A163073 -id:A163073 - cp's OEIS frontend

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

Mohammad K. Azarian

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

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

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

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).

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.

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

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

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 *)

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

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

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

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

A082761 Trinomial transform of the Fibonacci numbers (A000045).

Original entry on oeis.org

1, 4, 20, 104, 544, 2848, 14912, 78080, 408832, 2140672, 11208704, 58689536, 307302400, 1609056256, 8425127936, 44114542592, 230986743808, 1209462292480, 6332826779648, 33159111507968, 173623361929216, 909103725543424, 4760128905543680, 24924358531088384, 130505635564355584

Offset: 0

Emanuele Munarini, May 21 2003

Hankel transform of Sum_{k=0..n} (-1)^k*C(2k, k) (see A054108). - Paul Barry, Jan 13 2009
Hankel transform of A046748. - Paul Barry, Apr 14 2010
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(2)'s along the three central diagonals. - John M. Campbell, Jul 12 2011
The limiting ratio is: Lim_{n -> oo} a(n)/a(n-1) = 1 + phi^3. - Bob Selcoe, Mar 18 2014
Invert transform of A052984. Invert transform is A083066. Binomial transform of A033887. Binomial transform is A163073. - Michael Somos, May 26 2014

			a(5) = 2848 = 5*(544) + 4 + 20 + 104.
G.f. = 1 + 4*x + 20*x^2 + 104*x^3 + 544*x^4 + 2848*x^5 + 14912*x^6 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..150
László Németh, The trinomial transform triangle, J. Int. Seqs., Vol. 21 (2018), Article 18.7.3. Also arXiv:1807.07109 [math.NT], 2018.
Index entries for linear recurrences with constant coefficients, signature (6,-4).

Cf. A000045, A001519, A027907, A033887, A046748, A052984, A054108, A069429, A083066, A147703, A163073.

[2^n * Fibonacci(2*n+1): n in [0..40]]; // Vincenzo Librandi, Jul 15 2011

a[ n_] := 2^n Fibonacci[ 2 n + 1]; (* Michael Somos, May 26 2014 *)
a[ n_] := If[ n < 0, SeriesCoefficient[ (2 - x) / (4 - 6 x + x^2), {x, 0, -1 - n}], SeriesCoefficient[ (1 - 2 x) / (1 - 6 x + 4 x^2), {x, 0, n}]]; (* Michael Somos, Oct 22 2017 *)
LinearRecurrence[{6,-4},{1,4},30] (* Harvey P. Dale, Jul 11 2014 *)

a(n)=fibonacci(2*n+1)<Charles R Greathouse IV, Jul 15 2011

{a(n) = if( n<0, n = -1 - n; 2^(-1-2*n), 1) * polcoeff( (1 - 2*x) / (1 - 6*x + 4*x^2) + x * O(x^n), n)}; /* Michael Somos, Oct 22 2017 */

[2^n*fibonacci(2*n+1) for n in range(41)] # G. C. Greubel, Jul 28 2024

a(n) = Sum_{k=0..2*n} A027907(n, k)*A000045(k+1).
From Paul Barry, Jul 16 2003: (Start)
Third binomial transform of (1, 1, 5, 5, 25, 25, ....).
a(n) = ((1+sqrt(5))(3+sqrt(5))^n-(1-sqrt(5))*(3-sqrt(5))^n)/(2*sqrt(5)). (End)
From R. J. Mathar, Nov 04 2008: (Start)
G.f.: (1-2*x)/(1-6*x+4*x^2).
a(n) = 6*a(n-1) - 4*a(n-2). (End)
a(n) = Sum_{k=0..n} A147703(n,k)*3^k. - Philippe Deléham, Nov 14 2008
For n>=2: a(n) = 5*a(n-1) + Sum_{i=1..n-2} a(i). - Bob Selcoe, Mar 18 2014
a(n) = a(-1-n) * 2^(2*n+1) for all n in Z. - Michael Somos, Mar 18 2014
a(n) = 2^n*Fibonacci(2*n+1), or 2^n*A001519(n+1). - Bob Selcoe, May 25 2014
From Michael Somos, May 26 2014: (Start)
a(n) - a(n-1) = A069429(n).
a(n+1) * a(n-1) - a(n)^2 = 4^n.
G.f.: 1 / (1 - 4*x / (1 - x / (1 - x))). (End)
E.g.f.: exp(3*x)*(5*cosh(sqrt(5)*x) + sqrt(5)*sinh(sqrt(5)*x))/5. - Stefano Spezia, May 24 2024

A366858 Array read by ascending antidiagonals: A(n, k) = n! * [x^n] exp((k-1)x)(kcosh(sqrt(k)x) + sqrt(k)sinh(sqrt(k)x))/k, with 1 <= k <= n.

Original entry on oeis.org

1, 1, 2, 1, 5, 3, 1, 12, 11, 4, 1, 29, 41, 19, 5, 1, 70, 153, 94, 29, 6, 1, 169, 571, 469, 177, 41, 7, 1, 408, 2131, 2344, 1097, 296, 55, 8, 1, 985, 7953, 11719, 6829, 2181, 457, 71, 9, 1, 2378, 29681, 58594, 42565, 16186, 3889, 666, 89, 10, 1, 5741, 110771, 292969, 265401, 120421, 33415, 6413, 929, 109, 11

Offset: 1

Stefano Spezia, Oct 25 2023

			The array begins:
  1,   2,    3,     4,     5,      6, ...
  1,   5,   11,    19,    29,     41, ...
  1,  12,   41,    94,   177,    296, ...
  1,  29,  153,   469,  1097,   2181, ...
  1,  70,  571,  2344,  6829,  16186, ...
  1, 169, 2131, 11719, 42565, 120421, ...
  ...

Cf. A000012 (k=1), A000129 (k=2), A001835 (k=3), A083065 (k=4), A163073 (k=5).
Cf. A000027 (n=1), A028387 (n=2).
Cf. A366859 (antidiagonal sums).

A[n_,k_]:=n! SeriesCoefficient[E^((k-1) x)(k Cosh[Sqrt[k]x]+Sqrt[k]Sinh[Sqrt[k]*x])/k,{x,0,n}]; Table[A[n-k+1,k],{n,11},{k,n}]//Flatten (* or *)
A[n_,k_]:=(Sqrt[k]((k+Sqrt[k]-1)^n+(k-Sqrt[k]-1)^n)+(k+Sqrt[k]-1)^n-(k-Sqrt[k]-1)^n)/(2Sqrt[k]); Simplify[Table[A[n-k+1,k],{n,11},{k,n}]]//Flatten

A(n, k) = (sqrt(k)*(b(k)^n + c(k)^n) + b(k)^n - c(k)^n)/(2*sqrt(k)), with b(k) = k + sqrt(k) - 1 and c(k) = k - sqrt(k) - 1.
A(n, 2) = A000129(n+1).
A(2, n) = A028387(n-1).

cp's OEIS Frontend

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

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A082761 Trinomial transform of the Fibonacci numbers (A000045).

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A366858 Array read by ascending antidiagonals: A(n, k) = n! * [x^n] exp((k-1)*x)*(k*cosh(sqrt(k)*x) + sqrt(k)*sinh(sqrt(k)*x))/k, with 1 <= k <= n.

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Formula

A366858 Array read by ascending antidiagonals: A(n, k) = n! * [x^n] exp((k-1)x)(kcosh(sqrt(k)x) + sqrt(k)sinh(sqrt(k)x))/k, with 1 <= k <= n.