A066259 -id:A066259 - cp's OEIS frontend

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.

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

A056570 Third power of Fibonacci numbers (A000045).

Original entry on oeis.org

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

A066258 a(n) = Fibonacci(n)^2 * Fibonacci(n+1).

Original entry on oeis.org

0, 1, 2, 12, 45, 200, 832, 3549, 14994, 63580, 269225, 1140624, 4831488, 20466953, 86698690, 367262700, 1555747893, 6590256856, 27916771136, 118257348165, 500946152850, 2122041977276, 8989114033297, 38078498156832, 161303106585600, 683290924620625, 2894466804871682

Offset: 0

Len Smiley, Dec 09 2001

From Feryal Alayont, Apr 27 2023: (Start)
a(n) is the number of edge covers of a caterpillar graph with spine P_(3n-2), one pendant attached at vertex n counting from the left end of the spine and another pendant at vertex 2n-1. The caterpillar graph for n=3 is as follows:
        *     *
        |     |
  *--*--*--*--*--*--*
           v
Each pendant edge must be included in an edge cover. Every vertex except v is then incident with at least one edge. Therefore, of the remaining four edges in the spine, only two in the middle have restrictions. At least one of those two has to be in the edge cover to ensure v is incident with one edge in the edge cover, leaving us with 3*2*2 total edge covers. (End)

Harry J. Smith, Table of n, a(n) for n = 0..200
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.
David 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, A033887, A056570, A065563, A066259.

First differences of A001655.

[Fibonacci(n)^2*Fibonacci(n+1): n in [0..30]]; // G. C. Greubel, Feb 12 2024

#[[1]]^2 #[[2]]&/@Partition[Fibonacci[Range[0,30]],2,1] (* or *) LinearRecurrence[ {3,6,-3,-1},{0,1,2,12},30] (* Harvey P. Dale, Jul 28 2018 *)

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

[fibonacci(n)^2*fibonacci(n+1) for n in range(31)] # G. C. Greubel, Feb 12 2024

O.g.f.: x*(1-x) / ( (1-4*x-x^2)*(1+x-x^2) ).
a(n) = second term from right in M^(n+1) * [1 0 0 0], where M = the 4 X 4 upper Pascal's triangular matrix [1 3 3 1 / 1 2 1 0 / 1 1 0 0 / 1 0 0 0]. E.g., a(3) = 45 since M^4 * [1 0 0 0] = [125 75 45 27] where 125 = A056570(5), 75 = A066259(4) and 27 = A056570(4). - Gary W. Adamson, Oct 31 2004
a(n) = (1/5)*(Fibonacci(3n+1) - (-1)^n*Fibonacci(n+2)). - Ralf Stephan, Jul 26 2005
a(n) = 3*a(n-1) + 6*a(n-2) - 3*a(n-3) - a(n-4). - Zak Seidov, May 07 2015

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

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).

A220360 a(n) = Fibonacci(n-1) * Fibonacci(n+1) * Fibonacci(n+2).

Original entry on oeis.org

0, 6, 15, 80, 312, 1365, 5712, 24310, 102795, 435744, 1845360, 7817849, 33115680, 140282310, 594242103, 2517255280, 10663255848, 45170290605, 191344398960, 810547917686, 3433536019155, 14544692076096, 61612304191200, 260993909055025, 1105587940064832

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 A have sides A066259(n+1), a(n+1), A066259(n+1), a(n+1), A066259(n+1), and opposite diagonals A056570(n+2), A056570(n+2), A220361(n+2), A056570(n+2), A056570(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.

Cf. A000045.

Table[Fibonacci[n - 1]*Fibonacci[n + 1]*Fibonacci[n + 2], {n, 30}] (* T. D. Noe, Dec 13 2012 *)
#[[1]]#[[3]]#[[4]]&/@Partition[Fibonacci[Range[0,30]],4,1] (* Harvey P. Dale, Apr 08 2022 *)

a(n) = fibonacci(n-1) * fibonacci(n+1) * fibonacci(n+2); \\ Michel Marcus, Mar 26 2016

x='x+O('x^99); concat(0, Vec((6*x-3*x^2-x^3)/(1-3*x-6*x^2+3*x^3+x^4))) \\ Altug Alkan, Mar 26 2016

G.f.: (6*x - 3*x^2 - x^3)/(1 - 3*x - 6*x^2 + 3*x^3 + x^4); a(n) = 3*a(n-1) + 6*a(n-2) - 3*a(n-3) - a(n-4). [Ron Knott, Jun 27 2013]
Sum {n >= 2} 1/a(n) = 1/4. - Peter Bala, Nov 30 2013
a(n) = 4*(-1)^n*F(n-1)/5 + (-1)^n*F(n) + F(3*n+2)/5 with F=A000045. - Ehren Metcalfe, Mar 26 2016

A220361 a(n) = Fibonacci(n)^3 + (-1)^n*Fibonacci(n-2).

Original entry on oeis.org

1, 7, 28, 123, 515, 2192, 9269, 39291, 166396, 704935, 2986039, 12649248, 53582777, 226980767, 961505180, 4073002563, 17253513691, 73087060144, 309601749709, 1311494066355, 5555578003196, 23533806098447, 99690802365743, 422297015611968, 1788878864731825

Offset: 2

Michel Marcus, Dec 12 2012

An integral pentagon is a pentagon with integer sides and diagonals. There are two types of such pentagons. Type A have sides A066259(n+1), A220360(n+1), A066259(n+1), A220360(n+1), A066259(n+1), and opposite diagonals A056570(n+2), A056570(n+2), A220361(n+2), A056570(n+2), A056570(n+2), for n=1,2,...

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

Indranil Ghosh, Table of n, a(n) for n = 2..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).

with(combinat): A220361:=n->fibonacci(n)^3+(-1)^n*fibonacci(n-2): seq(A220361(n), n=2..30); # Wesley Ivan Hurt, Apr 26 2017

Table[Fibonacci[n]^3 + (-1)^n * Fibonacci[n - 2], {n, 2, 30}] (* T. D. Noe, Dec 13 2012 *)
LinearRecurrence[{3,6,-3,-1},{1,7,28,123},30] (* Harvey P. Dale, Jul 13 2021 *)

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

cp's OEIS Frontend

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.

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A056570 Third power of Fibonacci numbers (A000045).

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A066258 a(n) = Fibonacci(n)^2 * Fibonacci(n+1).

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

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

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A220360 a(n) = Fibonacci(n-1) * Fibonacci(n+1) * Fibonacci(n+2).

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A220361 a(n) = Fibonacci(n)^3 + (-1)^n*Fibonacci(n-2).

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