A081568 -id:A081568 - 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

A081567 Second binomial transform of F(n+1).

Original entry on oeis.org

1, 3, 10, 35, 125, 450, 1625, 5875, 21250, 76875, 278125, 1006250, 3640625, 13171875, 47656250, 172421875, 623828125, 2257031250, 8166015625, 29544921875, 106894531250, 386748046875, 1399267578125, 5062597656250, 18316650390625, 66270263671875, 239768066406250

Offset: 0

Paul Barry, Mar 22 2003

Binomial transform of F(2*n-1), index shifted by 1, where F is A000045. - corrected by Richard R. Forberg, Aug 12 2013
Case k=2 of family of recurrences a(n) = (2k+1)*a(n-1) - A028387(k-1)*a(n-2), a(0)=1, a(1)=k+1.
Number of (s(0), s(1), ..., s(2n+1)) such that 0 < s(i) < 10 and |s(i) - s(i-1)| = 1 for i = 1, 2, ..., 2*n+1, s(0) = 3, s(2*n+1) = 4.
a(n+1) gives the number of periodic multiplex juggling sequences of length n with base state <2>. - Steve Butler, Jan 21 2008
a(n) is also the number of idempotent order-preserving partial transformations (of an n-element chain) of waist n (waist(alpha) = max(Im(alpha))). - Abdullahi Umar, Sep 14 2008
Counts all paths of length (2*n+1), n>=0, starting at the initial node on the path graph P_9, see the Maple program. - Johannes W. Meijer, May 29 2010
Given the 3 X 3 matrix M = [1,1,1; 1,1,0; 1,1,3], a(n) = term (1,1) in M^(n+1). - Gary W. Adamson, Aug 06 2010
Number of nonisomorphic graded posets with 0 and 1 of rank n+2, with exactly 2 elements of each rank level between 0 and 1. Also the number of nonisomorphic graded posets with 0 of rank n+1, with exactly 2 elements of each rank level above 0. (This is by Stanley's definition of graded, that all maximal chains have the same length.) - David Nacin, Feb 26 2012
a(n) = 3^n a(n;1/3) = Sum_{k=0..n} C(n,k) * F(k-1) * (-1)^k * 3^(n-k), which also implies the Deleham formula given below and where a(n;d), n=0,1,...,d, denote the delta-Fibonacci numbers defined in comments to A000045 (see also the papers of Witula et al.). - Roman Witula, Jul 12 2012
The limiting ratio a(n)/a(n-1) is 1 + phi^2. - Bob Selcoe, Mar 17 2014
a(n) counts closed walks on K_2 containing 3 loops on the index vertex and 2 loops on the other. Equivalently the (1,1) entry of A^n where the adjacency matrix of digraph is A=(3,1; 1,2). - David Neil McGrath, Nov 18 2014

			a(4)=125: 35*(3 + (35 mod 10 - 10 mod 3)/(10-3)) = 35*(3 + 4/7) = 125. - _Bob Selcoe_, Mar 17 2014

R. P. Stanley, Enumerative Combinatorics, Vol. 1, Cambridge University Press, Cambridge, 1997, pages 96-100.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Santiago Alzate, Oscar Correa, and Rigoberto Flórez, Fibonacci identities from Jordan Identities, arXiv:2009.02639 [math.NT], 2020.
Carolina Benedetti, Christopher R. H. Hanusa, Pamela E. Harris, Alejandro H. Morales, and Anthony Simpson, Kostant's partition function and magic multiplex juggling sequences, arXiv:2001.03219 [math.CO], 2020. See Table 1 p. 12.
S. Butler and R. Graham, Enumerating (multiplex) juggling sequences, arXiv:0801.2597 [math.CO], 2008.
P. E. Harris, E. Insko, and L. K. Williams, The adjoint representation of a Lie algebra and the support of Kostant's weight multiplicity formula, arXiv preprint arXiv:1401.0055 [math.RT], 2013.
Edyta Hetmaniok, Bożena Piątek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Mathematics, 15(1) (2017), 477-485.
A. Laradji and A. Umar, A. Combinatorial results for semigroups of order-preserving partial transformations, Journal of Algebra 278, (2004), 342-359.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-decreasing partial transformations, J. Integer Seq. 7 (2004), #04.3.8.
Mircea Merca, A Note on Cosine Power Sums J. Integer Sequences, Vol. 15 (2012), Article 12.5.3.
D. Nacin, The Minimal Non-Koszul A(Gamma), arXiv preprint arXiv:1204.1534 [math.QA], 2012. - From _N. J. A. Sloane_, Oct 05 2012
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 (5,-5).

a(n) = 5*A052936(n-1), n > 1.
Row sums of A114164.
Cf. A000045, A007051 (INVERTi transform), A007598, A028387, A030191, A039717, A049310, A081568 (binomial transform), A086351 (INVERT transform), A090041, A093129, A094441, A111776, A147748, A178381, A189315.

I:=[1, 3]; [n le 2 select I[n] else 5*Self(n-1)-5*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 27 2012

with(GraphTheory):G:=PathGraph(9): A:= AdjacencyMatrix(G): nmax:=23; n2:=nmax*2+2: for n from 0 to n2 do B(n):=A^n; a(n):=add(B(n)[1,k],k=1..9); od: seq(a(2*n+1),n=0..nmax); # Johannes W. Meijer, May 29 2010

Table[MatrixPower[{{2,1},{1,3}},n][[2]][[2]],{n,0,44}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
LinearRecurrence[{5,-5},{1,3},30] (* Vincenzo Librandi, Feb 27 2012 *)

Vec((1-2*x)/(1-5*x+5*x^2)+O(x^99)) \\ Charles R Greathouse IV, Mar 18 2014

def a(n, adict={0:1, 1:3}):
    if n in adict:
        return adict[n]
    adict[n]=5*a(n-1) - 5*a(n-2)
    return adict[n] # David Nacin, Mar 04 2012

a(n) = 5*a(n-1) - 5*a(n-2) for n >= 2, with a(0) = 1 and a(1) = 3.
a(n) = (1/2 - sqrt(5)/10) * (5/2 - sqrt(5)/2)^n + (sqrt(5)/10 + 1/2) * (sqrt(5)/2 + 5/2)^n.
G.f.: (1 - 2*x)/(1 - 5*x + 5*x^2).
a(n-1) = Sum_{k=1..n} binomial(n, k)*F(k)^2. - Benoit Cloitre, Oct 26 2003
a(n) = A090041(n)/2^n. - Paul Barry, Mar 23 2004
The sequence 0, 1, 3, 10, ... with a(n) = (5/2 - sqrt(5)/2)^n/5 + (5/2 + sqrt(5)/2)^n/5 - 2(0)^n/5 is the binomial transform of F(n)^2 (A007598). - Paul Barry, Apr 27 2004
From Paul Barry, Nov 15 2005: (Start)
a(n) = Sum_{k=0..n} Sum_{j=0..n} binomial(n, j)*binomial(j+k, 2k);
a(n) = Sum_{k=0..n} Sum_{j=0..n} binomial(n, k+j)*binomial(k, k-j)2^(n-k-j);
a(n) = Sum_{k=0..n} Sum_{j=0..n-k} binomial(n+k-j, n-k-j)*binomial(k, j)(-1)^j*2^(n-k-j). (End)
a(n) = A111776(n, n). - Abdullahi Umar, Sep 14 2008
a(n) = Sum_{k=0..n} A094441(n,k)*2^k. - Philippe Deléham, Dec 14 2009
a(n+1) = Sum_{k=-floor(n/5)..floor(n/5)} ((-1)^k*binomial(2*n, n+5*k)/2). -Mircea Merca, Jan 28 2012
a(n) = A030191(n) - 2*A030191(n-1). - R. J. Mathar, Jul 19 2012
G.f.: Q(0,u)/x - 1/x, where u=x/(1-2*x), Q(k,u) = 1 + u^2 + (k+2)*u - u*(k+1 + u)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 07 2013
For n>=3: a(n) = a(n-1)*(3+(a(n-1) mod a(n-2) - a(n-2) mod a(n-3))/(a(n-2) - a(n-3))). - Bob Selcoe, Mar 17 2014
a(n) = sqrt(5)^(n-1)*(3*S(n-1, sqrt(5)) - sqrt(5)*S(n-2, sqrt(5))) with Chebyshev's S-polynomials (see A049310), where S(-1, x) = 0 and S(-2, x) = -1. This is the (1,1) entry of A^n with the matrix A=(3,1;1,2). See the comment by David Neil McGrath, Nov 18 2014. - Wolfdieter Lang, Dec 04 2014
Conjecture: a(n) = 2*a(n-1) + A039717(n). - Benito van der Zander, Nov 20 2015
a(n) = A189315(n+1) / 10. - Tom Copeland, Dec 08 2015
a(n) = A093129(n) + A030191(n-1). - Gary W. Adamson, Apr 24 2023
E.g.f.: exp(5*x/2)*(5*cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, Jun 03 2024

A094441 Triangular array T(n,k) = Fibonacci(n+1-k)*C(n,k), 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 3, 6, 3, 1, 5, 12, 12, 4, 1, 8, 25, 30, 20, 5, 1, 13, 48, 75, 60, 30, 6, 1, 21, 91, 168, 175, 105, 42, 7, 1, 34, 168, 364, 448, 350, 168, 56, 8, 1, 55, 306, 756, 1092, 1008, 630, 252, 72, 9, 1, 89, 550, 1530, 2520, 2730, 2016, 1050, 360, 90, 10, 1

Offset: 0

Clark Kimberling, May 03 2004

Triangle of coefficients of polynomials u(n,x) jointly generated with A209415; see the Formula section.
Column 1:  Fibonacci numbers:  F(n)=A000045(n)
Column 2:  n*F(n)
Row sums:  odd-indexed Fibonacci numbers
Alternating row sums: signed Fibonacci numbers
Coefficient of x^n in u(n,x):  1
Coefficient of x^(n-1) in u(n,x):  n
Coefficient of x^(n-2) in u(n,x):  n(n+1)
For a discussion and guide to related arrays, see A208510.
Subtriangle of the triangle given by (0, 1, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 27 2012
Row n shows the coefficients of the numerator of the n-th derivative of (1/n!)*(x+1)/(1-x-x^2); see the Mathematica program. - Clark Kimberling, Oct 22 2019

			First five rows:
  1;
  1,  1;
  2,  2,  1;
  3,  6,  3,  1;
  5, 12, 12,  4,  1;
First three polynomials v(n,x): 1, 1 + x, 2 + 2x + x^2.
From _Philippe Deléham_, Mar 27 2012: (Start)
(0, 1, 1, -1, 0, 0, 0, ...) DELTA (1, 0, 0, 1, 0, 0, 0, ...) begins:
  1;
  0,  1;
  0,  1,  1;
  0,  2,  2,  1;
  0,  3,  6,  3,  1;
  0,  5, 12, 12,  4,  1. (End)

G. C. Greubel, Rows n = 0..100 of triangle, flattened
E. Kiliç, H. Belbachir, Generalized double binomial sums families by generating functions, 2014.

Cf. A000045, A094435, A094436, A094437, A094438, A094439, A094440, A094442, A094443, A094444.

Flat(List([0..12], n-> List([0..n], k-> Binomial(n,k)*Fibonacci(n-k+1) ))); # G. C. Greubel, Oct 30 2019

[Binomial(n,k)*Fibonacci(n-k+1): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 30 2019

with(combinat); seq(seq(fibonacci(n-k+1)*binomial(n,k), k=0..n), n=0..12); # G. C. Greubel, Oct 30 2019

(* First program *)
u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
v[n_, x_] := u[n - 1, x] + (x + 1)*v[n - 1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A094441 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A094442 *)
(* Next program outputs polynomials having coefficients T(n,k) *)
g[x_, n_] := Numerator[(-1)^(n + 1) Factor[D[(x + 1)/(1 - x - x^2), {x, n}]]]
Column[Expand[Table[g[x, n]/n!, {n, 0, 12}]]] (* Clark Kimberling, Oct 22 2019 *)
(* Second program *)
Table[Fibonacci[n-k+1]*Binomial[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 30 2019 *)

T(n,k) = binomial(n,k)*fibonacci(n-k+1);
for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Oct 30 2019

[[binomial(n,k)*fibonacci(n-k+1) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Oct 30 2019

Sum_{k=0..n} T(n,k)*x^k = A039834(n-1), A000045(n+1), A001519(n+1), A081567(n), A081568(n), A081569(n), A081570(n), A081571(n) for x = -1, 0, 1, 2, 3, 4, 5, 6 respectively. - Philippe Deléham, Dec 14 2009
From Clark Kimberling, Mar 09 2012: (Start)
A094441 shows the coefficient of the polynomials u(n,x) which are jointly generated with polynomials v(n,x) by these rules:
u(n,x) = x*u(n-1,x) + v(n-1,x),
v(n,x) = u(n-1,x) + (x+1)*v(n-1,x),
where u(1,x)=1, v(1,x)=1.
(End)
T(n,k) = T(n-1,k) + 2*T(n-1,k-1) + T(n-2,k) - T(n-2,k-1) - T(n-2,k-2), T(1,0) = T(2,0) = T(2,1) = 1 and T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Mar 27 2012
G.f. (1-x*y)/(1 - 2*x*y - x - x^2 + x^2*y + x^2*y^2). - R. J. Mathar, Aug 11 2015
From G. C. Greubel, Oct 30 2019: (Start)
T(n,k) = binomial(n,k)*Fibonacci(n-k+1).
Sum_{k=0..n} T(n,k) = Fibonacci(2*n+1).
Sum_{k=0..n} (-1)^k * T(n,k) = (-1)^n * Fibonacci(n-1). (End)

A081569 Fourth binomial transform of F(n+1).

Original entry on oeis.org

1, 5, 26, 139, 757, 4172, 23165, 129217, 722818, 4050239, 22718609, 127512940, 715962889, 4020920141, 22584986378, 126867394723, 712691811325, 4003745802188, 22492567804517, 126361939999081, 709898671705906, 3988211185370615, 22405825905923321, 125876420631268204

Offset: 0

Paul Barry, Mar 22 2003

Binomial transform of A081568.
Case k = 4 of family of recurrences a(n) = (2*k+1)*a(n-1) - A028387(k-1)*a(n-2) for n >= 2, with a(0) = 1 and a(1) = k + 1.
a(n) = 5^n * a(n;1/5) = Sum_{k=0..n} binomial(n,k) * (-1)^k * F(k-1) * 5^(n-k), which implies also Deléham's formula given below and where a(n;d), n=0,1,...,d, denote the delta-Fibonacci numbers defined in comments to A000045 (see also Witula's et al. papers). - Roman Witula, Jul 12 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Edyta Hetmaniok, Bożena Piątek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Mathematics, 15(1) (2017), 477-485.
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 (9,-19).

Cf. A000045, A028387, A081568, A081574, A094441.

a:=[1,5];; for n in [3..30] do a[n]:=9*a[n-1]-19*a[n-2]; od; a; # G. C. Greubel, Aug 12 2019

I:=[1, 5]; [n le 2 select I[n] else 9*Self(n-1)-19*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Aug 09 2013

seq(coeff(series((1-4*x)/(1-9*x+19*x^2), x, n+1), x, n), n = 0 .. 30); # G. C. Greubel, Aug 12 2019

CoefficientList[Series[(1-4x)/(1 -9x +19x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 09 2013 *)

Vec((1-4*x)/(1-9*x+19*x^2) + O(x^30)) \\ Altug Alkan, Dec 10 2015

def A081569_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P((1-4*x)/(1-9*x+19*x^2)).list()
A081569_list(30) # G. C. Greubel, Aug 12 2019

a(n) = 9*a(n-1) - 19*a(n-2) for n >= 2, with a(0) = 1 and a(1) = 5.
a(n) = (1/2 - sqrt(5)/10)*(9/2 - sqrt(5)/2)^n + (sqrt(5)/10 + 1/2)*(sqrt(5)/2 + 9/2)^n.
G.f.: (1 - 4*x)/(1 - 9*x + 19*x^2).
a(n) = Sum_{k=0..n} A094441(n,k)*4^k. - Philippe Deléham, Dec 14 2009
a(n) = A081574(n) - 4*A081574(n-1). - R. J. Mathar, Jul 19 2012
E.g.f.: exp(9*x/2)*(5*cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, Jun 03 2024

A270863 Self-composition of the Fibonacci sequence.

Original entry on oeis.org

0, 1, 2, 6, 17, 50, 147, 434, 1282, 3789, 11200, 33109, 97878, 289354, 855413, 2528850, 7476023, 22101326, 65338038, 193158521, 571033600, 1688143881, 4990651642, 14753839486, 43616704857, 128943855250, 381196100507, 1126928202714, 3331532438042, 9848993360069

Offset: 0

Oboifeng Dira, Mar 24 2016

This sequence has the same relation to the Fibonacci numbers A000045 as A030267 has to the natural numbers A000027.
From Oboifeng Dira, Jun 28 2020: (Start)
This sequence can be generated from a family of composition pairs of generating functions g(f(x)), where k is an integer and where
f(x) = x/(1-k*x-x^2) and g(x) = (x+(k-1)*x^2)/(1-(3-2*k)*x-(3*k-k^2-1)*x^2).
Some cases of k values are:
k=-5, f(x) g.f. 0,A052918(-1)^n and g(x) g.f. 0,A081571
k=-4, f(x) g.f. A001076(-1)^(n+1) and g(x) g.f. 0,A081570
k=-3, f(x) g.f. A006190(-1)^(n+1) and g(x) g.f. 0,A081569
k=-2, f(x) g.f. A215936(n+2) and g(x) g.f. 0,A081568
k=-1, f(x) g.f. A039834(n+2) and g(x) g.f. 0,A081567
k=0,  f(x) g.f. A000035 and g(x) g.f. 0,A001519(n+1)
k=1,  f(x) g.f. A000045 and g(x) g.f. A000045
k=2,  f(x) g.f. A000129 and g(x) g.f. 0,A039834(n+1)
k=3,  f(x) g.f. A006190 and g(x) g.f. 0,A001519(-1)^n
k=4,  f(x) g.f. A001076 and g(x) g.f. 0,A093129(-1)^n
k=5,  f(x) g.f. 0,A052918 and g(x) g.f. 0,A192240(-1)^n
k=6,  f(x) g.f. A005668 and g(x)=(x+5*x^2)/(1+9*x+19*x^2)
k=7,  f(x) g.f. 0,A054413 and g(x)=(x+6*x^2)/(1+11*x+29*x^2).
(End)

			a(5) = 3*a(4)+a(3)-3*a(2)-a(1) = 51+6-6-1 = 50.

Colin Barker, Table of n, a(n) for n = 0..1000
Oboifeng Dira, A Note on Composition and Recursion, Southeast Asian Bulletin of Mathematics (2017), Vol. 41, Issue 6, 849-853.
Oboifeng Dira, Family of composition pairs g(f(x)) generating A270683
Index entries for linear recurrences with constant coefficients, signature (3,1,-3,-1).

Cf. A000027, A000045, A001622, A030267, A000035, A001519, A000129, A039834.

I:=[0, 1, 2, 6]; [m le 4 select I[m] else 3*Self(m-1)+Self(m-2)-3*Self(m-3)-Self(m-4): m in [1..30]]; // Marius A. Burtea, Aug 03 2019

f:= x-> x/(1-x-x^2):
a:= n-> coeff(series(f(f(x)), x, n+1), x, n):
seq(a(n), n=0..30);

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,-3,1,3]^(n-1)*[1;2;6;17])[1,1] \\ Charles R Greathouse IV, Mar 24 2016

concat(0, Vec(x*(1-x-x^2)/(1-3*x-x^2+3*x^3+x^4) + O(x^40))) \\ Colin Barker, Mar 24 2016

a(n) = 3*a(n-1)+a(n-2)-3*a(n-3)-a(n-4) for n > 3, a(0)=0, a(1)=1, a(2)=2, a(3)=6.
G.f.: x*(1-x-x^2) / (1-3*x-x^2+3*x^3+x^4). - Colin Barker, Mar 24 2016
G.f.: B(B(x)) where B(x) is the g.f. of A000045. - Joerg Arndt, Mar 25 2016
a(n) = (phi*((phi^2 + 5^(1/4)*sqrt(3*phi))^n - (phi^2 - 5^(1/4)*sqrt(3*phi))^n) + (psi^2 + 5^(1/4)*sqrt(3*psi))^n - (psi^2 - 5^(1/4)*sqrt(3*psi))^n)/(2^n * 5^(3/4) * sqrt(3*phi)), where phi = (sqrt(5) + 1)/2 is the golden ratio, and psi = 1/phi = (sqrt(5) - 1)/2. - Vladimir Reshetnikov, Aug 01 2019
0 = a(n)*(a(n) +6*a(n+1) -a(n+2)) +a(n+1)*(8*a(n+1) -9*a(n+2) +a(n+3)) +a(n+2)*(-8*a(n+2) +6*a(n+3)) +a(n+3)*(-a(n+3)) if n>=0. - Michael Somos, Feb 05 2022

A081572 Square array of binomial transforms of Fibonacci numbers, read by ascending antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 5, 3, 1, 4, 10, 13, 5, 1, 5, 17, 35, 34, 8, 1, 6, 26, 75, 125, 89, 13, 1, 7, 37, 139, 338, 450, 233, 21, 1, 8, 50, 233, 757, 1541, 1625, 610, 34, 1, 9, 65, 363, 1490, 4172, 7069, 5875, 1597, 55, 1, 10, 82, 535, 2669, 9633, 23165, 32532, 21250, 4181, 89

Offset: 0

Paul Barry, Mar 22 2003

Array rows are solutions of the recurrence a(n) = (2*k+1)*a(n-1) - A028387(k-1)*a(n-2), where a(0) = 1 and a(1) = k+1.

			The array rows begins as:
  1, 1,  2,   3,    5,     8,     13, ... A000045;
  1, 2,  5,  13,   34,    89,    233, ... A001519;
  1, 3, 10,  35,  125,   450,   1625, ... A081567;
  1, 4, 17,  75,  338,  1541,   7069, ... A081568;
  1, 5, 26, 139,  757,  4172,  23165, ... A081569;
  1, 6, 37, 233, 1490,  9633,  62753, ... A081570;
  1, 7, 50, 363, 2669, 19814, 148153, ... A081571;
Antidiagonal triangle begins as:
  1;
  1, 1;
  1, 2,  2;
  1, 3,  5,   3;
  1, 4, 10,  13,   5;
  1, 5, 17,  35,  34,    8;
  1, 6, 26,  75, 125,   89,   13;
  1, 7, 37, 139, 338,  450,  233,  21;
  1, 8, 50, 233, 757, 1541, 1625, 610, 34;

G. C. Greubel, Antidiagonal rows n = 0..50, flattened

Array row n: A000045 (n=0), A001519 (n=1), A081567 (n=2), A081568 (n=3), A081569 (n=4), A081570 (n=5), A081571 (n=6).
Array column k: A000027 (k=1), A002522 (k=2).
Different from A073133.
Cf. A028387.

A081572:= func< n,k | (&+[Binomial(k,j)*Fibonacci(j+1)*(n-k)^(k-j): j in [0..k]]) >;
[A081572(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, May 27 2021

T[n_, k_]:= If[n==0, Fibonacci[k+1], Sum[Binomial[k, j]*Fibonacci[j+1]*n^(k-j), {j, 0, k}]]; Table[T[n-k, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, May 26 2021 *)

def A081572(n,k): return sum( binomial(k,j)*fibonacci(j+1)*(n-k)^(k-j) for j in (0..k) )
flatten([[A081572(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 27 2021

Rows are successive binomial transforms of F(n+1).
T(n, k) = ((5+sqrt(5))/10)*( (2*n + 1 + sqrt(5))/2)^k + ((5-sqrt(5)/10)*( 2*n + 1 - sqrt(5))/2 )^k.
From G. C. Greubel, May 27 2021: (Start)
T(n, k) = Sum_{j=0..k} binomial(k,j)*n^(k-j)*Fibonacci(j+1) (square array).
T(n, k) = Sum_{j=0..k} binomial(k,j)*(n-k)^(k-j)*Fibonacci(j+1) (antidiagonal triangle). (End)

A106198 Triangle, columns = successive binomial transforms of Fibonacci numbers.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 3, 5, 3, 1, 5, 13, 10, 4, 1, 8, 34, 35, 17, 5, 1, 13, 89, 125, 75, 26, 6, 1, 21, 233, 450, 338, 139, 37, 7, 1, 34, 610, 1625, 1541, 757, 233, 50, 8, 1

Offset: 0

Gary W. Adamson, Apr 24 2005

Column 0 = Fibonacci numbers, column 1 = odd-indexed Fibonacci numbers (first binomial transform of 1, 1, 2, 3, 5, ...); column 2 = second binomial transform of Fibonacci numbers, etc.

			First few rows of the triangle are:
   1;
   1,   1;
   2,   2,   1;
   3,   5,   3,   1;
   5,  13,  10,   4,   1;
   8,  34,  35,  17,   5,   1;
  13,  89, 125,  75,  26,   6,   1;
  21, 233, 450, 338, 139,  37,   7,   1;
  ...
Column 2 = A081567, second binomial transform of Fibonacci numbers: 1, 3, 10, 35, 125, ...

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Cf. A000045, A081567, A081568, A081569, A081570.

T:= function(n,k)
    if k=0 then return Fibonacci(n+1);
    else return Sum([0..n-k], j-> Binomial(n-k,j)*Fibonacci(j+1)*k^(n-k-j));
    fi; end;
Flat(List([0..10], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Dec 11 2019

function T(n,k)
  if k eq 0 then return Fibonacci(n+1);
  else return (&+[Binomial(n-k,j)*Fibonacci(j+1)*k^(n-k-j): j in [0..n-k]]);
  end if; return T; end function;
[T(n,k): k in [0..n], n in [0..10]]; // G. C. Greubel, Dec 11 2019

with(combinat);
T:= proc(n, k) option remember;
      if k=0 then fibonacci(n+1)
    else add( binomial(n-k,j)*fibonacci(j+1)*k^(n-k-j), j=0..n-k)
      fi; end:
seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Dec 11 2019

Table[If[k==0, Fibonacci[n+1], Sum[Binomial[n-k, j]*Fibonacci[j+1]*k^(n-k-j), {j,0,n-k}]], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 11 2019 *)

T(n,k) = if(k==0, fibonacci(n+1), sum(j=0,n-k, binomial(n-k,j)*fibonacci( j+1)*k^(n-k-j)) ); \\ G. C. Greubel, Dec 11 2019

@CachedFunction
def T(n, k):
    if (k==0): return fibonacci(n+1)
    else: return sum(binomial(n-k,j)*fibonacci(j+1)*k^(n-k-j) for j in (0..n-k))
[[T(n, k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Dec 11 2019

Offset column k = k-th binomial transform of the Fibonacci numbers, given leftmost column = Fibonacci numbers.

cp's OEIS Frontend

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

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A081567 Second binomial transform of F(n+1).

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A094441 Triangular array T(n,k) = Fibonacci(n+1-k)*C(n,k), 0 <= k <= n.

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A081569 Fourth binomial transform of F(n+1).

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A270863 Self-composition of the Fibonacci sequence.

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A081572 Square array of binomial transforms of Fibonacci numbers, read by ascending antidiagonals.