A054413 a(n) = 7*a(n-1) + a(n-2), with a(0)=1 and a(1)=7.

1, 7, 50, 357, 2549, 18200, 129949, 927843, 6624850, 47301793, 337737401, 2411463600, 17217982601, 122937341807, 877779375250, 6267392968557, 44749530155149, 319514104054600, 2281348258537349, 16288951913816043, 116304011655249650, 830417033500563593

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.

For more information follow the Khovanova link and see A087130, A140455 and A178765.

(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

Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.

Index entries for linear recurrences with constant coefficients, signature (7,1).

Index entries for sequences related to Chebyshev polynomials.

Formula

a(3n) = A041091(5n), a(3n+1) = A041091(5n+3), a(3n+2) = 2*A041091(5n+4).

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) = -(7/106)*sqrt(53)*(7/2 - (1/2)*sqrt(53))^n + (1/2)*(7/2 + (1/2)*sqrt(53))^n + (1/2)*(7/2 - (1/2)*sqrt(53))^n + (7/106)*(7/2 + (1/2)*sqrt(53))^n*sqrt(53), with n >= 0. - Paolo P. Lava, Jun 25 2008

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)

a(2n+1) = 7*A097836(n), a(2n) = A097838(n).

Lim_{k->infinity} a(n+k)/a(k) = (A086902(n) + A054413(n-1)*sqrt(53))/2.

Lim_{n->infinity} A086902(n)/A054413(n-1) = sqrt(53).

(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

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

Cf. A000045, A000129, A001076, A005668, A006190, A052918, A243399.

Row n=7 of A073133, A172236 and A352361.

Cf. A099367 (squares).

Sequence in context: A096882 A033125 A022037 * A163458 A081571 A275827

Adjacent sequences: A054410 A054411 A054412 * A054414 A054415 A054416

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