A066258 -id:A066258 - cp's OEIS frontend

A056570 Third power of Fibonacci numbers (A000045).

0, 1, 1, 8, 27, 125, 512, 2197, 9261, 39304, 166375, 704969, 2985984, 12649337, 53582633, 226981000, 961504803, 4073003173, 17253512704, 73087061741, 309601747125, 1311494070536, 5555577996431, 23533806109393, 99690802348032, 422297015640625

Offset: 0

Wolfdieter Lang, Jul 10 2000

Divisibility sequence; that is, if n divides m, then a(n) divides a(m).
In general, cubing the terms of a Horadam sequence with signature (c,d) will result in a fourth-order recurrence with signature (c^3+2*c*d, c^4*d+3*(c*d)^2+2*d^3, -(c*d)^3-2*c*d^4, -d^6). - Gary Detlefs, Nov 12 2021
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using (1/3,2/3)-fences and third-squares (1/3 X 1 pieces, always placed so that the shorter sides are horizontal). A (w,g)-fence is a tile composed of two w X 1 pieces separated by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/6,1/3)-fences and (1/6,5/6)-fences. - Michael A. Allen, Jan 11 2022

			a(4) = 27 because the fourth Fibonacci number is 3 and 3^3 = 27.
a(5) = 125 because the fifth Fibonacci number is 5 and 5^3 = 125.

D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1969, Vol. 1, p. 85, (exercise 1.2.8. Nr. 30) and p. 492 (solution).

Vincenzo Librandi, Table of n, a(n) for n = 0..173
Feryal Alayont and Evan Henning, Edge Covers of Caterpillars, Cycles with Pendants, and Spider Graphs, J. Int. Seq. (2023) Vol. 26, Art. 23.9.4.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059.
A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
Andrej Dujella, A bijective proof of Riordan's theorem on powers of Fibonacci numbers, Discrete Math. 199 (1999), no. 1-3, 217--220. MR1675924 (99k:05016).
Kenneth Edwards and Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers cubed, Fib. Q. 58:5 (2020) 128-134.
D. Foata and G.-N. Han, Nombres de Fibonacci et polynomes orthogonaux
Mariana Nagy, Simon R. Cowell and Valeriu Beiu, Survey of Cubic Fibonacci Identities - When Cuboids Carry Weight, arXiv:1902.05944 [math.HO], 2019.
Hilary I. Okagbue, Muminu O. Adamu, Sheila A. Bishop and Abiodun A. Opanuga, Digit and Iterative Digit Sum of Fibonacci numbers, their identities and powers, International Journal of Applied Engineering Research ISSN 0973-4562 Volume 11, Number 6 (2016) pp 4623-4627.
J. Riordan, Generating functions for powers of Fibonacci numbers, Duke. Math. J. 29 (1962) 5-12.
E. L. Roettger and H. C. Williams, Appearance of Primes in Fourth-Order Odd Divisibility Sequences, J. Int. Seq., Vol. 24 (2021), Article 21.7.5.
H. C. Williams and R. K. Guy, Odd and even linear divisibility sequences of order 4, INTEGERS, 2015, #A33.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1).

Cf. A000045, A007598, A056588, A055870, A066259, A066258.
Cf. A346513 (first differences), A005968 (partial sums).
Third row of array A103323.

[Fibonacci(n)^3: n in [0..30]]; // Vincenzo Librandi, Jun 04 2011

A056570 := proc(n) combinat[fibonacci](n)^3 ; end proc:
seq(A056570(n),n=0..20) ;

Table[Fibonacci[n]^3, {n, 0, 30}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2008 *)

a(n)=fibonacci(n)^3 \\ Charles R Greathouse IV, Sep 24 2015

concat(0, Vec(x*(1-2*x-x^2)/((1+x-x^2)*(1-4*x-x^2)) + O(x^30))) \\ Colin Barker, Jun 04 2016

[fibonacci(n)^3 for n in (0..30)] # G. C. Greubel, Feb 20 2019

a(n) = A000045(n)^3.
G.f.: x*p(3, x)/q(3, x) with p(3, x) = Sum_{m=0..2} A056588(2, m)*x^m = 1 -2*x -x^2 and q(3, x) = Sum_{m=0..4} A055870(4, m)*x^m = 1 -3*x -6*x^2 +3*x^3 +x^4 = (1+x-x^2)*(1-4*x-x^2) (factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): 1*a(n) -3*a(n-1) -6*a(n-2) +3*a(n-3) +1*a(n-4) = 0, n >= 4, a(0)=0, a(1)=a(2)=1, a(3)=2^3. See 5th row of signed Fibonomial triangle for coefficients: A055870(4, m), m=0..4
a(n) = (Fibonacci(3n) - 3(-1)^n*Fibonacci(n))/5. - Ralf Stephan, May 14 2004
a(n) and a(n+1) are found as rightmost and leftmost terms (respectively) in M^n * [1 0 0 0] where M = the 4 X 4 upper triangular Pascal's triangle matrix [1 3 3 1 / 1 2 1 0 / 1 1 0 0 / 1 0 0 0]. E.g., Ma(4) = 27, a(5) = 125. M^4 * [1 0 0 0] = [125 75 45 27]; where 75 = A066259(4) and 45 = A066258(3). The characteristic polynomial of M = x^4 - 3x^3 - 6x^2 + 3x + 1. a(n)/a(n-1) of the sequence and companions tend to 2+sqrt(5) = 4.2360679..., an eigenvalue of M and a root of the polynomial. - Gary W. Adamson, Oct 31 2004
From R. J. Mathar, Oct 16 2006: (Start)
Sum_{j=0..n} binomial(n,j)*a(j) = (2^n*A001906(n) + 3*A000045(n))/5.
Sum_{j=0..n} (-1)^j*binomial(n,j)*a(j) = ((-2)^n*A000045(n) - 3*A001906(n))/5. (End)
G.f.: x*(1-2*x-x^2)/((1+x-x^2)*(1-4*x-x^2)). - Colin Barker, Feb 28 2012
a(n) = F(n-2)*F(n+1)^2 + F(n-1)*(-1)^n. - J. M. Bergot, Mar 17 2016
a(n) = ((-3*(1/2*(-1-sqrt(5)))^n-(2-sqrt(5))^n+3*(1/2*(-1+sqrt(5)))^n+(2+sqrt(5))^n)) / (5*sqrt(5)). - Colin Barker, Jun 04 2016
a(n) = F(n-1)*F(n)*F(n+1) + F(n)*(-1)^(n-1). - Tony Foster III, Apr 11 2018
5*a(n) = L(2*n-1)*F(n+2) - L(2*n+1)*F(n-2) - 7*(-1)^n*F(n), where L(n) = A000032(n). - Peter Bala, Nov 12 2019
F(n+1)*F(n)*F(n-1) = 2*Sum_{j=1..n-1} P(j)*a(n-j) for n>0, where Pell number P(n) = A000129(n). - Michael A. Allen, Jan 11 2022

A001655 Fibonomial coefficients: a(n) = F(n+1) * F(n+2) * F(n+3)/2, where F() = Fibonacci numbers A000045.

Original entry on oeis.org

1, 3, 15, 60, 260, 1092, 4641, 19635, 83215, 352440, 1493064, 6324552, 26791505, 113490195, 480752895, 2036500788, 8626757644, 36543528780, 154800876945, 655747029795, 2777789007071, 11766903040368, 49845401197200, 211148507782800, 894439432403425

Offset: 0

N. J. A. Sloane

In a triangle having sides of F(n+1), 2*F(n+2) and F(n+3), the product of the area and circumradius will be a(n). For example: a triangle having sides of 5, 16 and 13 will have an area of 4*sqrt(51), a circumradius of 65*sqrt(51)/51, and the product is 4*65 = 260. - Gary Detlefs, Dec 14 2010

Explanation of this comment: if a triangle with sides (a, b, c) has a circumradius R and an area A, then A*R = abc/4; here, with a = F(n+1), b=2*F(n+2) and c=F(n+3), this gives a(n)= A*R. - Bernard Schott, Jan 26 2023

			G.f. = 1 + 3*x + 15*x^2 + 60*x^3 + 260*x^4 + 1092*x^5 + 4641*x^6 + ...

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..200
A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
A. Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972, p. 74.
Kenneth Edwards and Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers cubed, Fib. Q. 58:5 (2020) 128-134.
Milan Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 13 2013
Ronald Orozco López, Generating Functions of Generalized Simplicial Polytopic Numbers and (s,t)-Derivatives of Partial Theta Function, arXiv:2408.08943 [math.CO], 2024. See p. 13.
Ronald Orozco López, Simplicial d-Polytopic Numbers Defined on Generalized Fibonacci Polynomials, arXiv:2501.11490 [math.CO], 2025. See p. 10.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
David Treeby, Further Physical Derivations of Fibonacci Summations, Fibonacci Quart. 54 (2016), no. 4, 327-334.
Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1).

Cf. A066258 (first differences), A215037 (partial sums), A363753 (alternating sums).

Cf. A000045, A055870, A065563, A079586, A114525, A256178.

[Fibonacci(n+3)*Fibonacci(n+2)*Fibonacci(n+1)/2: n in [0..30]]; // Vincenzo Librandi, May 09 2016

A001655:=1/(z**2-z-1)/(z**2+4*z-1); # Simon Plouffe in his 1992 dissertation.

Table[(Fibonacci[n+3]*Fibonacci[n+2]*Fibonacci[n+1])/2, {n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Nov 23 2009 *)
LinearRecurrence[{3, 6, -3, -1}, {1, 3, 15, 60}, 25] (* Jean-François Alcover, Sep 23 2017 *)

b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j)); vector(20, n, b(n-1, 3))  \\ Joerg Arndt, May 08 2016

G.f.: 1/(1-3*x-6*x^2+3*x^3+x^4) = 1/((1+x-x^2)*(1-4*x-x^2)) (see Comments to A055870).
a(n) = A010048(n+3, 3) = fibonomial(n+3, 3).
a(n) = (1/2) * A065563(n).
a(n) = 4*a(n-1) + a(n-2) + ((-1)^n)*F(n+1), n >= 2; a(0)=1, a(1)=3.
a(n) = (F(n+3)^3 - F(n+2)^3 - F(n+1)^3)/6. - Gary Detlefs, Dec 24 2010
a(n-1) = Sum_{k=0..n} F(k+1)*F(k)^2, n >= 1. - Wolfdieter Lang, Aug 01 2012
From Wolfdieter Lang, Aug 09 2012: (Start)
a(n-1)*(-1)^n =  Sum_{k=0..n} (-1)^k*F(k+1)^2*F(k), n >= 1. See the link under A215037, eq. (25).
a(n) = (F(3*(n+2)) + 2*(-1)^n*F(n+2))/10, n >= 0. See the same link, eq. (32). (End)
a(n) = -a(-4-n)*(-1)^n for all n in Z. - Michael Somos, Sep 19 2014
0 = a(n)*(-a(n+1) - a(n+2)) + a(n+1)*(-3*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014
O.g.f.: exp( Sum_{n >= 1} L(n)*L(2*n)*x^n/n ), where L(n) = A000032(n) is a Lucas number. Cf. A114525, A256178. - Peter Bala, Mar 18 2015
Sum_{n>=0} (-1)^n/a(n) = 2 * A079586 - 6. - Amiram Eldar, Oct 04 2020
The formula by Gary Detlefs above is valid for all sequences of the Fibonacci type f(n) = f(n-1) + f(n-2): 3*f(n+2)*f(n+1)*f(n) = f(n+2)^3 - f(n+1)^3 - f(n)^3. - Klaus Purath, Mar 25 2021
a(n) = sqrt(Sum_{j=1..n+1} F(j)^3*F(j+1)^3). See Treeby link. - Michel Marcus, Apr 10 2022
a(n) = Sum_{k=1..n+1} A000129(k)*A056570(n+2-k). - Michael A. Allen, Jan 25 2023
G.f.: exp( Sum_{k>=1} F(4*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

A066259 a(n) = Fibonacci(n)*Fibonacci(n+1)^2.

Original entry on oeis.org

1, 4, 18, 75, 320, 1352, 5733, 24276, 102850, 435655, 1845504, 7817616, 33116057, 140281700, 594243090, 2517253683, 10663258432, 45170286424, 191344405725, 810547906740, 3433536036866, 14544692047439, 61612304237568, 260993908980000, 1105587940186225, 4683345669678532

Offset: 1

Len Smiley, Dec 09 2001

Harry J. Smith, Table of n, a(n) for n = 1..200
D. Zeitlin, Generating Functions for Products of Recursive Sequences, Transactions A.M.S., 116, Apr. 1965, p. 304.
Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1).

Cf. A000045, A065563, A066258, A000045, A056570.

First[#]Last[#]^2&/@Partition[Fibonacci[Range[30]],2,1]  (* Harvey P. Dale, Mar 04 2011 *)

a(n) = { fibonacci(n) * fibonacci(n+1)^2 } \\ Harry J. Smith, Feb 07 2010

O.g.f.: (x+x^2)/(1-3x-6x^2+3x^3+x^4) = x(1+x)/((1+x-x^2)(1-4x-x^2)).
a(n) = second term from left in M^n * [1 0 0 0] where M = the 4 X 4 upper triangular Pascal's triangle matrix [1 3 3 1 / 1 2 1 0 / 1 1 0 0 / 1 0 0 0]. E.g., a(4) = 75 since M^4 * [1 0 0 0] = [125 75 45 27] = [A056570(5) a(4) A066258(3) A056570(4)]. - Gary W. Adamson, Oct 31 2004
a(n) = (1/5)*(F(3n+2) - (-1)^n*F(n-1)). - Ralf Stephan, Jul 26 2005
a(n) = (F(n+2)^3 - 2*F(n)^3 - F(n-1)^3)/6. - Greg Dresden, Aug 12 2022

A197424 Number of subsets of {1, 2, ..., 4*n + 2} which do not contain two numbers whose difference is 4.

Original entry on oeis.org

4, 36, 225, 1600, 10816, 74529, 509796, 3496900, 23961025, 164249856, 1125736704, 7716041281, 52886200900, 362488284900, 2484529385121, 17029223715904, 116720020119616, 800010960336225, 5483356589096100, 37583485459535236, 257601040852192129

Offset: 0

John W. Layman, Oct 14 2011

This sequence is an instance of a general result given in Math. Mag. Problem 1854 (see Links).
From Feryal Alayont, May 20 2023: (Start)
a(n) is the number of edge covers of a caterpillar graph with spine P_(4n+5), one pendant attached at vertex n+2 counting from the left end of the spine, a second one at 2n+3 and a third at 3n+4. The caterpillar graph for n=1 is as follows:
        *      *      *
        |      |      |
  *--*--*--v1--*--v2--*--*--*
Each pendant edge must be included in an edge cover leaving only the middle six edges flexible. Every vertex except v1 and v2 is incident with at least one of the pendant edges. Therefore, if we label the middle six edges in the spine with numbers 3, 1, 5, 2, 6, 4 (starting from the left), the edges have to be chosen so that both 1,5 and 2,6 cannot be missing. This corresponds to choosing subsets of {1, 2, ..., 6} which do not contain two numbers whose difference is 4. (End)

F. Alayont and E. Henning, Edge Covers of Caterpillars, Cycles with Pendants, and Spider Graphs; submitted.

Colin Barker, Table of n, a(n) for n = 0..1000
Mathramz Problem Solving Group, Solution for Problem 1854 in Mathematics Magazine Vol. 83, No.4, October 2010, February 28, 2011
Marian Tetiva, Problem 1854, Mathematics Magazine, 84 (2011) 300.
Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Cf. A000045, A001654, A060635, A066258.

Table[(1/25) (LucasL[2 (2 n + 5)] - 2 (-1)^n LucasL[2 n + 5] - 1), {n, 0, 20}] (* Michael De Vlieger, Mar 27 2016 *)

Vec((4+16*x-15*x^2-5*x^3+x^4) / ((1-x)*(1-7*x+x^2)*(1+3*x+x^2)) + O(x^30)) \\ Colin Barker, Mar 26 2016

a(n) = (fibonacci(n+2)*fibonacci(n+3))^2; \\ Altug Alkan, Mar 26 2016

a(n) = F(n+2)^2*F(n+3)^2 = A001654(n+2)^2, where F(n) denotes the n-th Fibonacci number A000045(n).
G.f.: ( -4-16*x+15*x^2+5*x^3-x^4 ) / ( (x-1)*(x^2+3*x+1)*(x^2-7*x+1) ). - R. J. Mathar, Oct 15 2011
Empirical: a(n) = A189145(2n+3). - R. J. Mathar, Oct 15 2011
For L=Lucas, a(n) = (1/25)*(L(2*(2*n+5)) - 2*(-1)^n*L(2*n+5) - 1), an instance of (F(n+p)*F(n+q))^2 = (1/25)*(L(2*(2*n+p+q)) - 2*(-1)^(n+q)*L(p-q)*L(2*n+p+q) + L(2*(p-q)) + 4*(-1)^(p-q)) which follows from squaring a specialization of identity 17b in the Vajda reference at A000045, F(n+p)*F(n+q) = (1/5)*(L(2*n+p+q) - (-1)*(n+q)*L(p-q)), then applying Vajda 17c, L(n)^2 = L(2*n) + 2*(-1)^n, to the expansion. - Ehren Metcalfe, Mar 26 2016
a(n) = A060635(n+2)/2. - Alois P. Heinz, Jul 03 2025

A099014 a(n) = Fibonacci(n)*(Fibonacci(n-1)^2 + Fibonacci(n+1)^2).

Original entry on oeis.org

0, 1, 5, 20, 87, 365, 1552, 6565, 27825, 117844, 499235, 2114729, 8958240, 37947545, 160748653, 680941780, 2884516383, 12219006325, 51760543280, 219261176861, 928805254905, 3934482189716, 16666734024715, 70601418270865

Offset: 0

Paul Barry, Sep 22 2004

Form the matrix A=[1,1,1,1;3,2,1,0;3,1,0,0;1,0,0,0]=(binomial(3-j,i)). Then a(n)=(2,3)-element of A^n.

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..173 from Vincenzo Librandi)
Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1).

Cf. A000045, A056570, A066258, A066259.

[Fibonacci(n)*(Fibonacci(n-1)^2+Fibonacci(n+1)^2): n in [0..30]]; // Vincenzo Librandi, Jun 05 2011

CoefficientList[Series[x*(1 + 2*x - x^2)/(1 - 3*x - 6*x^2 + 3*x^3 + x^4), {x, 0, 50}], x] (* G. C. Greubel, Dec 31 2017 *)
Join[{0},#[[2]](#[[1]]^2+#[[3]]^2)&/@Partition[Fibonacci[ Range[ 0,30]],3,1]] (* or *) LinearRecurrence[{3,6,-3,-1},{0,1,5,20},30] (* Harvey P. Dale, Oct 17 2021 *)

a(n)=fibonacci(n)*(fibonacci(n-1)^2+fibonacci(n+1)^2) \\ Charles R Greathouse IV, Jun 05 2011

G.f.: x*(1+2*x-x^2)/(1-3*x-6*x^2+3*x^3+x^4) = x*(1+2*x-x^2)/((1+x-x^2)*(1-4*x-x^2)).
a(n) = Sum_{k=0..n} (-1)^(k+1)*Fib(k)*(0^(n-k) + 6*A001076(n-k)).
a(n) = ((-1)^n*Fib(n) + 3*Fib(3*n))/5. - Ehren Metcalfe, May 21 2016

A220362 a(n) = Fibonacci(n-1) * Fibonacci(n) * Fibonacci(n+2).

Original entry on oeis.org

0, 3, 10, 48, 195, 840, 3536, 15015, 63546, 269280, 1140535, 4831632, 20466720, 86699067, 367262090, 1555748880, 6590255259, 27916773720, 118257343984, 500946159615, 2122041966330, 8989114051008, 38078498128175, 161303106631968, 683290924545600

Offset: 1

Michel Marcus, Dec 12 2012

nonn

An integral pentagon is a pentagon with integer sides and diagonals. There are two types of such pentagons.

Type B have sides A056570(n+2), A056570(n+2), A220363(n+2), A056570(n+2), A056570(n+2), and opposite diagonals a(n+2), A066258(n+2), A066258(n+2), A066258(n+2), a(n+2), for n=1,2,...

R. K. Guy, Unsolved Problems in Number Theory, D20.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
J. H. Jordan, B. E. Peterson, Almost regular integer Fibonacci pentagons, Rocky Mountain J. Math. Volume 23, Number 1 (1993), 243-247.

Table[Fibonacci[n - 1] * Fibonacci[n] * Fibonacci[n + 2], {n, 30}] (* T. D. Noe, Dec 13 2012 *)
#[[1]]*#[[2]]*#[[4]]&/@Partition[Fibonacci[Range[0,30]],4,1] (* Harvey P. Dale, Jan 16 2016 *)

x='x+O('x^99); concat(0, Vec(x*(3+x)/((x^2-x-1)*(x^2+4*x-1)))) \\ Altug Alkan, Mar 26 2016

a(n) = 3*a(n-1) + 6*a(n-2) -3*a(n-3) - a(n-4); g.f.: (3x+x^2)/(1-3x-6x^2+3x^2+x^4) = x(3+x)/( (x^2-x-1)(x^2+4x-1) ). [Ron Knott, Jun 27 2013]

a(n) = 2*(-1)^n*Lucas(n-3)/25 + 3*(-1)^n*Lucas(n+1)/25 + Fibonacci(3*n+1)/5 = (1/5)*(-(-1)^n*Fibonacci(n-1) + 3*(-1)^n*Fibonacci(n) + Fibonacci(3*n+1)). - Ehren Metcalfe, Mar 26 2016

A099015 a(n) = Fib(n+1)(2Fib(n)^2 + Fib(n)*Fib(n-1) + Fib(n-1)^2).

Original entry on oeis.org

1, 2, 8, 33, 140, 592, 2509, 10626, 45016, 190685, 807764, 3421728, 14494697, 61400482, 260096680, 1101787113, 4667245276, 19770767984, 83750317589, 354772037730, 1502838469496, 6366125914117, 26967342128548

Offset: 0

Paul Barry, Sep 22 2004

Form the matrix A=[1,1,1,1;3,2,1,0;3,1,0,0;1,0,0,0]=(binomial(3-j,i)). Then a(n)=(2,2)-element of A^n.

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..172 from Vincenzo Librandi)
Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1).

Cf. A000045, A056570, A066258, A066259, A099014, A033887.

[Fibonacci(n+1)*(2*Fibonacci(n)^2 + Fibonacci(n)*Fibonacci(n-1) + Fibonacci(n-1)^2): n in [0..30]]; // Vincenzo Librandi, Jun 05 2011

LinearRecurrence[{3,6,-3,-1},{1,2,8,33},30] (* Harvey P. Dale, Nov 28 2015 *)
CoefficientList[Series[(1-x-4*x^2)/((1+x-x^2)*(1-4*x-x^2)), {x, 0, 50}], x] (* G. C. Greubel, Dec 31 2017 *)

a(n)=my(e=fibonacci(n-1),f=fibonacci(n));(e+f)*(2*f^2+f*e+e^2) \\ Charles R Greathouse IV, Jun 05 2011

first(n) = Vec((1 - x - 4*x^2)/(1 - 3*x - 6*x^2 + 3*x^3 + x^4) + O(x^n)) \\ Iain Fox, Dec 31 2017

G.f.: (1-x-4*x^2)/((1+x-x^2)*(1-4*x-x^2)).
G.f.: (1-x-4*x^2)/(1-3*x-6*x^2+3*x^3+x^4).
a(n) = (3*Fib(3*n+1) + (-1)^n*Fib(n-3))/5.
a(n) = (2+sqrt(5))^n*(3/10 + 3*sqrt(5)/50) + (2-sqrt(5))^n*(3/10 - 3*sqrt(5)/50) + (-1)^n*((1/2 - sqrt(5)/2)^n*(1/5 + 2*sqrt(5)/25) + (1/5 - 2*sqrt(5)/25)*(1/2 + sqrt(5)/2)^n).

A220363 a(n) = Fibonacci(n)^3 + (-1)^n*Fibonacci(n+2).

Original entry on oeis.org

1, -1, 4, 3, 35, 112, 533, 2163, 9316, 39215, 166519, 704736, 2986361, 12648727, 53583620, 226979403, 961507387, 4072998992, 17253519469, 73087050795, 309601764836, 1311494041879, 5555578042799, 23533806034368, 99690802469425, 422297015444207

Offset: 0

Michel Marcus, Dec 12 2012

An integral pentagon is a pentagon with integer sides and diagonals. There are two types of such pentagons.

Type B have sides A056570(n+2), A056570(n+2), a(n+2), A056570(n+2), A056570(n+2), and opposite diagonals A220362(n+2), A066258(n+2), A066258(n+2), A066258(n+2), A220362(n+2), for n=1,2,...

R. K. Guy, Unsolved Problems in Number Theory, D20.

Indranil Ghosh, Table of n, a(n) for n = 0..1593
J. H. Jordan, B. E. Peterson, Almost regular integer Fibonacci pentagons, Rocky Mountain J. Math. Volume 23, Number 1 (1993), 243-247.
Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1).

Table[Fibonacci[n]^3 + (-1)^n * Fibonacci[n + 2], {n, 0, 30}] (* T. D. Noe, Dec 13 2012 *)
LinearRecurrence[{3,6,-3,-1},{1,-1,4,3},30] (* Harvey P. Dale, Mar 19 2022 *)

Vec((x^2-4*x+1)/((x^2-x-1)*(x^2+4*x-1)) + O(x^100)) \\ Colin Barker, Sep 23 2014

a(n) = fibonacci(n)^3 + (-1)^n*fibonacci(n+2) \\ Charles R Greathouse IV, Feb 14 2017

a(n) = 3*a(n-1)+6*a(n-2)-3*a(n-3)-a(n-4). G.f.: (x^2-4*x+1) / ((x^2-x-1)*(x^2+4*x-1)). - Colin Barker, Sep 23 2014

A177727 a(0)=1; a(n) = a(n-1) * Fibonacci(3+n) * Fibonacci(1+n) / (Fibonacci(n))^2, n > 1.

Original entry on oeis.org

1, 3, 30, 180, 1300, 8736, 60333, 412335, 2829310, 19384200, 132882696, 910735488, 6242420665, 42785803515, 293259265950, 2010026277756, 13776931957468, 94428478367520, 647222466507045, 4436128656563175, 30405678471399166, 208403619747957648, 1428419662108160400

Offset: 0

Roger L. Bagula, May 12 2010

Similar recurrences a(n) = a(n-1)*F(a0+n-1)*F(b0+n-1)/(F(n)*F(c0+n-1)) are:
{a0,b0,c0} = {3,2,1} in A066258.
{a0,b0,c0} = {3,1,1} in A001654.
{a0,b0,c0} = {4,1,1} in A001655 and next for 5,6 as well.

Harry Hochstadt, The Functions of Mathematical Physics, Dover, New York, 1986, p. 93.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1)

Cf. A066258, A001654, A001655, A001656, A001657.

I:=[1, 3, 30, 180, 1300]; [n le 5 select I[n] else 5*Self(n-1)+15*Self(n-2)-15*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..30]]; // Vincenzo Librandi, Nov 18 2011

with (combinat):
A177727 := proc(n)
   if n = 0 then
           1;
   else
           procname(n-1)*fibonacci(3+n)*fibonacci(1+n)/fibonacci(n)^2 ;
   end if;
end proc:
seq(A177727(n),n=0..10) ; # R. J. Mathar, Nov 17 2011

a0 = 4; b0 = 2; c0 = 1;
a[0] = 1;
a[n_] := a[n] = (Fibonacci[(a0 + n - 1)]*Fibonacci[( b0 + n - 1)]/(Fibonacci[n]*Fibonacci[(c0 + n - 1)]))*a[n - 1];
Table[a[n], {n, 0, 30}]
LinearRecurrence[{5,15,-15,-5,1},{1,3,30,180,1300},30] (* Vincenzo Librandi, Nov 18 2011 *)

G.f.: ( -1+2*x ) / ( (x-1)*(x^2+3*x+1)*(x^2-7*x+1) ). - R. J. Mathar, Nov 17 2011

a(n) = A001656(n) - 2*A001656(n-1). - R. J. Mathar, Nov 17 2011

A215038 Partial sums of A066259: a(n) = Sum_{k=0..n} F(k+1)^2*F(k), n>=0, with the Fibonacci numbers F=A000045.

Original entry on oeis.org

0, 1, 5, 23, 98, 418, 1770, 7503, 31779, 134629, 570284, 2415788, 10233404, 43349461, 183631161, 777874251, 3295127934, 13958386366, 59128672790, 250473078515, 1061020985255, 4494557022121, 19039249069560, 80651553307128

Offset: 0

Wolfdieter Lang, Aug 09 2012

For a derivation of the explicit form of this sum see the link under A215308 on the partial summation formula, eq. (7).

			a(2) = 0 + 1^2*1 + 2^2*1 = 1 + 4 = 5.

Index entries for linear recurrences with constant coefficients, signature (4,3,-9,2,1).

Cf. A000045, A001655, A008346, A066258, A066259, A215308, A215037.

a(n) = Sum_{k=0..n} A066259(k) = Sum_{k=0..n} F(k+1)^2*F(k), n >= 0, with A066259(0)=0.
a(n) = (F(n+2)*F(n+1)^2 - (-1)^n*F(n) - 1)/2 = (A066258(n+1) - (-1)^n*A008346(n))/2, n >= 0.
O.g.f.: x*(1+x)/((1+x-x^2)*(1-4*x-x^2)*(1-x)) (from A066259).
E.g.f.: (2*exp(-x/2)*(5*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2)) + exp(2*x)*(15*cosh(sqrt(15)*x) + 7*sqrt(5)*sinh(sqrt(5)*x)) - 25*exp(x))/50. - Stefano Spezia, Oct 28 2024

cp's OEIS Frontend

A056570 Third power of Fibonacci numbers (A000045).

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A001655 Fibonomial coefficients: a(n) = F(n+1) * F(n+2) * F(n+3)/2, where F() = Fibonacci numbers A000045.

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A066259 a(n) = Fibonacci(n)*Fibonacci(n+1)^2.

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A197424 Number of subsets of {1, 2, ..., 4*n + 2} which do not contain two numbers whose difference is 4.

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A099014 a(n) = Fibonacci(n)*(Fibonacci(n-1)^2 + Fibonacci(n+1)^2).

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A220362 a(n) = Fibonacci(n-1) * Fibonacci(n) * Fibonacci(n+2).

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A099015 a(n) = Fib(n+1)*(2*Fib(n)^2 + Fib(n)*Fib(n-1) + Fib(n-1)^2).

A099015 a(n) = Fib(n+1)(2Fib(n)^2 + Fib(n)*Fib(n-1) + Fib(n-1)^2).