
FORMULA
G.f.: 1/(1 - 7x - x^2).
a(n) = U(n, 7*i/2)*(-i)^n with i^2=-1 and Chebyshev's U(n, x/2) = S(n, x) polynomials. See A049310.
a(n) = F(n, 7), the n-th Fibonacci polynomial evaluated at x=7. - T. D. Noe, Jan 19 2006
From Sergio Falcon, Sep 24 2007: (Start)
a(n) = (sigma^n - (-sigma)^(-n))/(sqrt(53)) with sigma = (7+sqrt(53))/2;
a(n) = Sum_{i=0..floor((n-1)/2)} binomial(n-1-i,i)*7^(n-1-2i). (End)
a(n) = ((7 + sqrt(53))^n - (7 - sqrt(53))^n)/(2^n*sqrt(53)). Offset 1. a(3)=50. - Al Hakanson (hawkuu(AT)gmail.com), Jan 17 2009
From Johannes W. Meijer, Jun 12 2010: (Start)
(End)
Sum_{n>=0} (-1)^n/(a(n)*a(n+1)) = (sqrt(53)-7)/2. - Vladimir Shevelev, Feb 23 2013
From Kai Wang, Feb 24 2020: (Start)
Sum_{m>=0} 1/(a(m)*a(m+2)) = 1/49.
Sum_{m>=0} 1/(a(2*m)*a(2*m+2)) = (sqrt(53)-7)/14.
In general, for sequences with recurrence f(n)= k*f(n-1)+f(n-2) and f(0)=1,
Sum_{m>=0} 1/(f(m)*f(m+2)) = 1/(k^2).
Sum_{m>=0} 1/(f(2*m)*f(2*m+2)) = (sqrt(k^2+4) - k)/(2*k). (End)
E.g.f.: (1/53)*exp(7*x/2)*(53*cosh(sqrt(53)*x/2) + 7*sqrt(53)*sinh(sqrt(53)*x/2)). - Stefano Spezia, Feb 26 2020
G.f.: x/(1 - 7*x - x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (m*k + 7 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024
MATHEMATICA
LinearRecurrence[{7, 1}, {1, 7}, 30] (* Vincenzo Librandi, Feb 23 2013 *)
PROG
(Sage) [lucas_number1(n, 7, -1) for n in range(1, 19)] # Zerinvary Lajos, Apr 24 2009
(Magma) I:=[1, 7]; [n le 2 select I[n] else 7*Self(n-1)+Self(n-2): n in [1..25]]; // Vincenzo Librandi, Feb 23 2013
(PARI) a(n)=([0, 1; 1, 7]^n*[1; 7])[1, 1] \\ Charles R Greathouse IV, Apr 08 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Henry Bottomley, May 10 2000
EXTENSIONS
Formula corrected by Johannes W. Meijer, May 30 2010, Jun 02 2010
Extended by T. D. Noe, May 23 2011
STATUS
approved