A056570 -id:A056570 - cp's OEIS frontend

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

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).

A216699 Digital root of cubes of Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, 8, 9, 8, 8, 1, 9, 1, 1, 8, 9, 8, 8, 1, 9, 1, 1, 8, 9, 8, 8, 1, 9, 1, 1, 8, 9, 8, 8, 1, 9, 1, 1, 8, 9, 8, 8, 1, 9, 1, 1, 8, 9, 8, 8, 1, 9, 1, 1, 8, 9, 8, 8, 1, 9, 1, 1, 8, 9, 8, 8, 1, 9, 1, 1, 8, 9, 8, 8, 1, 9, 1, 1, 8, 9, 8, 8, 1, 9, 1, 1, 8, 9, 8, 8

Offset: 0

Ravi Bhandari, Sep 15 2012

This sequence repeats after every 8 terms, hence this is periodic with period 8.

Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 1).

Cf. A004090, A056570, A216676.

Table[NestWhile[Total[IntegerDigits[#]] &, Fibonacci[n]^3, # > 9 &], {n, 0, 86}] (* T. D. Noe, Oct 15 2012 *)
Join[{0},LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 1},{1, 1, 8, 9, 8, 8, 1, 9},86]] (* Ray Chandler, Aug 25 2015 *)

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), a(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 and 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).

Cf. A066259, A220360, A056570.

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

a(n) = F(n-1)*(F(n+1)*F(n) - (-1)^n). - Greg Dresden, Sep 04 2025

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

A244309 a(n) = F(n)^3 - F(n)^2, where F(n) is the n-th Fibonacci number (A000045).

Original entry on oeis.org

0, 0, 0, 4, 18, 100, 448, 2028, 8820, 38148, 163350, 697048, 2965248, 12595048, 53440504, 226608900, 960530634, 4070452764, 17246835648, 73069580980, 309555981900, 1311374255620, 5555264316910, 23532984885744, 99688652356608, 422291386890000

Offset: 0

Colin Barker, Jun 25 2014

			a(4) is 18 because F(4)^3 - F(4)^2 = 3^3 - 3^2 = 18.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,2,-22,-4,14,-1,-1).

Cf. A000045, A045991, A244310, A056570, A007598.

[Fibonacci(n)^3 - Fibonacci(n)^2: n in [0..30]]; // Vincenzo Librandi, Jun 26 2014

CoefficientList[Series[2 x^3 (x^2 - x + 2)/((x + 1) (x^2 - 3 x + 1) (x^2 - x - 1) (x^2 + 4 x - 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 26 2014 *)
Table[#^3 - #^2 &@ Fibonacci@ n, {n, 0, 25}] (* Michael De Vlieger, Mar 27 2016 *)
LinearRecurrence[{5,2,-22,-4,14,-1,-1},{0,0,0,4,18,100,448},30] (* Harvey P. Dale, Aug 22 2020 *)

vector(50, n, fibonacci(n-1)^3-fibonacci(n-1)^2)

G.f.: 2*x^3*(x^2-x+2) / ((x+1)*(x^2-3*x+1)*(x^2-x-1)*(x^2+4*x-1)).
a(n) = A045991(A000045(n)). - Michel Marcus, Jun 25 2014
a(n) = (F(3*n) - 3*(-1)^n*F(n))/5 - (L(2*n) - 2*(-1)^n)/5, where F=A000045 and L=A000032. - Ehren Metcalfe, Mar 26 2016

A292278 a(n) = (Fibonacci(3*n-1) + 1)/2 for n >= 1.

Original entry on oeis.org

1, 3, 11, 45, 189, 799, 3383, 14329, 60697, 257115, 1089155, 4613733, 19544085, 82790071, 350704367, 1485607537, 6293134513, 26658145587, 112925716859, 478361013021, 2026369768941, 8583840088783, 36361730124071, 154030760585065, 652484772464329

Offset: 1

Vincenzo Librandi, Sep 13 2017

Problem B-1211 proposed by Hideyuki Ohtsuka (see Links section): For n >= 1, prove that Fibonacci(n-1)^3 + Sum_{k=1..n} Fibonacci(k)^3 = (Fibonacci(3*n-1) + 1)/2.

Proof. Let F(n-1)^3 = (F(3*n-3) + 3*(-1)^n*F(n-1))/5 (see Ralf Stephan's formula in A056570) and Sum_{k=1..n} F(k)^3 = (F(3*n+2) - 6*(-1)^(n)*F(n-1) + 5)/10 (see Benjamin & Timothy's formula in A005968), where F=A000045, n>0. Therefore, (F(3*n-3) + 3*(-1)^n*F(n-1))/5 + (F(3*n+2) - 6*(-1)^(n)*F(n-1) + 5)/10 = (2*F(3*n-3) + F(3*n+2) + 5)/10 = (2*(F(3*n-1) - F(3*n-2)) + (3*F(3*n-1) + 2*F(3*n-2)) + 5)/10 = (5*F(3*n-1) + 5)/10 = a(n). - Bruno Berselli, Sep 14 2017

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Hideyuki Ohtsuka, Problem B-1211, The Fibonacci Quarterly, Volume 55, Number 3 (August 2017), p. 276 (see Comments section).
Index entries for linear recurrences with constant coefficients, signature (5,-3,-1).

Cf. A000045; A005968, A056570.

[(Fibonacci(3*n-1)+1)/2: n in [1..30]];

Table[(Fibonacci[3 n - 1] + 1) / 2, {n, 40}]
LinearRecurrence[{5,-3,-1},{1,3,11},30] (* Harvey P. Dale, Mar 06 2024 *)

a(n) = (fibonacci(3*n-1)+1)/2; \\ Altug Alkan, Sep 13 2017

G.f.: x*(1 - 2*x - x^2)/((1 - x)*(1 - 4*x -x^2)).

a(n) = 5*a(n-1) - 3*a(n-2) - a(n-3).

Edited by Bruno Berselli, Sep 14 2017

A056587 Tenth power of Fibonacci numbers A000045.

Original entry on oeis.org

0, 1, 1, 1024, 59049, 9765625, 1073741824, 137858491849, 16679880978201, 2064377754059776, 253295162119140625, 31181719929966183601, 3833759992447475122176, 471584161164422542970449

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..96
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 (89,4895,-83215,-582505,1514513,1514513,-582505,-83215,4895,89,-1).

Cf. A000045, A007598, A056570, A056571, A056572, A056573, A056574, A056585, A056586, A056588, A055870.

[Fibonacci(n)^10: n in [0..20]]; // Vincenzo Librandi, Jun 04 2011

Fibonacci[Range[0,15]]^10 (* Harvey P. Dale, Jul 29 2018 *)

a(n) = fibonacci(n)^10; \\ Michel Marcus, Sep 06 2017

a(n) = F(n)^10, F(n)=A000045(n).
G.f.: x*p(10, x)/q(10, x) with p(10, x) := sum_{m=0..9} A056588(9, m)*x^m = (1-x)*(1 - 87*x - 4047*x^2 + 42186*x^3 + 205690*x^4 + 42186*x^5 - 4047*x^6 - 87*x^7 + x^8) and q(10, x) := sum_{m=0..11} A055870(11, 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)*(1 - 123*x + x^2) (denominator factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): sum_{m=0..11} A055870(11, m)*a(n-m) = 0, n >= 11; inputs: a(n), n=0..10.  a(n) = 89*a(n-1) + 4895*a(n-2) - 83215*a(n-3) - 582505*a(n-4) + 1514513*a(n-5) + 1514513*a(n-6) - 582505*a(n-7) -83215*a(n-8) + 4895*a(n-9) + 89*a(n-10) - a(n-11).

More terms from Larry Reeves (larryr(AT)acm.org), Jul 17 2001

A133054 Cubes of Motzkin numbers.

Original entry on oeis.org

1, 1, 8, 64, 729, 9261, 132651, 2048383, 33698267, 582182875, 10474708672, 194910229592, 3731808877831, 73218245857875, 1467320155924104, 29956212068269248, 621670415400598563, 13090510375656917139, 279262277830549534968, 6027856524470650538304

Offset: 0

Omar E. Pol, Oct 30 2007

nonn

G. C. Greubel, Table of n, a(n) for n = 0..700

Cf. A000578, A030078, A056570.

Motzkin numbers: A001006.

a1006:= gfun:-rectoproc({(-3*x-6)*a(x+1)+(-2*x-7)*a(x+2)+(x+5)*a(x+3), a(0) = 1, a(1) = 1, a(2) = 2},a(x),remember):
map(a1006^3, [$0..60]); # Robert Israel, Oct 03 2017

CoefficientList[Series[(1 - x - (1 - 2 x - 3 x^2)^(1/2))/(2 x^2), {x, 0, 50}], x]^3 (* G. C. Greubel, Oct 03 2017 *)

a(n) = A001006(n)^3.

A215040 a(n) = F(2*n+1)^3, n>=0, with F = A000045 (Fibonacci).

Original entry on oeis.org

1, 8, 125, 2197, 39304, 704969, 12649337, 226981000, 4073003173, 73087061741, 1311494070536, 23533806109393, 422297015640625, 7577812474746632, 135978327528030989, 2440032083025183109, 43784599166913148552, 785682752921379769625, 14098504953417839657513

Offset: 0

Wolfdieter Lang, Aug 10 2012

Bisection (odd part) of A056570. From this follows the o.g.f., and its partial fraction decomposition leads to the explicit formula given below. For the computation see A215039 for a comment on Chebyshev's S-polynomials.

Harvey P. Dale, Table of n, a(n) for n = 0..797
Kenny B. Davenport, proposer, Problem B-1353, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 62, No. 3 (2024), p. 259.

Cf. A000045, A056570, A163200 (partial sums).

Fibonacci[2*Range[0,20]+1]^3 (* Harvey P. Dale, Jan 24 2013 *)

a(n) = F(2*n+1)^3, n>=0, with F=A000045.
O.g.f.: (1-x)*(1-12*x+x^2)/((1-3*x+x^2)*(1-18*x+x^2)) (from the bisection (odd part) of A056570).
a(n) = (12*F(2*n+1) + F(6*(n+1)) - F(6*n))/20, n>=0.
a(n) = 1 + (4*Sum_{k=1..n} F(6*k) + 3*Sum_{k=1..n} F(2*k)) / 5 (Davenport, 2024). - Amiram Eldar, Aug 29 2024

cp's OEIS Frontend

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

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A216699 Digital root of cubes of Fibonacci numbers.

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

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A244309 a(n) = F(n)^3 - F(n)^2, where F(n) is the n-th Fibonacci number (A000045).

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A292278 a(n) = (Fibonacci(3*n-1) + 1)/2 for n >= 1.

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