A074872 -id:A074872 - 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

A033887 a(n) = Fibonacci(3*n + 1).

Original entry on oeis.org

1, 3, 13, 55, 233, 987, 4181, 17711, 75025, 317811, 1346269, 5702887, 24157817, 102334155, 433494437, 1836311903, 7778742049, 32951280099, 139583862445, 591286729879, 2504730781961, 10610209857723, 44945570212853, 190392490709135, 806515533049393, 3416454622906707

Offset: 0

N. J. A. Sloane

Binomial transform of A063727, and second binomial transform of (1,1,5,5,25,25,...), which is A074872 with offset 0. - Paul Barry, Jul 16 2003
Equals INVERT transform of A104934: (1, 2, 8, 28, 100, 356, ...) and INVERTi transform of A005054: (1, 4, 20, 100, 500, ...). - Gary W. Adamson, Jul 22 2010
a(n) is the number of compositions of n when there are 3 types of 1 and 4 types of other natural numbers. - Milan Janjic, Aug 13 2010
F(3*n+1) = 3^n*a(n;2/3), where a(n;d), n = 0, 1, ..., d, denote the delta-Fibonacci numbers defined in comments to A000045 (see also the papers by Witula et al.). - Roman Witula, Jul 12 2012
We note that the remark above by Paul Barry can be easily obtained from the following scaling identity for delta-Fibonacci numbers y^n a(n;x/y) = Sum_{k=0..n} binomial(n,k) (y-1)^(n-k) a(k;x) and the fact that a(n;2)=5^floor(n/2). Indeed, for x=y=2 we get 2^n a(n;1) = Sum_{k=0..n} binomial(n,k) a(k;2) and, by A000045: Sum_{k=0..n} binomial(n,k) 2^k a(k;1) = Sum_{k=0..n} binomial(n,k) F(k+1) 2^k = 3^n a(n;2/3) = F(3n+1). - Roman Witula, Jul 12 2012
Except for the first term, this sequence can be generated by Corollary 1 (iv) of Azarian's paper in the references for this sequence. - Mohammad K. Azarian, Jul 02 2015
Number of 1’s in the substitution system {0 -> 110, 1 -> 11100} at step n from initial string "1" (1 -> 11100 -> 111001110011100110110 -> ...). - Ilya Gutkovskiy, Apr 10 2017
The o.g.f. of {F(m*n + 1)}A000045%20and%20L%20=%20A000032.%20-%20_Wolfdieter%20Lang">{n>=0}, for m = 1, 2, ..., is G(m,x) = (1 - F(m-1)*x) / (1 - L(m)*x + (-1)^m*x^2), with F = A000045 and L = A000032. - _Wolfdieter Lang, Feb 06 2023

			a(5) = Fibonacci(3*5 + 1) = Fibonacci(16) = 987. - _Indranil Ghosh_, Feb 04 2017

Indranil Ghosh, Table of n, a(n) for n = 0..1592
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, 7(38) (2012), 1871-1876.
Paul Barry and A. Hennessy, The Euler-Seidel Matrix, Hankel Matrices and Moment Sequences, J. Int. Seq. 13 (2010), #10.8.2, Example 13.
Gary Detlefs and Wolfdieter Lang, Improved Formula for the Multi-Section of the Linear Three-Term Recurrence Sequence, arXiv:2304.12937 [math.CO], 2023.
I. M. Gessel and Ji Li, Compositions and Fibonacci identities, J. Int. Seq. 16 (2013), #13.4.5.
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.
Roman Witula, Binomials transformation formulae of scaled Lucas numbers, Demonstratio Mathematica, 46(1) (2013), 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 (4,1).

Cf. A000032, A000045, A104934, A005054, A063727 (inverse binomial transform), A082761 (binomial transform), A001076, A001077.

Cf. A016777, A049310, A049651, A074872, A122070, A167808, A185384.

[Fibonacci(3*n+1): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

Fibonacci[Range[1,5!,3]] (* Vladimir Joseph Stephan Orlovsky, May 18 2010 *)

a(n)=fibonacci(3*n+1) \\ Charles R Greathouse IV, Feb 03 2014

Vec((1-x)/(1-4*x-x^2) + O(x^100)) \\ Altug Alkan, Dec 10 2015

a(n) = A001076(n) + A001077(n) = A001076(n+1) - A001076(n).
a(n) = 2*A049651(n) + 1.
a(n) = 4*a(n-1) + a(n-2) for n>1, a(0)=1, a(1)=3;
G.f.: (1 - x)/(1 - 4*x - x^2).
a(n) = ((1 + sqrt(5))*(2 + sqrt(5))^n - (1 - sqrt(5))*(2 - sqrt(5))^n)/(2*sqrt(5)).
a(n) = Sum_{k=0..n} Sum_{j=0..n-k} C(n,j)*C(n-j,k)*F(n-j+1). - Paul Barry, May 19 2006
First differences of A001076. - Al Hakanson (hawkuu(AT)gmail.com), May 02 2009
a(n) = A167808(3*n+1). - Reinhard Zumkeller, Nov 12 2009
a(n) = Sum_{k=0..n} C(n,k)*F(n+k+1). - Paul Barry, Apr 19 2010
Let p[1]=3, p[i]=4, (i>1), and A be a Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1] (i <= j), A[i,j]=-1 (i = j+1), and A[i,j] = 0 otherwise. Then, for n >= 1, a(n) = det A. - Milan Janjic, Apr 29 2010
a(n) = Sum_{i=0..n} C(n,n-i)*A063727(i). - Seiichi Kirikami, Mar 06 2012
a(n) = Sum_{k=0..n} A122070(n,k) = Sum_{k=0..n} A185384(n,k). - Philippe Deléham, Mar 13 2012
a(n) = A000045(A016777(n)). - Michel Marcus, Dec 10 2015
a(n) = F(2*n)*L(n+1) + F(n-1)*(-1)^n for n > 0. - J. M. Bergot, Feb 09 2016
a(n) = Sum_{k=0..n} binomial(n,k)*5^floor(k/2)*2^(n-k). - Tony Foster III, Sep 03 2017
2*a(n) = Fibonacci(3*n) + Lucas(3*n). - Bruno Berselli, Oct 13 2017
a(n)^2 is the denominator of continued fraction [4,...,4, 2, 4,...,4], which has n 4's before, and n 4's after, the middle 2. - Greg Dresden and Hexuan Wang, Aug 30 2021
a(n) = i^n*(S(n, -4*i) + i*S(n-1, -4*i)), with i = sqrt(-1), and the Chebyshev S-polynomials (see A049310) with S(n, -1) = 0. From the simplified trisection formula. See the first entry above with A001076. - Gary Detlefs and Wolfdieter Lang, Mar 06 2023
E.g.f.: exp(2*x)*(5*cosh(sqrt(5)*x) + sqrt(5)*sinh(sqrt(5)*x))/5. - Stefano Spezia, May 24 2024

A152198 Triangle read by rows, A007318 rows repeated.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 5, 10, 10, 5, 1, 1, 6, 15, 20, 15, 6, 1, 1, 6, 15, 20, 15, 6, 1, 1, 7, 21, 35, 35, 21, 7, 1, 1, 7, 21, 35, 35, 21, 7, 1, 1, 8, 28, 56, 70, 56, 28, 8, 1, 1, 8, 28, 56, 70, 56, 28, 8, 1

Offset: 0

Gary W. Adamson, Nov 28 2008

Eigensequence of the triangle = A051163: (1, 2, 5, 12, 30, 76,...)
Another version of A152815. - Philippe Deléham, Dec 13 2008
Row sums : A016116(n); Diagonal sums: A000931(n+5). - Philippe Deléham, Dec 13 2008
Triangle, with zeros omitted, given by (1, 0, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Jan 16 2012
Sums along rising diagonals are A134816. - John Molokach, Jul 09 2013

			The triangle starts
1;
1;
1, 1;
1, 1;
1, 2, 1;
1, 2, 1;
1, 3, 3, 1;
1, 3, 3, 1;
1, 4, 6, 4, 1;
1, 4, 6, 4, 1;
1, 5, 10, 10, 5, 1;
1, 5, 10, 10, 5, 1;
...
Triangle (1,0,-1,0,0,...) DELTA (0,1,-1,0,0,...) begins:
1
1, 0
1, 1, 0
1, 1, 0, 0
1, 2, 1, 0, 0
1, 2, 1, 0, 0, 0
1, 3, 3, 1, 0, 0, 0
1, 3, 3, 1, 0, 0, 0, 0
1, 4, 6, 4, 1, 0, 0, 0, 0
1, 4, 6, 4, 1, 0, 0, 0, 0, 0
1, 5, 10, 10, 5, 1, 0, 0, 0, 0, 0...

Cf. A007318, A133156, A051163.

t[n_, k_] := Binomial[ Floor[n/2], k]; Table[t[n, k], {n, 0, 17}, {k, 0, Floor[n/2]}] // Flatten (* Jean-François Alcover, Sep 13 2012 *)

Triangle read by rows, Pascal's triangle rows repeated.
Equals inverse binomial transform of A133156 unsigned.
G.f. : (1+x)/(1-(1+y)*x^2). - Philippe Deléham, Jan 16 2012
Sum_{k, 0<=k<=n} T(n,k)*x^k = A057077(n), A019590(n+1), A000012(n), A016116(n), A108411(n), A074872(n+1) for x = -2, -1, 0, 1, 2, 4 respectively. - Philippe Deléham, Jan 16 2012
T(n,k) = A065941(n-k, n-2*k) = abs(A108299(n-k, n-2*k)). - Johannes W. Meijer, Sep 05 2013

More terms from Philippe Deléham, Dec 14 2008

A152815 Triangle T(n,k), read by rows given by [1,0,-1,0,0,0,0,0,0,...] DELTA [0,1,-1,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 2, 1, 0, 0, 1, 2, 1, 0, 0, 0, 1, 3, 3, 1, 0, 0, 0, 1, 3, 3, 1, 0, 0, 0, 0, 1, 4, 6, 4, 1, 0, 0, 0, 0, 1, 4, 6, 4, 1, 0, 0, 0, 0, 0, 1, 5, 10, 10, 5, 1, 0, 0, 0, 0, 0, 1, 5, 10, 10, 5, 1, 0, 0, 0, 0, 0, 0, 1, 6, 15, 20, 15, 6, 1, 0, 0, 0, 0, 0, 0, 1, 6, 15, 20, 15, 6, 1, 0, 0, 0

Offset: 0

Philippe Deléham, Dec 13 2008

Triangle read by rows, Pascal's triangle (A007318) rows repeated.

Riordan array (1/(1-x), x^2/(1-x^2)). - Philippe Deléham, Feb 27 2012

			Triangle begins:
  1;
  1, 0;
  1, 1, 0;
  1, 1, 0, 0;
  1, 2, 1, 0, 0;
  1, 2, 1, 0, 0, 0;
  1, 3, 3, 1, 0, 0, 0;
  1, 3, 3, 1, 0, 0, 0, 0;
  1, 4, 6, 4, 1, 0, 0, 0, 0; ...

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened

Cf. A007318, A064861, A152198 (another version), A000931 (diagonal sums), A016116 (row sums).

a152815 n k = a152815_tabl !! n !! k
a152815_row n = a152815_tabl !! n
a152815_tabl = [1] : [1,0] : t [1,0] where
   t ys = zs : zs' : t zs' where
     zs' = zs ++ [0]; zs = zipWith (+) ([0] ++ ys) (ys ++ [0])
-- Reinhard Zumkeller, Feb 28 2012

m = 13;
(* DELTA is defined in A084938 *)
DELTA[Join[{1, 0, -1}, Table[0, {m}]], Join[{0, 1, -1}, Table[0, {m}]], m] // Flatten (* Jean-François Alcover, Feb 19 2020 *)
T[n_, k_] := If[n<0, 0, Binomial[Floor[n/2], k]]; (* Michael Somos, Oct 01 2022 *)

{T(n, k) = if(n<0, 0, binomial(n\2, k))}; /* Michael Somos, Oct 01 2022 */

T(n,k) = T(n-1,k) + ((1+(-1)^n)/2)*T(n-1,k-1).
G.f.: (1+x)/(1-(1+y)*x^2).
Sum_{k=0..n} T(n,k)*x^k = A000012(n), A016116(n), A108411(n), A213173(n), A074872(n+1) for x = 0,1,2,3,4 respectively. - Philippe Deléham, Nov 26 2011, Apr 22 2013

Example corrected by Philippe Deléham, Dec 13 2008

A163073 a(n) = ((5+sqrt(5))(4+sqrt(5))^n + (5-sqrt(5))(4-sqrt(5))^n)/10.

Original entry on oeis.org

1, 5, 29, 177, 1097, 6829, 42565, 265401, 1654993, 10320533, 64359341, 401348865, 2502838169, 15607867837, 97331722837, 606967236489, 3785088940705, 23604071924261, 147196597046333, 917927985203793, 5724261314120681, 35696882675723725, 222608186950462309, 1388199786170737497

Offset: 0

Al Hakanson (hawkuu(AT)gmail.com), Jul 20 2009

Binomial transform of A082761. Fourth binomial transform of A074872.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-11).

Cf. A082761, A074872 (1,1,5,5,25,25,...).

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-5); S:=[ ((5+r)*(4+r)^n+(5-r)*(4-r)^n)/10: n in [0..20] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Jul 24 2009

LinearRecurrence[{8,-11},{1,5},30] (* Harvey P. Dale, Dec 11 2017 *)

x='x+O('x^30); Vec((1-3*x)/(1-8*x+11*x^2)) \\ G. C. Greubel, Jan 08 2018

a(n) = 8*a(n-1)-11*a(n-2) for n > 1; a(0) = 1, a(1) = 5.
G.f.: (1-3*x)/(1-8*x+11*x^2).
E.g.f.: exp(4*x)*(5*cosh(sqrt(5)*x) + sqrt(5)*sinh(sqrt(5)*x))/5. - Stefano Spezia, Oct 25 2023

Edited and extended beyond a(5) by Klaus Brockhaus, Jul 24 2009

A163310 a(n) = 20a(n-1) - 95a(n-2) for n > 1; a(0) = 1, a(1) = 11.

Original entry on oeis.org

1, 11, 125, 1455, 17225, 206275, 2489125, 30186375, 367260625, 4477506875, 54660378125, 667844409375, 8164152265625, 99837826421875, 1221162063203125, 14938647753984375, 182762559075390625, 2236079644879296875

Offset: 0

Klaus Brockhaus, Jul 24 2009

Binomial transform of A163309. Tenth binomial transform of A074872.

G. C. Greubel, Table of n, a(n) for n = 0..915
Index entries for linear recurrences with constant coefficients, signature (20,-95).

Cf. A163309, A074872.

[ n le 2 select 10*n-9 else 20*Self(n-1)-95*Self(n-2): n in [1..18] ];

LinearRecurrence[{20,-95}, {1,11}, 50] (* G. C. Greubel, Dec 18 2016 *)

Vec((1-9*x)/(1-20*x+95*x^2) + O(x^50)) \\ G. C. Greubel, Dec 18 2016

a(n) = ((5+sqrt(5))*(10+sqrt(5))^n + (5-sqrt(5))*(10-sqrt(5))^n)/10.
G.f.: (1-9*x)/(1-20*x+95*x^2).
E.g.f.: (1/5)*exp(10*x)*(5*cosh(sqrt(5)*x) + sqrt(5)*sinh(sqrt(5)*x)). - G. C. Greubel, Dec 18 2016

A162962 a(n) = 5*a(n-2) for n > 2; a(1) = 1, a(2) = 5.

Original entry on oeis.org

1, 5, 5, 25, 25, 125, 125, 625, 625, 3125, 3125, 15625, 15625, 78125, 78125, 390625, 390625, 1953125, 1953125, 9765625, 9765625, 48828125, 48828125, 244140625, 244140625, 1220703125, 1220703125, 6103515625, 6103515625, 30517578125

Offset: 1

Klaus Brockhaus, Jul 19 2009

Apparently a(n) = A074872(n+1), a(n) = A056451(n-1) for n > 1.

Binomial transform is A084057 without initial 1, second binomial transform is A048876, third binomial transform is A082762, fourth binomial transform is A162769, fifth binomial transform is A093145 without initial 0.

Index entries for linear recurrences with constant coefficients, signature (0,5).

Cf. A000351 (powers of 5), A074872 (powers of 5 repeated), A056451 (5^floor((n+1)/2)), A084057, A048876, A082762, A162769, A093145.

[ n le 2 select 4*n-3 else 5*Self(n-2): n in [1..30] ];

LinearRecurrence[{0,5},{1,5},30] (* Harvey P. Dale, Mar 18 2023 *)

a(n) = 5^((1/4)*(2*n-1+(-1)^n)).

G.f.: x*(1+5*x)/(1-5*x^2).

A242763 a(n) = 1 for n <= 7; a(n) = a(n-5) + a(n-7) for n>7.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 4, 4, 4, 5, 5, 7, 7, 8, 9, 9, 12, 12, 15, 16, 17, 21, 21, 27, 28, 32, 37, 38, 48, 49, 59, 65, 70, 85, 87, 107, 114, 129, 150, 157, 192, 201, 236, 264, 286, 342, 358, 428, 465, 522, 606, 644, 770, 823, 950, 1071, 1166, 1376

Offset: 1

Vicente Jesús Maniega Granado, Oct 03 2016

Generalized Fibonacci growth sequence using i = 2 as maturity period, j = 5 as conception period, and k = 2 as growth factor.

Maturity period is the number of periods that a Fibonacci tree node needs for being able to start developing branches. Conception period is the number of periods in a Fibonacci tree node needed to develop new branches since its maturity. Growth factor is the number of additional branches developed by a Fibonacci tree node, plus 1, and equals the base of the exponential series related to the given tree if maturity factor would be zero. Standard Fibonacci would use 1 as maturity period, 1 as conception period, and 2 as growth factor as the series becomes equal to 2^n with a maturity period of 0. Related to Lucas sequences.

			For n = 13 the a(13) = a(8) + a(6) = 2 + 1 = 3.

Colin Barker, Table of n, a(n) for n = 1..1000
Julia Collins, Fibonacci Tree
Fractal Foundation, Fibonacci Fractals
D. H. Lehmer, An extended theory of Lucas' functions, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,1,0,1).

Cf. A000079 (i = 0, j = 1, k = 2), A000244 (i = 0, j = 1, k = 3), A000302 (i = 0, j = 1, k = 4), A000351 (i = 0, j = 1, k = 5), A000400 (i = 0, j = 1, k = 6), A000420 (i = 0, j = 1, k = 7), A001018 (i = 0, j = 1, k = 8), A001019 (i = 0, j = 1, k = 9), A011557 (i = 0, j = 1, k = 10), A001020 (i = 0, j = 1, k = 11), A001021 (i = 0, j = 1, k = 12), A016116 (i = 0, j = 2, k = 2), A108411 (i = 0, j = 2, k = 3), A213173 (i = 0, j = 2, k = 4), A074872 (i = 0, j = 2, k = 5), A173862 (i = 0, j = 3, k = 2), A127975 (i = 0, j = 3, k = 3), A200675 (i = 0, j = 4, k = 2), A111575 (i = 0, j = 4, k = 3), A000045 (i = 1, j = 1, k = 2), A001045 (i = 1, j = 1, k = 3), A006130 (i = 1, j = 1, k = 4), A006131 (i = 1, j = 1, k = 5), A015440 (i = 1, j = 1, k = 6), A015441 (i = 1, j = 1, k = 7), A015442 (i = 1, j = 1, k = 8), A015443 (i = 1, j = 1, k = 9), A015445 (i = 1, j = 1, k = 10), A015446 (i = 1, j = 1, k = 11), A015447 (i = 1, j = 1, k = 12), A000931 (i = 1, j = 2, k = 2), A159284 (i = 1, j = 2, k = 3), A238389 (i = 1, j = 2, k = 4), A097041 (i = 1, j = 2, k = 10), A079398 (i = 1, j = 3, k = 2), A103372 (i = 1, j = 4, k = 2), A103373 (i = 1, j = 5, k = 2), A103374 (i = 1, j = 6, k = 2), A000930 (i = 2, j = 1, k = 2), A077949 (i = 2, j = 1, k = 3), A084386 (i = 2, j = 1, k = 4), A089977 (i = 2, j = 1, k = 5), A178205 (i = 2, j = 1, k = 11), A103609 (i = 2, j = 2, k = 2), A077953 (i = 2, j = 2, k = 3), A226503 (i = 2, j = 3, k = 2), A122521 (i = 2, j = 6, k = 2), A003269 (i = 3, j = 1, k = 2), A052942 (i = 3, j = 1, k = 3), A005686 (i = 3, j = 2, k = 2), A237714 (i = 3, j = 2, k = 3), A238391 (i = 3, j = 2, k = 4), A247049 (i = 3, j = 3, k = 2), A077886 (i = 3, j = 3, k = 3), A003520 (i = 4, j = 1, k = 2), A108104 (i = 4, j = 2, k = 2), A005708 (i = 5, j = 1, k = 2), A237716 (i = 5, j = 2, k = 3), A005709 (i = 6, j = 1, k = 2), A122522 (i = 6, j = 2, k = 2), A005710 (i = 7, j = 1, k = 2), A237718 (i = 7, j = 2, k = 3), A017903 (i = 8, j = 1, k = 2).

[n le 7 select 1 else Self(n-5)+Self(n-7): n in [1..70]]; // Vincenzo Librandi, Nov 30 2016

LinearRecurrence[{0, 0, 0, 0, 1, 0, 1}, {1, 1, 1, 1, 1, 1, 1}, 70] (*  or *)
CoefficientList[ Series[(1+x+x^2+x^3+x^4)/(1-x^5-x^7), {x, 0, 70}], x] (* Robert G. Wilson v, Nov 25 2016 *)
nxt[{a_,b_,c_,d_,e_,f_,g_}]:={b,c,d,e,f,g,a+c}; NestList[nxt,{1,1,1,1,1,1,1},70][[;;,1]] (* Harvey P. Dale, Oct 22 2024 *)

Vec(x*(1+x+x^2+x^3+x^4)/((1-x+x^2)*(1+x-x^3-x^4-x^5)) + O(x^100)) \\ Colin Barker, Oct 27 2016

@CachedFunction # a = A242763
def a(n): return 1 if n<8 else a(n-5) +a(n-7)
[a(n) for n in range(1,76)] # G. C. Greubel, Oct 23 2024

Generic a(n) = 1 for n <= i+j; a(n) = a(n-j) + (k-1)*a(n-(i+j)) for n>i+j where i = maturity period, j = conception period, k = growth factor.
G.f.: x*(1+x+x^2+x^3+x^4) / ((1-x+x^2)*(1+x-x^3-x^4-x^5)). - Colin Barker, Oct 09 2016
Generic g.f.: x*(Sum_{l=0..j-1} x^l) / (1-x^j-(k-1)*x^(i+j)), with i > 0, j > 0 and k > 1.

A123220 a(n)=the (1,1)-term of M^(n-1), where M=matrix(5,5, [3,-1,-1,-1,-1; 1,3,-1,-1,-1; 1,1,3,-1,-1; 1,1,1,3,-1; 1,1,1,1,3]).

Original entry on oeis.org

1, 3, 5, -9, -99, -297, 389, 8655, 46573, 122823, -120491, -3003393, -15885315, -40246281, 50400229, 1040606127, 5296630541, 12512952615, -22872751243, -368600380833, -1789336379619, -3926384911017, 9502037022725, 129579396089871, 602116408170541, 1219711972804743

Offset: 1

Roger L. Bagula, Oct 05 2006

sign

Cf. A000952, A074872, A085340.

a[1]:=1:a[2]:=3:a[3]:=5:a[4]:=-9:a[5]:=-99: for n from 6 to 26 do a[n]:=15*a[n-1]-100*a[n-2]+360*a[n-3]-680*a[n-4]+528*a[n-5] od: seq(a[n],n=1..26); with(linalg): M[1]:=matrix(5,5,[3, -1, -1, -1, -1, 1, 3, -1, -1, -1, 1, 1, 3, -1, -1, 1, 1, 1, 3, -1, 1, 1, 1, 1, 3]): for n from 2 to 25 do M[n]:=multiply(M[1],M[n-1]) od: 1,seq(M[n][1,1],n=1..25);

M = {{3, -1, -1, -1, -1}, {1, 3, -1, -1, -1}, {1, 1, 3, -1, -1}, {1, 1, 1, 3, -1}, {1, 1, 1, 1, 3}}; w[1] = {1, 0, 0, 0, 0}; w[n_] := w[n] = M.w[n - 1]; a = Table[w[n][[1]], {n, 1, 30}]

a(1) = 1; a(2) = 3; a(3) = 5; a(4) = -9; a(5) = -99; a(n) = 15a(n-1)-100a(n-2)+360a(n-3)-680a(n-4)+528a(n-5) for n>= 6. The minimal polynomial of M is x^5-15x^4+100x^3-360x^2+680x-528, the coefficients of which yield the coefficients of the recurrence relation.

O.g.f.: -x*(1-12*x+60*x^2-144*x^3+136*x^4)/((3*x-1)*(176*x^4-168*x^3+64*x^2-12*x+1)). - R. J. Mathar, Dec 05 2007

Edited by N. J. A. Sloane, Oct 15 2006

A123222 Expansion of -x * (x-1) * (3x^2-1) / (9x^4-8x^3+4x-1).

Original entry on oeis.org

1, 3, 9, 31, 109, 391, 1397, 4995, 17833, 63675, 227313, 811543, 2897269, 10343647, 36928061, 131837979, 470678161, 1680380979, 5999172633, 21417807055, 76464283837, 272987183095, 974598829637, 3479441311347, 12422046335161

Offset: 1

Roger L. Bagula, Oct 05 2006

Index entries for linear recurrences with constant coefficients, signature (4,0,-8,9).

Cf. A000952, A074872, A085340, A003432.

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(-x*(x-1)*(3*x^2-1)/(9*x^4-8*x^3+4*x-1))); // G. C. Greubel, Oct 12 2018

seq(coeff(series(-x*(x-1)*(3*x^2-1)/(9*x^4-8*x^3+4*x-1),x,n+1), x, n), n = 1 .. 25); # Muniru A Asiru, Oct 13 2018

LinearRecurrence[{4,0,-8,9},{1,3,9,31},30] (* Harvey P. Dale, Jul 26 2018 *)

x='x+O('x^30); Vec(-x*(x-1)*(3*x^2-1)/(9*x^4-8*x^3+4*x-1)) \\ G. C. Greubel, Oct 12 2018

From Colin Barker, Oct 19 2012: (Start)
a(n) = 4*a(n-1) -8*a(n-3) +9*a(n-4).
G.f.: -x*(x-1)*(3*x^2-1)/(9*x^4-8*x^3+4*x-1). (End)

Sequence edited by Joerg Arndt and Colin Barker, Oct 19 2012

cp's OEIS Frontend

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

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A033887 a(n) = Fibonacci(3*n + 1).

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A152198 Triangle read by rows, A007318 rows repeated.

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A152815 Triangle T(n,k), read by rows given by [1,0,-1,0,0,0,0,0,0,...] DELTA [0,1,-1,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

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A163073 a(n) = ((5+sqrt(5))*(4+sqrt(5))^n + (5-sqrt(5))*(4-sqrt(5))^n)/10.

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A163310 a(n) = 20*a(n-1) - 95*a(n-2) for n > 1; a(0) = 1, a(1) = 11.

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A163073 a(n) = ((5+sqrt(5))(4+sqrt(5))^n + (5-sqrt(5))(4-sqrt(5))^n)/10.

A163310 a(n) = 20a(n-1) - 95a(n-2) for n > 1; a(0) = 1, a(1) = 11.

A123222 Expansion of -x * (x-1) * (3x^2-1) / (9x^4-8x^3+4x-1).