A056570 -id:A056570 - cp's OEIS frontend

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

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

A105317 Powers of Fibonacci numbers.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 8, 9, 13, 16, 21, 25, 27, 32, 34, 55, 64, 81, 89, 125, 128, 144, 169, 233, 243, 256, 377, 441, 512, 610, 625, 729, 987, 1024, 1156, 1597, 2048, 2187, 2197, 2584, 3025, 3125, 4096, 4181, 6561, 6765, 7921, 8192, 9261, 10946, 15625, 16384, 17711

Offset: 1

Reinhard Zumkeller, Apr 25 2005

nonn

The subset of nontrivial Fibonacci powers [numbers A000045(k)^n which are not in A000045] starts 4, 9, 16, 25, 27, 32, 64, 81, 125, 128, 169, 243, 256, 441, 512, 625, 729, 1024, 1156... - R. J. Mathar, Jan 26 2015. These are the initial terms of A254719. - Reinhard Zumkeller, Feb 06 2015

			2197 = 13^3 = A000045(7)^3, therefore 2197 is a term.

R. Zumkeller, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
Eric Weisstein's World of Mathematics, Fibonacci Number

Subsequences: A056570-A056574, A007598, A000045, A000079, A000244, A000351, A001018, A001022 and A009965.

Cf. A103323, A254719.

import Data.Set (singleton, deleteFindMin, insert)
a105317 n = a105317_list !! (n-1)
a105317_list = 0 : 1 : h 1 (drop 4 a000045_list) (singleton (2, 2)) where
  h y xs'@(x:xs) s
    | x < ff    = h y xs (insert (x, x) s)
    | ff == y   = h y xs' s'
    | otherwise = ff : h ff xs' (insert (f * ff, f) s')
    where ((ff, f), s') = deleteFindMin s
-- Reinhard Zumkeller, Feb 06 2015

N:= 10^6: # to get all terms <= N
select(`<=`,{0,1,seq(seq(combinat:-fibonacci(i)^j, i = 3 ..floor(log[phi](sqrt(5)*N^(1/j)+1))),j=1..ilog2(N))},N);
# if using Maple 11 or earlier, uncomment the next line
# sort(convert(%,list)); # Robert Israel, Jan 26 2015

lim = 10^5; t = Table[f = Fibonacci[n]; f^Range[Floor[Log[lim]/Log[f]]], {n, 3, Ceiling[Log[GoldenRatio, lim] + 1]}]; Union[{0, 1}, Flatten[t]] (* T. D. Noe, Sep 27 2011 *)

list(lim)=my(v=List([0]),k=1,f,t); while(k<=lim, listput(v,k); k*=2); k=3; while(k<=lim, listput(v,k); k*=3); k=5; while(k<=lim, listput(v,k); k*=5); k=6; while((f=fibonacci(k++))<=lim, t=1; while((t*=f)<=lim, listput(v,t))); Set(v) \\ Charles R Greathouse IV, Oct 03 2016

A056574 Seventh power of Fibonacci numbers A000045.

Original entry on oeis.org

0, 1, 1, 128, 2187, 78125, 2097152, 62748517, 1801088541, 52523350144, 1522435234375, 44231334895529, 1283918464548864, 37281334283719577, 1082404156823183753, 31427428360210000000, 912473096871571914483

Offset: 0

Wolfdieter Lang, Jul 10 2000

A divisibility sequence; that is, if n divides m, then a(n) divides a(m).

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..115
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876. Mathematical Reviews, MR2959001. Zentralblatt MATH, Zbl 1255.05003.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059. Mathematical Reviews, MR2980853. Zentralblatt MATH, Zbl 1255.05004.
A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
J. Riordan, Generating functions for powers of Fibonacci numbers, Duke. Math. J. 29 (1962) 5-12.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (21,273,-1092,-1820,1092,273,-21,-1).

Cf. A000045, A007598, A056570, A056571, A056572, A056573, A056588, A055870.

Seventh row of array A103323.

[Fibonacci(n)^7: n in [0..20]]; // Vincenzo Librandi, Jun 04 2011

f[n_]:=Fibonacci[n]^7; lst={}; Do[AppendTo[lst,f[n]],{n,0,5!}]; lst (* Vladimir Joseph Stephan Orlovsky, Feb 12 2010 *)

a(n)=fibonacci(n)^7 \\ Charles R Greathouse IV, Jan 30 2012

a(n) = F(n)^7, where F(n) = A000045(n).
G.f.: x*p(7, x)/q(7, x) with p(7, x) := sum_{m=0..6} A056588(6, m)*x^m = 1 - 20*x - 166*x^2 + 318*x^3 + 166*x^4 - 20*x^5 - x^6 and q(7, x) := sum_{m=0..8} A055870(8, m)*x^m = (1 + x - x^2)*(1 - 4*x - x^2)*(1 + 11*x - x^2)*(1 - 29*x - x^2) (factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): sum_{m=0..8} A055870(8, m)*a(n-m) = 0, n >= 8; inputs: a(n), n=0..7. a(n) = 21*a(n-1) + 273*a(n-2) - 1092*a(n-3) - 1820*a(n-4) + 1092*a(n-5) + 273*a(n-6) - 21*a(n-7) - a(n-8).
a(n+1) = F(n)^7+F(n+1)^7+7*F(n)*F(n+1)*F(n+2)*[2*F(n+1)^2-(-1)^n]^2 = {Sum(0 <= j <= [n/2]; binomial(n-j, j))}^7, for n>=0 (This is Theorem 2.3 (iv) of Azarian's second paper in the references for this sequence). - Mohammad K. Azarian, Jun 29 2015

A110224 a(n) = Fibonacci(n)^3 + Fibonacci(n+1)^3.

Original entry on oeis.org

1, 2, 9, 35, 152, 637, 2709, 11458, 48565, 205679, 871344, 3690953, 15635321, 66231970, 280563633, 1188485803, 5034507976, 21326515877, 90340574445, 382688808866, 1621095817661, 6867072066967, 29089384105824

Offset: 0

Paul Barry, Jul 16 2005

Vincenzo Librandi, Table of n, a(n) for n = 0..172
Diego Marques and Alain Togbé, On the sum of powers of two consecutive Fibonacci numbers, Proc. Japan Acad. Ser. A Math. Sci., Volume 86, Number 10 (2010), 174-176.
Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1).

Cf. A056570, A000045, A061646.

[Fibonacci(n)^3 + Fibonacci(n+1)^3: n in [0..30]]; // Vincenzo Librandi, Jun 05 2011

Total/@Partition[Fibonacci[Range[0,30]]^3,2,1] (* or *) LinearRecurrence [{3,6,-3,-1},{1,2,9,35},30] (* Harvey P. Dale, May 29 2013 *)

a(n)=fibonacci(n)^3+fibonacci(n+1)^3 \\ Charles R Greathouse IV, Jun 05 2011

[sum(fibonacci(n+k)^3 for k in (0..1)) for n in (0..30)] # G. C. Greubel, Mar 18 2019

G.f.: (1 - x - 3*x^2 - x^3)/(1 - 3*x - 6*x^2 + 3*x^3 + x^4) = (1 - x - 3*x^2 - x^3)/((1 + x - x^2)*(1 - 4x - x^2)).
a(n) = 3*a(n-1) + 6*a(n-2) - 3*a(n-3) - a(n-4).
a(n) = (3*(-1)^n*Fibonacci(n-1) + 2*Fibonacci(3*n+2))/5.
a(n) = Fibonacci(n+2)*A061646(n). - Greg Dresden, Sep 04 2025

A215039 a(n) = Fibonacci(2*n)^3, n>=0.

Original entry on oeis.org

0, 1, 27, 512, 9261, 166375, 2985984, 53582633, 961504803, 17253512704, 309601747125, 5555577996431, 99690802348032, 1788878864685457, 32100128763082731, 576013438873664000, 10336141770970357629, 185474538438612378103

Offset: 0

Wolfdieter Lang, Aug 10 2012

Bisection (even part) of A056570. From this follows the o.g.f., and its partial fraction decomposition leads to the explicit formula given below. The recurrences for F(2*n) and F(6*n)/8 are used in this computation. They follow from the fact that F(2*n) = S(n-1,3), and F(6*n)/8 = S(n-1,18), with Chebyshev's S(n,x) = U(n,x/2) polynomial of the second kind (see A001906 and A049660, respectively).

G. C. Greubel, Table of n, a(n) for n = 0..790
E. Kilic, Y. T. Ulutas, N. Omur, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011) #11.5.6, Table 1, k=3.
Index entries for linear recurrences with constant coefficients, signature (21,-56,21,-1).

Cf. A000045, A056570, A163198 (partial sums).

List([0..20], n-> Fibonacci(2*n)^3 ); # G. C. Greubel, Dec 22 2019

[Fibonacci(2*n)^3: n in [0..20]]; // G. C. Greubel, Dec 22 2019

with(combinat); seq( fibonacci(2*n)^3, n=0..20); # G. C. Greubel, Dec 22 2019

Fibonacci[2*(Range[21]-1)]^3 (* G. C. Greubel, Dec 22 2019 *)

vector(21, n, fibonacci(2*(n-1)) ); \\ G. C. Greubel, Dec 22 2019

[fibonacci(2*n)^3 for n in (0..20)] # G. C. Greubel, Dec 22 2019

a(n) = F(2*n)^3, n>=0, with F=A000045.
O.g.f.: x*(1+6*x+x^2)/((1-3*x+x^2)*(1-18*x+x^2)) (from the bisection (even part) of A056570).
a(n) = (F(6*n) - 3*F(2*n))/5, n>=0.
a(n+2) - 18*a(n+1) + a(n) - 9*F(2*(n+1)) = 0, n>=0. From the F_n^3 recurrence (see a comment and references on A055870, use row n=4) together with the recurrence appearing in the solution of exercise 6.58, p. 315, on p. 556 of the second edition of the Graham-Knuth-Patashnik book (reference given on A007318), both with n -> 2*n. See also Koshy's book (reference given on A065563) p. 87, 1. and p. 89, 32. (with a - sign) and 33. - Wolfdieter Lang, Aug 11 2012

A056585 Eighth power of Fibonacci numbers A000045.

Original entry on oeis.org

0, 1, 1, 256, 6561, 390625, 16777216, 815730721, 37822859361, 1785793904896, 83733937890625, 3936588805702081, 184884258895036416, 8686550888106661441, 408066367122340274881, 19170731299728100000000

Offset: 0

Wolfdieter Lang, Jul 10 2000

A divisibility sequence; that is, if n divides m, then a(n) divides a(m).

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..107
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876. Mathematical Reviews, MR2959001. Zentralblatt MATH, Zbl 1255.05003.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059. Mathematical Reviews, MR2980853. Zentralblatt MATH, Zbl 1255.05004.
A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
J. Riordan, Generating functions for powers of Fibonacci numbers, Duke. Math. J. 29 (1962) 5-12.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (34,714,-4641,-12376,12376,4641,-714,-34,1).

Cf. A000045, A007598, A056570, A056571, A056572, A056573, A056574, A056588, A055870.

[Fibonacci(n)^8: n in [0..20]]; // Vincenzo Librandi, Jun 04 2011

lst={};Do[f=Fibonacci[n];AppendTo[lst, f^8], {n, 0, 4!}];lst (* Vladimir Joseph Stephan Orlovsky, Sep 27 2008 *)
Fibonacci[Range[0,20]]^8 (* Harvey P. Dale, Jul 03 2017 *)

a(n)=fibonacci(n)^8 \\ Charles R Greathouse IV, Jun 30 2015

a(n) = F(n)^8, F(n)=A000045(n).
G.f.: x*p(8, x)/q(8, x) with p(8, x) := sum_{m=0..7} A056588(7, m)*x^m = (1+x)*(1 - 34*x - 458*x^2 + 2242*x^3 - 458*x^4 - 34*x^5 + x^6) and q(8, x) := sum_{m=0..9} A055870(9, m)*x^m = (1-x)*(1 + 3*x + x^2)*(1 - 7*x + x^2)*(1 + 18*x + x^2)*(1 - 47*x + x^2) (denominator factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): sum_{m=0..9} A055870(9, m)*a(n-m) = 0, n >= 9; inputs: a(n), n=0..8. a(n) = 34*a(n-1) + 714*a(n-2) - 4641*a(n-3) - 12376*a(n-4) + 12376*a(n-5) + 4641*a(n-6) - 714*a(n-7) - 34*a(n-8) + a(n-9).
a(n+1) = 8*F(n)^2*F(n+1)^2*[F(n)^4+F(n+1)^4+4*F(n)^2*F(n+1)^2+3*F(n)*F(n+1)*F(n+2)]-[F(n)^8+F(n+2)^8]+2*[2*F(n+1)^2-(-1)^n]^4 = {Sum(0 <= j <= [n/2]; binomial(n-j, j))}^8, for n>=0 (This is Theorem 2.2 (vii) of Azarian's second paper in the references for this sequence). - Mohammad K. Azarian, Jun 29 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

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

A346513 a(n) = Fibonacci(n+1)^3 - Fibonacci(n)^3.

Original entry on oeis.org

1, 0, 7, 19, 98, 387, 1685, 7064, 30043, 127071, 538594, 2281015, 9663353, 40933296, 173398367, 734523803, 3111498370, 13180509531, 55833549037, 236514685384, 1001892323411, 4244083925895, 17978228112962, 76156996238639, 322606213292593, 1366581849044832

Offset: 0

Lamine Ngom, Jul 21 2021

The version related to sum of consecutive Fibonacci numbers cubed is given by A110224.

a(n+1) is divisible by Fibonacci(n). The related quotient sequence is provided by A061646, from its 3rd term.

Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1).

Cf. A056570 (partial sums).

Cf. A000045, A110224, A061646.

Differences[Fibonacci[Range[0, 26]]^3] (* Amiram Eldar, Jul 22 2021 *)

a(n) = fibonacci(n+1)^3 - fibonacci(n)^3; \\ Michel Marcus, Jul 22 2021

a(n) = F(n-1)*(2*F(n+1)^2+(-1)^(n+1)), n>0.
a(n) = F(n-1)*A061646(n+1).
G.f.: (x-1)*(x^2+2*x-1)/((x^2+4*x-1)*(x^2-x-1)). - Alois P. Heinz, Jul 21 2021
For n >= 2, a(n) is the numerator of the continued fraction [1,...,1, 3 ,1,...,1, 2 ,1,...,1] with three runs of 1's each of length n-2. For example, a(5)=387 which is the numerator of the continued fraction [1,1,1, 3 ,1,1,1, 2 ,1,1,1]. - Greg Dresden, Jan 01 2022

A056586 Ninth power of Fibonacci numbers A000045.

Original entry on oeis.org

0, 1, 1, 512, 19683, 1953125, 134217728, 10604499373, 794280046581, 60716992766464, 4605366583984375, 350356403707485209, 26623333280885243904, 2023966356928852115753, 153841020405122283630137

Offset: 0

Wolfdieter Lang, Jul 10 2000

Divisibility sequence; that is, if n divides m, then a(n) divides a(m).

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..101
A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
J. Riordan, Generating functions for powers of Fibonacci numbers, Duke. Math. J. 29 (1962) 5-12.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (55,1870,-19635,-85085,136136,85085,-19635,-1870,55,1).

Cf. A000045, A007598, A056570, A056571, A056572, A056573, A056574, A056585, A056588, A055870.

[Fibonacci(n)^9: n in [0..20]]; // Vincenzo Librandi, Jun 04 2011

a(n) = fibonacci(n)^9; \\ Michel Marcus, Sep 06 2017

a(n) = F(n)^9, F(n)=A000045(n).
G.f.: x*p(9, x)/q(9, x) with p(9, x) := sum_{m=0..8} A056588(8, m)*x^m = 1 - 54*x - 1413*x^2 + 9288*x^3 + 17840*x^4 - 9288*x^5 - 1413*x^6 + 54*x^7 + x^8 and q(9, x) := sum_{m=0..10} A055870(10, m)*x^m = (1 - x - x^2)*(1 + 4*x - x^2)*(1 - 11*x - x^2)*(1 + 29*x - x^2)*(1 - 76*x - x^2) (factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): sum_{m=0..10} A055870(10, m)*a(n-m) = 0, n >= 10; inputs: a(n), n=0..9. a(n) = 55*a(n-1) + 1870*a(n-2) - 19635*a(n-3) - 85085*a(n-4) + 136136*a(n-5) + 85085*a(n-6) - 19635*a(n-7) - 1870*a(n-8) + 55*a(n-9) + a(n-10).

cp's OEIS Frontend

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

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A105317 Powers of Fibonacci numbers.

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A056574 Seventh power of Fibonacci numbers A000045.

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Magma

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A110224 a(n) = Fibonacci(n)^3 + Fibonacci(n+1)^3.

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Magma

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Sage

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A215039 a(n) = Fibonacci(2*n)^3, n>=0.

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GAP

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Sage

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A056585 Eighth power of Fibonacci numbers A000045.

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Magma

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

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