A014445 Even Fibonacci numbers; or, Fibonacci(3*n).

0, 2, 8, 34, 144, 610, 2584, 10946, 46368, 196418, 832040, 3524578, 14930352, 63245986, 267914296, 1134903170, 4807526976, 20365011074, 86267571272, 365435296162, 1548008755920, 6557470319842, 27777890035288, 117669030460994, 498454011879264, 2111485077978050

Offset: 0

Author

Mohammad K. Azarian

Keywords

• nonn
• easy
• nice

Comments

a(n) = 3^n*b(n;2/3) = -b(n;-2), but we have 3^n*a(n;2/3) = F(3n+1) = A033887 and a(n;-2) = F(3n-1) = A015448, where a(n;d) and b(n;d), n=0,1,...,d, denote the so-called delta-Fibonacci numbers (the argument "d" of a(n;d) and b(n;d) is abbreviation of the symbol "delta") defined by the following equivalent relations: (1 + d*((sqrt(5) - 1)/2))^n = a(n;d) + b(n;d)*((sqrt(5) - 1)/2) equiv. a(0;d)=1, b(0;d)=0, a(n+1;d) = a(n;d) + d*b(n;d), b(n+1;d) = d*a(n;d) + (1-d)b(n;d) equiv. a(0;d)=a(1;d)=1, b(0;1)=0, b(1;d)=d, and x(n+2;d) + (d-2)*x(n+1;d) + (1-d-d^2)*x(n;d) = 0 for every n=0,1,...,d, and x=a,b equiv. a(n;d) = Sum_{k=0..n} C(n,k)*F(k-1)*(-d)^k, and b(n;d) = Sum_{k=0..n} C(n,k)*(-1)^(k-1)*F(k)*d^k equiv. a(n;d) = Sum_{k=0..n} C(n,k)*F(k+1)*(1-d)^(n-k)*d^k, and b(n;d) = Sum_{k=1..n} C(n;k)*F(k)*(1-d)^(n-k)*d^k. The sequences a(n;d) and b(n;d) for special values d are connected with many known sequences: A000045, A001519, A001906, A015448, A020699, A033887, A033889, A074872, A081567, A081568, A081569, A081574, A081575, A163073 (see also the papers of Witula et al.). - Roman Witula, Jul 12 2012

For any odd k, Fibonacci(k*n) = sqrt(Fibonacci((k-1)*n) * Fibonacci((k+1)*n) + Fibonacci(n)^2). - Gary Detlefs, Dec 28 2012

The ratio of consecutive terms approaches the continued fraction 4 + 1/(4 + 1/(4 +...)) = A098317. - Hal M. Switkay, Jul 05 2020

Examples

			G.f. = 2*x + 8*x^2 + 34*x^3 + 144*x^4 + 610*x^5 + 2584*x^6 + 10946*x^7 + ...
		

References

• Arthur T. Benjamin and Jennifer J. Quinn,, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, id. 232.

Links

• T. D. Noe, Table of n, a(n) for n = 0..200
• Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38 (2012), pp. 1871-1876 (See Corollary 1 (v)).
• H. H. Ferns, Problem B-115, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 5, No. 2 (1967), p. 202; Identities for F_{kn} and L{kn}, Solution to Problem B-115 by Stanley Rabinowitz, ibid., Vol. 6, No. 1 (1968), pp. 92-93.
• Ira M. Gessel and Ji Li, Compositions and Fibonacci identities, J. Int. Seq., Vol. 16 (2013), Article 13.4.5.
• Edyta Hetmaniok, Bozena Piatek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Math., Vol. 15 (2017), pp. 477-485.
• Tanya Khovanova, Recursive Sequences.
• Ron Knott, Mathematics of the Fibonacci Series.
• Bahar Kuloğlu, Engin Özkan, and Marin Marin, Fibonacci and Lucas Polynomials in n-gon, An. Şt. Univ. Ovidius Constanţa (Romania 2023) Vol. 31, No 2, 127-140.
• Peter McCalla and Asamoah Nkwanta, Catalan and Motzkin Integral Representations, arXiv:1901.07092 [math.NT], 2019.
• Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.
• Roberto Tauraso, Some Congruences for Central Binomial Sums Involving Fibonacci and Lucas Numbers, JIS 19 (2016) # 16.5.4
• Roman Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math., Vol. 3, No. 2 (2009), pp. 310-329, MR2555042.
• Roman Witula, Binomials transformation formulae of scaled Lucas numbers, Demonstratio Math., Vol. 46 (2013), pp. 15-27.
• Index entries for linear recurrences with constant coefficients, signature (4,1).

Crossrefs

Cf. A000032, A000045, A001076, A049310, A111125.

Cf. A001519, A001906, A015448, A020699, A033887, A033889, A074872, A081567, A081568, A081569, A081574, A081575, A098317, A163073.

First differences of A099919.

Third column of array A102310.

Programs

Magma

[Fibonacci(3*n): n in [0..30]]; // Vincenzo Librandi, Apr 18 2011

Maple

a:= n-> (<<0|1>, <1|1>>^(3*n))[1,2]:
seq(a(n), n=0..25);  # Alois P. Heinz, Feb 03 2023

Mathematica

Table[Fibonacci[3n], {n,0,30}] (* Stefan Steinerberger, Apr 07 2006 *)
LinearRecurrence[{4,1},{0,2},30] (* Harvey P. Dale, Nov 14 2021 *)
Table[ SeriesCoefficient[2*x/(1 - 4*x - x^2), {x, 0, n}], {n, 0, 20}] (* Nikolaos Pantelidis, Feb 02 2023 *)

MuPAD

numlib::fibonacci(3*n) $ n = 0..30; // Zerinvary Lajos, May 09 2008

PARI

a(n)=fibonacci(3*n) \\ Charles R Greathouse IV, Oct 25 2012

Sage

[fibonacci(3*n) for n in range(0, 30)] # Zerinvary Lajos, May 15 2009

Formula

a(n) = Sum_{k=0..n} binomial(n, k)*F(k)*2^k. - Benoit Cloitre, Oct 25 2003

From Lekraj Beedassy, Jun 11 2004: (Start)

a(n) = 4*a(n-1) + a(n-2), with a(-1) = 2, a(0) = 0.

a(n) = 2*A001076(n).

a(n) = (F(n+1))^3 + (F(n))^3 - (F(n-1))^3. (End)

a(n) = Sum_{k=0..floor((n-1)/2)} C(n, 2*k+1)*5^k*2^(n-2*k). - Mario Catalani (mario.catalani(AT)unito.it), Jul 22 2004

a(n) = Sum_{k=0..n} F(n+k)*binomial(n, k). - Benoit Cloitre, May 15 2005

O.g.f.: 2*x/(1 - 4*x - x^2). - R. J. Mathar, Mar 06 2008

a(n) = second binomial transform of (2,4,10,20,50,100,250). This is 2* (1,2,5,10,25,50,125) or 5^n (offset 0): *2 for the odd numbers or *4 for the even. The sequences are interpolated. Also a(n) = 2*((2+sqrt(5))^n - (2-sqrt(5))^n)/sqrt(20). - Al Hakanson (hawkuu(AT)gmail.com), May 02 2009

a(n) = 3*F(n-1)*F(n)*F(n+1) + 2*F(n)^3, F(n)=A000045(n). - Gary Detlefs, Dec 23 2010

a(n) = (-1)^n*3*F(n) + 5*F(n)^3, n >= 0. See the D. Jennings formula given in a comment on A111125, where also the reference is given. - Wolfdieter Lang, Aug 31 2012

With L(n) a Lucas number, F(3*n) = F(n)*(L(2*n) + (-1)^n) = (L(3*n+1) + L(3*n-1))/5 starting at n=1. - J. M. Bergot, Oct 25 2012

a(n) = sqrt(Fibonacci(2*n)*Fibonacci(4*n) + Fibonacci(n)^2). - Gary Detlefs, Dec 28 2012

For n > 0, a(n) = 5*F(n-1)*F(n)*F(n+1) - 2*F(n)*(-1)^n. - J. M. Bergot, Dec 10 2015

a(n) = -(-1)^n * a(-n) for all n in Z. - Michael Somos, Nov 15 2018

a(n) = (5*Fibonacci(n)^3 + Fibonacci(n)*Lucas(n)^2)/4 (Ferns, 1967). - Amiram Eldar, Feb 06 2022

a(n) = 2*i^(n-1)*S(n-1,-4*i), with i = sqrt(-1), and the Chebyshev S-polynomials (see A049310) with S(-1, x) = 0. From the simplified trisection formula. - Gary Detlefs and Wolfdieter Lang, Mar 04 2023

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

a(n) = 2*F(n) + 3*Sum_{k=0..n-1} F(3*k)*F(n-k). - Yomna Bakr and Greg Dresden, Jun 10 2024