OFFSET
0,3
COMMENTS
Define the sequence T(a_0,a_1) by a_{n+2} is the greatest integer such that a_{n+2}/a_{n+1}<a_{n+1}/a_n for n >= 0 . This sequence (with initial 0 omitted) is T(1,7).
This is a divisibility sequence.
For n>=2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 7's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
a(n) and b(n) := A056854(n) are the proper and improper nonnegative solutions of the Pell equation b(n)^2 - 5*(3*a(n))^2 = +4. see the cross-reference to A056854 below. - Wolfdieter Lang, Jun 26 2013
For n>=1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,2,3,4,5,6}. - Milan Janjic, Jan 25 2015
The digital root is A253298, which shares its digital root with A253368. - Peter M. Chema, Jul 04 2016
Lim_{n->oo} a(n+1)/a(n) = 2 + 3*phi = 1+ A090550 = 6.854101... - Wolfdieter Lang, Nov 16 2023
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
Kasper K. S. Andersen, Lisa Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, J. Num. Theory Comb. 2(3) (2011), 245-278. See Section 9.
Daniel Birmajer, Juan B. Gil, and Michael D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3 , example 12
David W. Boyd, Linear recurrence relations for some generalized Pisot sequences, Advances in Number Theory (Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993.
Zvonko Cerin, Some alternating sums of Lucas numbers, Centr. Eur. J. Math. 3(1) (2005), 1-13.
Rigoberto Flórez, Robinson A. Higuita, and Antara Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, J. Int. Seq. 17 (2014), 14.9.5.
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case a=0,b=1; p=7, q=-1.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, J. Int. Seq. 18 (2015), Article 15.4.7.
Vitaly M. Khamitov, Dmitriy Dmitrishin, Alexander Stokolos, and Daniel Gray, Convolved Numbers of k-sections of the Fibonacci Sequence: Properties, Consequences, arXiv:2603.08636 [math.CA], 2026. See p. 4, (3) and after.
Tanya Khovanova, Recursive Sequences.
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), lhs, m=9.
Index entries for linear recurrences with constant coefficients, signature (7,-1).
FORMULA
G.f.: x/(1-7*x+x^2).
a(n) = S(2*n-1, sqrt(9))/sqrt(9) = S(n-1, 7); S(n, x) := U(n, x/2), Chebyshev polynomials of the 2nd kind, A049310.
a(n) = Sum_{i = 0..n-1} C(2*n-1-i, i)*5^(n-i-1). - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
[A049685(n-1), a(n)] = [1,5; 1,6]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = A167816(4*n). - Reinhard Zumkeller, Nov 13 2009
a(n) = (((7+sqrt(45))/2)^n-((7-sqrt(45))/2)^n)/sqrt(45). - Noureddine Chair, Aug 31 2011
a(n+1) = Sum_{k = 0..n} A101950(n,k)*6^k. - Philippe Deléham, Feb 10 2012
a(n) = (A081072(n)/3)-1. - Martin Ettl, Nov 11 2012
From Peter Bala, Dec 23 2012: (Start)
Product {n >= 1} (1 + 1/a(n)) = (1/5)*(5 + 3*sqrt(5)).
Product {n >= 2} (1 - 1/a(n)) = (1/14)*(5 + 3*sqrt(5)). (End)
From Peter Bala, Apr 02 2015: (Start)
Sum_{n >= 1} a(n)*x^(2*n) = -A(x)*A(-x), where A(x) = Sum_{n >= 1} Fibonacci(2*n)* x^n.
1 + 5*Sum_{n >= 1} a(n)*x^(2*n) = F(x)*F(-x) = G(x)*G(-x), where F(x) = 1 + A(x) and G(x) = 1 + 5*A(x).
1 + Sum_{n >= 1} a(n)*x^(2*n) = H(x)*H(-x) = I(x)*I(-x), where H(x) = 1 + Sum_{n >= 1} Fibonacci(2*n + 3)*x^n and I(x) = 1 + x + x*Sum_{n >= 1} Fibonacci(2*n - 1)*x^n. (End)
E.g.f.: 2*exp(7*x/2)*sinh(3*sqrt(5)*x/2)/(3*sqrt(5)). - Ilya Gutkovskiy, Jul 03 2016
a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*9^k*binomial(n+k, 2*k+1). - Peter Bala, Jul 17 2023
a(n) = Sum_{k = 0..floor(n/2)} (-1)^k*7^(n-2*k)*binomial(n-k, k). - Greg Dresden, Aug 03 2024
From Peter Bala, Jul 22 2025: (Start)
The following products telescope:
Product {n >= 2} (1 + (-1)^n/a(n)) = (3/14)*(3 + sqrt(5)).
Product {n >= 1} (1 - (-1)^n/a(n)) = (1/3)*(3 + sqrt(5)).
Product_{n >= 1} (a(2*n) + 1)/(a(2*n) - 1) = (3/5)*sqrt(5). (End)
Sum_{n >= 1} a(n)*x^(2*n)/(2*n)! = 2/(3*sqrt(5)) * sinh(3*x/2) * sinh(sqrt(5)*x/2). - Peter Bala, Sep 24 2025
EXAMPLE
a(2) = 7*a(1) - a(0) = 7*7 - 1 = 48. - Michael B. Porter, Jul 04 2016
MAPLE
seq(combinat:-fibonacci(4*n)/3, n = 0 .. 30); # Robert Israel, Jan 26 2015
MATHEMATICA
LinearRecurrence[{7, -1}, {0, 1}, 30] (* Harvey P. Dale, Jul 13 2011 *)
CoefficientList[Series[x/(1 - 7*x + x^2), {x, 0, 50}], x] (* Vincenzo Librandi, Dec 23 2012 *)
PROG
(MuPAD) numlib::fibonacci(4*n)/3 $ n = 0..25; // Zerinvary Lajos, May 09 2008
(SageMath) [lucas_number1(n, 7, 1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008
(SageMath) [fibonacci(4*n)/3 for n in range(0, 21)] # Zerinvary Lajos, May 15 2009
(Magma) [Fibonacci(4*n)/3 : n in [0..30]]; // Vincenzo Librandi, Jun 07 2011
(PARI) a(n)=fibonacci(4*n)/3 \\ Charles R Greathouse IV, Mar 09 2012
(PARI) concat(0, Vec(x/(1-7*x+x^2) + O(x^99))) \\ Altug Alkan, Jul 03 2016
(Maxima)
a[0]:0$ a[1]:1$ a[n]:=7*a[n-1] - a[n-2]$ A004187(n):=a[n]$ makelist(A004187(n), n, 0, 30); /* Martin Ettl, Nov 11 2012 */
(Magma) /* By definition: */ [n le 2 select n-1 else 7*Self(n-1)-Self(n-2): n in [1..23]]; // Bruno Berselli, Dec 24 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
EXTENSIONS
Entry improved by comments from Michael Somos and Wolfdieter Lang, Aug 02 2000
STATUS
approved
