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A054413
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a(n) = 7*a(n-1) + a(n-2), with a(0)=1 and a(1)=7.
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39
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1, 7, 50, 357, 2549, 18200, 129949, 927843, 6624850, 47301793, 337737401, 2411463600, 17217982601, 122937341807, 877779375250, 6267392968557, 44749530155149, 319514104054600, 2281348258537349, 16288951913816043, 116304011655249650, 830417033500563593
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OFFSET
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0,2
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COMMENTS
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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
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LINKS
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Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Sergio Falcon and Ángel Plaza, On the Fibonacci k-numbers, Chaos, Solitons & Fractals 2007; 32(5): 1615-24.
Sergio Falcon and Ángel Plaza, The k-Fibonacci sequence and the Pascal 2-triangle Chaos, Solitons & Fractals 2007; 33(1): 38-49.
M. 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.
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FORMULA
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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
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MATHEMATICA
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LinearRecurrence[{7, 1}, {1, 7}, 30] (* Vincenzo Librandi, Feb 23 2013 *)
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PROG
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(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
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CROSSREFS
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Cf. A000045, A000129, A001076, A005668, A006190, A052918, A099367, A243399.
Sequence in context: A096882 A033125 A022037 * A163458 A081571 A275827
Adjacent sequences: A054410 A054411 A054412 * A054414 A054415 A054416
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KEYWORD
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nonn,easy
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AUTHOR
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Henry Bottomley, May 10 2000
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EXTENSIONS
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Formula corrected by Johannes W. Meijer, May 30 2010, Jun 02 2010
Extended by T. D. Noe, May 23 2011
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STATUS
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approved
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