A033889 a(n) = Fibonacci(4*n + 1).

1, 5, 34, 233, 1597, 10946, 75025, 514229, 3524578, 24157817, 165580141, 1134903170, 7778742049, 53316291173, 365435296162, 2504730781961, 17167680177565, 117669030460994, 806515533049393, 5527939700884757, 37889062373143906, 259695496911122585, 1779979416004714189

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

For positive n, a(n) equals (-1)^n times the permanent of the (4n) X (4n) tridiagonal matrix with sqrt(i)'s along the three central diagonals, where i is the imaginary unit. - John M. Campbell, Jul 12 2011

a(n) = 5^n*a(n; 3/5) = (16/5)^n*a(2*n; 3/4), and F(4*n) = 5^n*b(n; 3/5) = (16/5)^n*b(2*n; 3/4), where a(n; d) and b(n; d), n=0, 1, ..., d in C, denote the delta-Fibonacci numbers defined in comments to A014445. Two of these identities from the following relations follows: F(k+1)^n*a(n; F(k)/F(k+1)) = F(k*n+1) and F(k+1)^n*b(n; F(k)/F(k+1)) = F(k*n) (see also Witula's et al. papers). - Roman Witula, Jul 24 2012

Links

• Bruno Berselli, Table of n, a(n) for n = 0..500
• Edyta Hetmaniok, Bożena Piątek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Mathematics, 15(1) (2017), 477-485.
• Tanya Khovanova, Recursive Sequences.
• Kai Wan, Problem H-90, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 60, No. 3 (2022), p. 282; Solution to Problem H-90, by Ángel Plaza, ibid., Vol. 62, No. 1 (2024), pp. 95-96.
• Roman Wituła, Binomials Transformation Formulae of Scaled Lucas Numbers, Demonstratio Mathematica, Vol. XLVI, No. 1 (2013), pp. 15-27.
• Roman Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math 3 (2009), 310-329, MR2555042.
• Index entries for linear recurrences with constant coefficients, signature (7,-1).

Crossrefs

Cf. A000032, A000045, A004187, A014445, A016813, A122070, A167816, A094567.

Cf. A003881, A049684, A081018, A081016, A081068, A172968.

Programs

Magma

[Fibonacci(4*n+1): n in [0..100]]; // Vincenzo Librandi, Apr 16 2011

Mathematica

Table[Fibonacci[4*n+1], {n,0,14}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2008 *)

PARI

a(n)=fibonacci(4*n+1) \\ Charles R Greathouse IV, Jul 15 2011

PARI

Vec((1-2*x)/(1-7*x+x^2) + O(x^100)) \\ Altug Alkan, Dec 10 2015

Formula

a(n) = 7*a(n-1) - a(n-2) for n >= 2. - Floor van Lamoen, Dec 10 2001

From R. J. Mathar, Jan 17 2008: (Start)

O.g.f.: (1 - 2*x)/(1 - 7*x + x^2).

a(n) = A004187(n+1) - 2*A004187(n). (End); corrected by Klaus Purath, Jul 29 2020

a(n) = A167816(4*n+1). - Reinhard Zumkeller, Nov 13 2009

a(n) = sqrt(1 + 2 * Fibonacci(2*n) * Fibonacci(2*n + 1) + 5 * (Fibonacci(2*n) * Fibonacci(2*n + 1))^2). - Artur Jasinski, Feb 06 2010

a(n) = Sum_{k=0..n} A122070(n,k)*2^k. - Philippe Deléham, Mar 13 2012

a(n) = Fibonacci(2*n)^2 + Fibonacci(2*n)*Fibonacci(2*n+2) + 1. - Gary Detlefs, Apr 18 2012

a(n) = Fibonacci(2*n)^2 + Fibonacci(2*n+1)^2. - Bruno Berselli, Apr 19 2012

a(n) = Sum_{k = 0..n} A238731(n,k)*4^k. - Philippe Deléham, Mar 05 2014

a(n) = A000045(A016813(n)). - Michel Marcus, Mar 05 2014

2*a(n) = Fibonacci(4*n) + Lucas(4*n). - Bruno Berselli, Oct 13 2017

a(n) = A094567(n-1) + A094567(n), assuming A094567(-1) = 0. - Klaus Purath, Jul 29 2020

Sum_{n>=0} (-1)^n * arctan(3/a(n)) = Pi/4 (A003881) (Wan, 2022). - Amiram Eldar, Mar 01 2024

E.g.f.: exp(7*x/2)*(5*cosh(3*sqrt(5)*x/2) + sqrt(5)*sinh(3*sqrt(5)*x/2))/5. - Stefano Spezia, Jun 03 2024