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

Author

Henry Bottomley, May 10 2000

Keywords

• nonn
• easy

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-1) == 53^((p-1)/2) (mod p), for odd primes p. - Gary W. Adamson, Feb 22 2009 [See A087475 for more info about this congruence. - Jason Yuen, Apr 05 2025]

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.

Crossrefs

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

Row n=7 of A073133, A172236 and A352361.

Cf. A099367 (squares).

Programs

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

Mathematica

LinearRecurrence[{7, 1}, {1, 7}, 30] (* Vincenzo Librandi, Feb 23 2013 *)

PARI

a(n)=([0,1; 1,7]^n*[1;7])[1,1] \\ Charles R Greathouse IV, Apr 08 2016

Sage

[lucas_number1(n,7,-1) for n in range(1, 19)] # Zerinvary Lajos, Apr 24 2009

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

Lim_{n->oo} 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

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

Extensions

Formula corrected by Johannes W. Meijer, May 30 2010, Jun 02 2010

Extended by T. D. Noe, May 23 2011