%I #108 May 17 2024 10:19:25

%S 1,7,50,357,2549,18200,129949,927843,6624850,47301793,337737401,

%T 2411463600,17217982601,122937341807,877779375250,6267392968557,

%U 44749530155149,319514104054600,2281348258537349,16288951913816043,116304011655249650,830417033500563593

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

%C 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).]

%C a(p) == 53^((p-1)/2)) (mod p), for odd primes p. - _Gary W. Adamson_, Feb 22 2009

%C From _Johannes W. Meijer_, Jun 12 2010: (Start)

%C For the sequence given above k=7 which implies that it is associated with A041091.

%C 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.

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

%C (End)

%C 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

%C 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

%C From _Michael A. Allen_, Feb 21 2023: (Start)

%C Also called the 7-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.

%C 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)

%H Vincenzo Librandi, <a href= "/A054413/b054413.txt" >Table of n, a(n) for n = 0..1000</a>

%H Michael A. Allen and Kenneth Edwards, <a href= "https://www.fq.math.ca/Papers1/60-5/allen.pdf" >Fence tiling derived identities involving the metallonacci numbers squared or cubed</a>, Fib. Q. 60:5 (2022) 5-17.

%H Sergio Falcón and Ángel Plaza, <a href= "http://dx.doi.org/10.1016/j.chaos.2006.09.022" >On the Fibonacci k-numbers</a>, Chaos, Solitons & Fractals 2007; 32(5): 1615-24.

%H Sergio Falcón and Ángel Plaza, <a href= "http://dx.doi.org/10.1016/j.chaos.2006.10.022" >The k-Fibonacci sequence and the Pascal 2-triangle</a> Chaos, Solitons & Fractals 2007; 33(1): 38-49.

%H Milan Janjic, <a href= "https://cs.uwaterloo.ca/journals/JIS/VOL18/Janjic/janjic63.html" >On Linear Recurrence Equations Arising from Compositions of Positive Integers</a>, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.

%H Tanya Khovanova, <a href= "http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html" >Recursive Sequences</a>

%H Kai Wang, <a href= "https://www.researchgate.net/publication/339487198_On_k-Fibonacci_Sequences_And_Infinite_Series_List_of_Results_and_Examples" >On k-Fibonacci Sequences And Infinite Series List of Results and Examples</a>, 2020.

%H <a href= "/index/Rec#order_02" >Index entries for linear recurrences with constant coefficients</a>, signature (7,1).

%H <a href= "/index/Ch#Cheby" >Index entries for sequences related to Chebyshev polynomials.</a>

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

%F G.f.: 1/(1 - 7x - x^2).

%F 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.

%F a(n) = F(n, 7), the n-th Fibonacci polynomial evaluated at x=7. - _T. D. Noe_, Jan 19 2006

%F From _Sergio Falcon_, Sep 24 2007: (Start)

%F a(n) = (sigma^n - (-sigma)^(-n))/(sqrt(53)) with sigma = (7+sqrt(53))/2;

%F a(n) = Sum_{i=0..floor((n-1)/2)} binomial(n-1-i,i)*7^(n-1-2i). (End)

%F 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

%F From _Johannes W. Meijer_, Jun 12 2010: (Start)

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

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

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

%F (End)

%F Sum_{n>=0} (-1)^n/(a(n)*a(n+1)) = (sqrt(53)-7)/2. - _Vladimir Shevelev_, Feb 23 2013

%F From _Kai Wang_, Feb 24 2020: (Start)

%F Sum_{m>=0} 1/(a(m)*a(m+2)) = 1/49.

%F Sum_{m>=0} 1/(a(2*m)*a(2*m+2)) = (sqrt(53)-7)/14.

%F In general, for sequences with recurrence f(n)= k*f(n-1)+f(n-2) and f(0)=1,

%F Sum_{m>=0} 1/(f(m)*f(m+2)) = 1/(k^2).

%F Sum_{m>=0} 1/(f(2*m)*f(2*m+2)) = (sqrt(k^2+4) - k)/(2*k). (End)

%F 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

%F 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

%t LinearRecurrence[{7, 1}, {1, 7}, 30] (* _Vincenzo Librandi_, Feb 23 2013 *)

%o (Sage) [lucas_number1(n,7,-1) for n in range(1, 19)] # _Zerinvary Lajos_, Apr 24 2009

%o (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

%o (PARI) a(n)=([0,1; 1,7]^n*[1;7])[1,1] \\ _Charles R Greathouse IV_, Apr 08 2016

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

%Y Row n=7 of A073133, A172236 and A352361.

%Y Cf. A099367 (squares).

%K nonn,easy

%O 0,2

%A _Henry Bottomley_, May 10 2000

%E Formula corrected by _Johannes W. Meijer_, May 30 2010, Jun 02 2010

%E Extended by _T. D. Noe_, May 23 2011