A056570 -id:A056570 - cp's OEIS frontend

A007598 Squared Fibonacci numbers: a(n) = F(n)^2 where F = A000045.

0, 1, 1, 4, 9, 25, 64, 169, 441, 1156, 3025, 7921, 20736, 54289, 142129, 372100, 974169, 2550409, 6677056, 17480761, 45765225, 119814916, 313679521, 821223649, 2149991424, 5628750625, 14736260449, 38580030724, 101003831721, 264431464441, 692290561600

Offset: 0

N. J. A. Sloane, Robert G. Wilson v

a(n)*(-1)^(n+1) = (2*(1-T(n,-3/2))/5), n>=0, with Chebyshev's polynomials T(n,x) of the first kind, is the r=-1 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found. - Wolfdieter Lang, Oct 18 2004
From Giorgio Balzarotti, Mar 11 2009: (Start)
Determinant of power series with alternate signs of gamma matrix with determinant 1!.
a(n) = Determinant(A - A^2 + A^3 - A^4 + A^5 - ... - (-1)^n*A^n) where A is the submatrix A(1..2,1..2) of the matrix with factorial determinant.
A = [[1,1,1,1,1,1,...], [1,2,1,2,1,2,...], [1,2,3,1,2,3,...], [1,2,3,4,1,2,...], [1,2,3,4,5,1,...], [1,2,3,4,5,6,...], ...]; note: Determinant A(1..n,1..n) = (n-1)!.
a(n) is even with respect to signs of power of A.
See A158039...A158050 for sequence with matrix 2!, 3!, ... (End)
Equals the INVERT transform of (1, 3, 2, 2, 2, ...). Example: a(7) = 169 = (1, 1, 4, 9, 25, 64) dot (2, 2, 2, 2, 3, 1) = (2 + 2 + 8 + 18 + 75 + 64). - Gary W. Adamson, Apr 27 2009
This is a divisibility sequence.
a(n+1)*(-1)^n, n>=0, is the sequence of the alternating row sums of the Riordan triangle A158454. - Wolfdieter Lang, Dec 18 2010
a(n+1) is the number of tilings of a 2 X 2n rectangle with n tetrominoes of any shape, cf. A230031. - Alois P. Heinz, Nov 29 2013
This is the case P1 = 1, P2 = -6, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 31 2014
Differences between successive golden rectangle numbers A001654. - Jonathan Sondow, Nov 05 2015
a(n+1) is the number of 2 X n matrices that can be obtained from a 2 X n matrix by moving each element to an adjacent position, horizontally or vertically. This is because F(n+1) is the number of domino tilings of that matrix, therefore with a checkerboard coloring and two domino tilings we can move the black element of each domino of the first tiling to the white element of the same domino and similarly move the white element of each domino of the second tiling to the black element of the same domino. - Fabio Visonà, May 04 2022
In general, squaring the terms of a second-order linear recurrence with signature (c,d) will result in a third-order linear recurrence with signature (c^2+d,(c^2+d)*d,-d^3). - Gary Detlefs, Jan 05 2023

			G.f. = x + x^2 + 4*x^3 + 9*x^4 + 25*x^5 + 64*x^6 + 169*x^7 + 441*x^8 + ...

Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 8.
Ross Honsberger, Mathematical Gems III, M.A.A., 1985, p. 130.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Richard P. Stanley, Enumerative Combinatorics I, Example 4.7.14, p. 251.

Indranil Ghosh, Table of n, a(n) for n = 0..2389 (terms 0..200 from T. D. Noe)
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.
R. C. Alperin, A family of nonlinear recurrences and their linear solutions, Fib. Q., 57:4 (2019), 318-321.
Paul S. Bruckman, Problem B-1023: And a cubic as a sum of two squares, Fibonacci Quarterly, Vol. 45, Number 2; May 2007; p. 186.
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).
Sergio Falcon, Some series of reciprocal k-Fibonacci numbers, Asian Journal of Mathematics and Computer Research, Vol. 11, No. 3 (2016), pp. 184-191; ResearchGate link.
Dominique Foata and Guo-Niu Han, Nombres de Fibonacci et polynômes orthogonaux, in: Marcello Morelli and Marco Tangheroni (eds.), Leonardo Fibonacci : il tempo, le opere, l'ereditá scientifica, Pisa, 23-25 Marzo 1994, Pisa, Pacini Editore, 1994, pp. 179-200.
Svenja Huntemann and Neil A. McKay, Counting Domineering Positions, arXiv:1909.12419 [math.CO], 2019.
Jong Hyun Kim, Hadamard products and tilings, JIS 12 (2009) 09.7.4.
Toufik Mansour, A note on sum of k-th power of Horadam's sequence, arXiv:math/0302015 [math.CO], 2003.
Toufik Mansour, Squaring the terms of an l-th order linear recurrence, arXiv:math/0303138 [math.CO], 2003.
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.
Pantelimon Stanica, Generating functions, weighted and non-weighted sums of powers of second-order recurrence sequences, arXiv:math/0010149 [math.CO], 2000.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).
Index entries for two-way infinite sequences.
Index to divisibility sequences.

Cf. A000032, A000045, A001254, A001622, A001654, A047946, A056570, A059260, A061646, A065885, A079291, A080097, A105393.
Bisection of A006498 and A074677. First differences of A001654.
Second row of array A103323.
Half of A175395.

List([0..30], n -> Fibonacci(n)^2); # G. C. Greubel, Dec 10 2018

a007598 = (^ 2) . a000045  -- Reinhard Zumkeller, Sep 01 2013

[Fibonacci(n)^2: n in [0..30]]; // Vincenzo Librandi, Apr 14 2011

with(combinat): seq(fibonacci(n)^2, n=0..27); # Zerinvary Lajos, Sep 21 2007

f[n_] := Fibonacci[n]^2; Array[f, 4!, 0] (* Vladimir Joseph Stephan Orlovsky, Oct 25 2009 *)
LinearRecurrence[{2,2,-1},{0,1,1},41] (* Harvey P. Dale, May 18 2011 *)

PARI
```
{a(n) = fibonacci(n)^2};
```

concat(0, Vec(x*(1-x)/((1+x)*(1-3*x+x^2)) + O(x^30))) \\ Altug Alkan, Nov 06 2015

from sympy import fibonacci
def A007598(n): return fibonacci(n)**2 # Chai Wah Wu, Apr 14 2025

[(fibonacci(n))^2 for n in range(0, 28)]# Zerinvary Lajos, May 15 2009

[fibonacci(n)^2 for n in range(30)] # G. C. Greubel, Dec 10 2018

G.f.: x*(1-x)/((1+x)*(1-3*x+x^2)).
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3), n > 2. a(0)=0, a(1)=1, a(2)=1.
a(-n) = a(n) for all n in Z.
a(n) = A080097(n-2) + 1.
L.g.f.: 1/5*log((1+3*x+x^2)/(1-6*x+x^2)) = Sum_{n>=0} a(n)/n*x^n; special case of l.g.f. given in A079291. - Joerg Arndt, Apr 13 2011
a(0) = 0, a(1) = 1; a(n) = a(n-1) + Sum(a(n-i)) + k, 0 <= i < n where k = 1 when n is odd, or k = -1 when n is even. E.g., a(2) = 1 = 1 + (1 + 1 + 0) - 1, a(3) = 4 = 1 + (1 + 1 + 0) + 1, a(4) = 9 = 4 + (4 + 1 + 1 + 0) - 1, a(5) = 25 = 9 + (9 + 4 + 1 + 1 + 0) + 1. - Sadrul Habib Chowdhury (adil040(AT)yahoo.com), Mar 02 2004
a(n) = (2*Fibonacci(2*n+1) - Fibonacci(2*n) - 2*(-1)^n)/5. - Ralf Stephan, May 14 2004
a(n) = F(n-1)*F(n+1) - (-1)^n = A059929(n-1) - A033999(n).
Sum_{j=0..2*n} binomial(2*n,j)*a(j) = 5^(n-1)*A005248(n+1) for n >= 1 [P. Stanica]. Sum_{j=0..2*n+1} binomial(2*n+1,j)*a(j) = 5^n*A001519(n+1) [P. Stanica]. - R. J. Mathar, Oct 16 2006
a(n) = (A005248(n) - 2*(-1)^n)/5. - R. J. Mathar, Sep 12 2010
a(n) = (-1)^k*(Fibonacci(n+k)^2-Fibonacci(k)*Fibonacci(2*n+k)), for any k. - Gary Detlefs, Dec 13 2010
a(n) = 3*a(n-1) - a(n-2) + 2*(-1)^(n+1), n > 1. - Gary Detlefs, Dec 20 2010
a(n) = Fibonacci(2*n-2) + a(n-2). - Gary Detlefs, Dec 20 2010
a(n) = (Fibonacci(3*n) - 3*(-1)^n*Fibonacci(n))/(5*Fibonacci(n)), n > 0. - Gary Detlefs, Dec 20 2010
a(n) = (Fibonacci(n)*Fibonacci(n+4) - 3*Fibonacci(n)*Fibonacci(n+1))/2. - Gary Detlefs, Jan 17 2011
a(n) = (((3+sqrt(5))/2)^n + ((3-sqrt(5))/2)^n - 2*(-1)^n)/5; without leading zero we would have a(n) = ((3+sqrt(5))*((3+sqrt(5))/2)^n + (3-sqrt(5))*((3-sqrt(5))/2)^n + 4*(-1)^n)/10. - Tim Monahan, Jul 17 2011
E.g.f.: (exp((phi+1)*x) + exp((2-phi)*x) - 2*exp(-x))/5, with the golden section phi:=(1+sqrt(5))/2. From the Binet-de Moivre formula for F(n). - Wolfdieter Lang, Jan 13 2012
Starting with "1" = triangle A059260 * the Fibonacci sequence as a vector. - Gary W. Adamson, Mar 06 2012
a(0) = 0, a(1) = 1; a(n+1) = (a(n)^(1/2) + a(n-1)^(1/2))^2. - Thomas Ordowski, Jan 06 2013
a(n) + a(n-1) = A001519(n), n > 0. - R. J. Mathar, Mar 19 2014
From Peter Bala, Mar 31 2014: (Start)
a(n) = ( T(n,alpha) - T(n,beta) )/(alpha - beta), where alpha = 3/2 and beta = -1 and T(n,x) denotes the Chebyshev polynomial of the first kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, 3/2; 1, 1/2].
a(n) = U(n-1,i/2)*U(n-1,-i/2), where U(n,x) denotes the Chebyshev polynomial of the second kind.
See the remarks in A100047 for the general connection between Chebyshev polynomials and 4th-order linear divisibility sequences. (End)
a(n) = (F(n+2)*F(n+3) - L(n)*L(n+1))/3 for F = A000045 and L = A000032. - J. M. Bergot, Jun 02 2014
0 = a(n)*(+a(n) - 2*a(n+1) - 2*a(n+2)) + a(n+1)*(+a(n+1) - 2*a(n+2)) + a(n+2)*(+a(n+2)) for all n in Z. - Michael Somos, Jun 03 2014
(F(n)*b(n+2))^2 + (F(n+1)*b(n-1))^2 = F(2*n+1)^3 = A001519(n+1)^3, with b(n) = a(n) + 2*(-1)^n and F(n) = A000045(n) (see Bruckman link). - Michel Marcus, Jan 24 2015
a(n) = 1/4*( a(n-2) - a(n-1) - a(n+1) + a(n+2) ). The same recurrence holds for A001254. - Peter Bala, Aug 18 2015
a(n) = F(n)*F(n+1) - F(n-1)*F(n). - Jonathan Sondow, Nov 05 2015
For n>2, a(n) = F(n-2)*(3*F(n-1) + F(n-3)) + F(2*n-5). Also, for n>2 a(n)=2*F(n-3)*F(n) + F(2*n-3) -(2)*(-1)^n. - J. M. Bergot, Nov 05 2015
a(n) = (F(n+2)^2 + L(n+1)^2) - 2*F(n+2)*L(n+1). - J. M. Bergot, Nov 08 2015
a(n) = F(n+3)^2 - 4*F(n+1)*F(n+2). - J. M. Bergot, Mar 17 2016
a(n) = (F(n-2)*F(n+2) + F(n-1)*F(n+1))/2. - J. M. Bergot, May 25 2017
4*a(n) = L(n+1)*L(n-1) - F(n+2)*F(n-2), where L = A000032. - Bruno Berselli, Sep 27 2017
a(n) = F(n+k)*F(n-k) + (-1)^(n+k)*a(k), for every integer k >= 0. - Federico Provvedi, Dec 10 2018
From Peter Bala, Nov 19 2019: (Start)
Sum_{n >= 3} 1/(a(n) - 1/a(n)) = 4/9.
Sum_{n >= 3} (-1)^n/(a(n) - 1/a(n)) =  (10 - 3*sqrt(5))/18.
Conjecture: Sum_{n >= 1, n != 2*k+1} 1/(a(n) + (-1)^n*a(2*k+1)) = 1/a(4*k+2) for k = 0,1,2,.... (End)
Sum_{n>=1} 1/a(n) = A105393. - Amiram Eldar, Oct 22 2020
Product_{n>=2} (1 + (-1)^n/a(n)) = phi (A001622) (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A056588 Coefficient triangle of certain polynomials.

Original entry on oeis.org

1, 1, -1, 1, -2, -1, 1, -4, -4, 1, 1, -7, -16, 7, 1, 1, -12, -53, 53, 12, -1, 1, -20, -166, 318, 166, -20, -1, 1, -33, -492, 1784, 1784, -492, -33, 1, 1, -54, -1413, 9288, 17840, -9288, -1413, 54, 1, 1, -88, -3960, 46233, 163504, -163504, -46233, 3960, 88, -1

Offset: 0

Wolfdieter Lang, Jul 10 2000

G.f. for column m: see column sequences: A000012, A000071, A056589-91, for m=0..4.
The row polynomials p(n,x) := sum(a(n,m)*x^m) occur as numerators of the g.f. for the (n+1)-th power of Fibonacci numbers A000045. The corresponding denominator polynomials are the row polynomials q(n+2,x) = Sum_{m=0..n+2} A055870(n+2, m)*x^m (signed Fibonomial triangle).
The row polynomials p(n,x) and the companion denominator polynomials q(n,x) can be deduced from Riordan's recursion result.
The explicit formula is found from the recursion relation for powers of Fibonacci numbers (see Knuth's exercise with solution). - Roger L. Bagula, Apr 03 2010

			Row polynomial for n=4: p(4,x) = 1 - 7*x - 16*x^2 + 7*x^3 + x^4. x*p(4,x) is the numerator of the g.f. for A056572(n), n >= 0 (fifth power of Fibonacci numbers) {0,1,1,32,243,...}. The denominator polynomial is Sum_{m=0..6} A055870(6,m)*x^m (n=6 row polynomial of signed fibonomial triangle).
From _Roger L. Bagula_, Apr 03 2010: (Start)
1;
1,  -1;
1,  -2,    -1;
1,  -4,    -4,     1;
1,  -7,   -16,     7,      1;
1, -12,   -53,    53,     12,      -1;
1, -20,  -166,   318,    166,     -20,     -1;
1, -33,  -492,  1784,   1784,    -492,    -33,    1;
1, -54, -1413,  9288,  17840,   -9288,  -1413,   54,  1;
1, -88, -3960, 46233, 163504, -163504, -46233, 3960, 88, -1; (End)

D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1969, Vol. 1, p. 84, (exercise 1.2.8. Nr. 30) and p. 492 (solution).

A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
S. Falcon, On The Generating Functions of the Powers of the K-Fibonacci Numbers, Scholars Journal of Engineering and Technology (SJET), 2014; 2 (4C):669-675.
J. Riordan, Generating functions for powers of Fibonacci numbers, Duke. Math. J. 29 (1962) 5-12.

Cf. A055870, A000012, A000071, A056589-91, A056592 (row sums), A000045, A007598, A056570-4, A056585-7.

A056588 := proc(n,k)
    if k = 0 then
        1;
    elif k >n then
        0;
    else
        combinat[fibonacci](k+1)^(n+1)+add( A055870(n+2, j)*(combinat[fibonacci](k+1-j)^(n+1)), j=1..k) ;
    end if;
end proc: # R. J. Mathar, Jun 14 2015

p[x_, n_] = Sum[(((1 + Sqrt[5])^k - (1 - Sqrt[5])^k)/(2^k*Sqrt[5]))^n*x^k, {k, 0, Infinity}];
a = Table[CoefficientList[FullSimplify[Numerator[p[ x, n]]/x], x]/2^(1 + Floor[n/2]), {n, 1, 10}];
Table[a[[n]]/a[[n]][[1]], {n, 1, 10}];
Flatten[%] (* Roger L. Bagula, Apr 03 2010 *)

S(n, k) = (-1)^floor((k+1)/2)*(prod(j=0, k-1, fibonacci(n-j))/prod(j=1, k, fibonacci(j)));
T(n, k) = sum(j=0, k, fibonacci(k+1-j)^(n+1) * S(n+2, j));
tabl(m) = for (n=0, m, for (k=0, n, print1(T(n, k), ", ")); print);
tabl(9); \\ Tony Foster III, Aug 20 2018

a(n, m)=0 if nA000045(n) (Fibonacci) and sfibonomial(n, m) := A055870(n, m).
From Roger L. Bagula, Apr 03 2010: (Start)
p(x,n) = Sum_{k>=0} (((1 + sqrt(5))^k - (1 - sqrt(5))^k)/(2^k*sqrt(5)))^n*x^k;
t(n,m) = Numerator_coefficients(p(x,n)/x)/2^(1 + floor(n/2));
out(n,m) = t(n,m)/t(n,1). (End)
T(n, k) = Sum_{j=0..k} Fibonacci(k+1-j)^(n+1) * A055870(n+2, j). - Tony Foster III, Aug 20 2018
Sum_{j=0..n-1} a(n-1, n-1-j)*A010048(k+j, n) = Fibonacci(k)^n. - Tony Foster III, Jul 24 2018

A005968 Sum of cubes of first n Fibonacci numbers.

Original entry on oeis.org

0, 1, 2, 10, 37, 162, 674, 2871, 12132, 51436, 217811, 922780, 3908764, 16558101, 70140734, 297121734, 1258626537, 5331629710, 22585142414, 95672204155, 405273951280, 1716768021816, 7272346018247, 30806152127640, 130496954475672, 552793970116297, 2341672834801754

Offset: 0

N. J. A. Sloane

From Alexander Adamchuk, Aug 07 2006: (Start)
The only two prime terms are a(2) = 2 and a(4) = 37.
The prime p divides a(p-1) iff p is in A045468.
The prime p divides a((p-1)/2) iff p is in A047650.
3^4 divides a(p) iff p is in A003628.
3^5 divides a(p) for p = {37,53,109,181,197,269,397,431,541,...}.
3^6 divides a(p) for p = {109,541,...}.
3^7 divides a(p) for p = {557,...}. (End)

Art Benjamin, Timothy A. Carnes, and Benoit Cloitre, Recounting the Sums of Cubes of Fibonacci Numbers, Congressus Numerantium, Proceedings of the Eleventh International Conference on Fibonacci Numbers and their Applications, (William Webb, ed.), Vol 194, pp. 45-51, 2009.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 14.
A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 18.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Art Benjamin, Timothy A. Carnes, and Benoit Cloitre, Counting the Sums of Cubes of Fibonacci Numbers.
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, Hidden Formulas in Fibonacci Tilings, Fibonacci Quart. 54 (2016), no. 1, 23-30.
Index entries for linear recurrences with constant coefficients, signature (4,3,-9,2,1).

Partial sums of A056570.  Cf. A119284 (alternating sum).
Cf. A045468, A047650, A003628.
Sums of other powers: A000071, A001654, A005969, A098531, A098532, A098533, A128697.

[(1/10)*( Fibonacci(3*n+2)-(-1)^(n)*6*Fibonacci(n-1)+5 ): n in [0..30]]; // G. C. Greubel, Jan 17 2018

with(combinat): l[0] := 0: for i from 1 to 50 do l[i] := l[i-1]+fibonacci(i)^3; printf(`%d,`,l[i]) od: # James Sellers, May 29 2000
A005968:=(-1+2*z+z**2)/(z-1)/(z**2+4*z-1)/(z**2-z-1); # conjectured by Simon Plouffe in his 1992 dissertation

f[n_]:=(Fibonacci[n]*Fibonacci[n+1]^2+(-1)^(n-1)*Fibonacci[n-1]+1)/2;Table[f[n],{n,0,5!}] (* Vladimir Joseph Stephan Orlovsky, Nov 22 2010 *)
Accumulate[Fibonacci[Range[0,20]]^3]
CoefficientList[Series[x*(1-2*x-x^2)/((1-x)*(1+x-x^2)*(1-4*x-x^2)), {x, 0, 50}], x] (* Vincenzo Librandi, Jun 09 2013 *)

a(n)=(fibonacci(n)*fibonacci(n+1)^2+(-1)^(n-1)*fibonacci(n-1)+1)/2

a(n)=(fibonacci(3*n+2)-(-1)^(n)*6*fibonacci(n-1)+5)/10

PARI
```
a(n)=sum(i=1,n,fibonacci(i)^3)
```

G.f.: x*(1-2*x-x^2)/((1-x)*(1+x-x^2)*(1-4*x-x^2)). - Ralf Stephan, Apr 23 2004
a(n) = (1/2)*(F(n)*F(n+1)^2 + (-1)^(n-1)*F(n-1) + 1). - Benoit Cloitre, Aug 06 2004
a(n) = Sum_{i=1..n} A000045(i)^3.
a(n) = (1/10)*(F(3*n+2) - (-1)^(n)*6*F(n-1) + 5). - Art Benjamin and Timothy A. Carnes
a(n+5) = 4*a(n+4) + 3*a(n+3) - 9*a(n+2) + 2*a(n+1) + a(n). - Benoit Cloitre, Sep 12 2004

More terms from James Sellers, May 29 2000

A056572 Fifth power of Fibonacci numbers A000045.

Original entry on oeis.org

0, 1, 1, 32, 243, 3125, 32768, 371293, 4084101, 45435424, 503284375, 5584059449, 61917364224, 686719856393, 7615646045657, 84459630100000, 936668172433707, 10387823949447757, 115202670521319424, 1277617458486664901

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..135
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.
Hilary I. Okagbue, Muminu O. Adamu, Sheila A. Bishop, 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.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (8,40,-60,-40,8,1).

Cf. A000045, A007598, A056570-1, A056588, A055870.

Fifth row of array A103323.

[Fibonacci(n)^5: n in [0..30]]; // Vincenzo Librandi, Jun 04 2011

Table[f=Fibonacci[n];f^5,{n,0,12}] (* Vladimir Joseph Stephan Orlovsky, Jul 22 2008 *)
Fibonacci[Range[0,20]]^5 (* Harvey P. Dale, Aug 07 2022 *)

a(n)=fibonacci(n)^5 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = F(n)^5, F(n)=A000045(n).
G.f.: x*p(5, x)/q(5, x) with p(5, x) := sum(A056588(4, m)*x^m, m=0..4)= 1-7*x-16*x^2+7*x^3+x^4 and q(5, x) := sum(A055870(6, m)*x^m, m=0..6)= 1-8*x-40*x^2+60*x^3+40*x^4-8*x^5-x^6 = (1-x-x^2)*(1+4*x-x^2)*(1-11*x-x^2) (factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): sum(A055870(6, m)*a(n-m), m=0..6) = 0, n >= 6; inputs: a(n), n=0..5. a(n) = +8*a(n-1) +40*a(n-2) -60*a(n-3) -40*a(n-4) +8*a(n-5) +a(n-6).
a(n) = (10*F(n) + 5*(-1)^(n+1)*F(3*n) + F(5*n))/25, n >= 0. See the general comment on A111418 regarding the Ozeki reference; here the row 10, 5, 1 of that triangle applies. - Wolfdieter Lang, Aug 25 2012
a(n) = (F(n)^2*(F(3*n)-(-1)^n*3*F(n)))/5. - Gary Detlefs, Jan 07 2013
a(n) = F(n-2)*F(n-1)*F(n)*F(n+1)*F(n+2) + F(n). - Tony Foster III, Apr 11 2018

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

A056571 Fourth power of Fibonacci numbers A000045.

Original entry on oeis.org

0, 1, 1, 16, 81, 625, 4096, 28561, 194481, 1336336, 9150625, 62742241, 429981696, 2947295521, 20200652641, 138458410000, 949005240561, 6504586067281, 44583076827136, 305577005139121, 2094455819300625, 14355614096087056

Offset: 0

Wolfdieter Lang, Jul 10 2000

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

a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using (1/4,3/4)-fences and quarter-squares (1/4 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 (1/8,3/8)-fences and (1/8,7/8)-fences. - Michael A. Allen, Jan 11 2022

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 31.
Donald 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).
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.

Vincenzo Librandi, Table of n, a(n) for n = 0..151
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.
Alfred Brousseau, A sequence of power formulas, Fib. Quart., Vol. 6, No. 1 (1968), pp. 81-83.
Andrej Dujella, A bijective proof of Riordan's theorem on powers of Fibonacci numbers, Discrete Math., Vol. 199, No. 1-3 (1999), pp. 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.
Hideyuki Ohtsuka, Problem N-1220, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 55, No. 4 (2017), p. 368; Gelin-Cesàro Identity Yields a Telescoping Product, Solution to Problem H-790 by Ramya Dutta, ibid., Vol. 56, No. 4 (2018), p. 372.
John Riordan, Generating functions for powers of Fibonacci numbers, Duke. Math. J., Vol. 29, No. 1 (1962), pp. 5-12.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Cf. A000045, A001622, A007598, A056570, A056588, A055870.
First differences of A005969.
Fourth row of array A103323.

[Fibonacci(n)^4: n in [0..30]]; // Vincenzo Librandi, Jun 04 2011

A056571 := proc(n)
    combinat[fibonacci](n)^4 ;
end proc:
seq(A056571(n),n=0..10) ; # R. J. Mathar, Jan 23 2022

Fibonacci[Range[0, 25]]^4 (* Wesley Ivan Hurt, Jan 11 2022 *)

a(n)=fibonacci(n)^4 \\ Charles R Greathouse IV, Oct 07 2016

a(n) = F(n)^4 = A007598(n)^2, F(n) = A000045(n).
G.f.: x*p(4, x)/q(4, x) with p(4, x) := sum(A056588(3, m)*x^m, m=0..3) = 1 - 4*x - 4*x^2 + x^3 = (1+x)*(1 - 5*x + x^2) and q(4, x) := Sum_{m=0..5} A055870(5, m)*x^m = 1 - 5*x - 15*x^2 + 15*x^3 + 5*x^4 - x^5 = (1-x)*(1 + 3*x + x^2)*(1 - 7*x + x^2) (denominator factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): 1*a(n) - 5*a(n-1) - 15*a(n-2) + 15*a(n-3) + 5*a(n-4) - 1*a(n-5) = 0, n >= 5; inputs: a(n), n=0..4.
(1/25)*((-1)^n*(2*F(2*n-2) - 6*F(2*n+1)) + 2*F(4*n-1) + F(4*n) + 6). - Ralf Stephan, May 14 2004
a(n) = F(n-2)*F(n-1)*F(n+1)*F(n+2) + 1 = A244855(n)+1.
Sum_{j=0..n} binomial(n,j)*a(j) = (3^n*A005248(n) - 4*(-1)^n*A000032(n) + 6*2^n)/25. Sum_{j=0..n} (-1)^j*binomial(n,j)*a(j) = -5^((n+1)/2-2)*(A001906(n) + 4*A000045(n)) if n odd. - R. J. Mathar, Oct 16 2006
a(n) = (F(n)*F(3n) - 3*F(n)^2*(-1)^n)/5. - Gary Detlefs, Dec 26 2010
Product_{n>=3} (1 - 1/a(n)) = phi^5/12, where phi is the golden ratio (A001622)(Ohtsuka, 2017). - Amiram Eldar, Dec 02 2021

A103323 Square array T(n,k) read by antidiagonals: powers of Fibonacci numbers.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 8, 9, 5, 1, 1, 16, 27, 25, 8, 1, 1, 32, 81, 125, 64, 13, 1, 1, 64, 243, 625, 512, 169, 21, 1, 1, 128, 729, 3125, 4096, 2197, 441, 34, 1, 1, 256, 2187, 15625, 32768, 28561, 9261, 1156, 55, 1, 1, 512, 6561, 78125, 262144, 371293, 194481, 39304, 3025, 89

Offset: 1

Ralf Stephan, Feb 02 2005

Number of ways to create subsets S(1), S(2),..., S(k-1) such that S(1) is in [n] and for 2<=i<=k-1, S(i) is in [n] and S(i) is disjoint from S(i-1).

			Square array T(n,k) begins:
  1, 1,  2,   3,     5,      8, ...
  1, 1,  4,   9,    25,     64, ...
  1, 1,  8,  27,   125,    512, ...
  1, 1, 16,  81,   625,   4096, ...
  1, 1, 32, 243,  3125,  32768, ...
  1, 1, 64, 729, 15625, 262144, ...
  ...

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, identity 138.

Alois P. Heinz, Antidiagonals n = 1..100, flattened

Rows include A000045, A007598, A056570, A056571, A056572, A056573, A056574.
Main diagonal gives A100399.
Cf. A244003.
Cf. A105317, A254719.

A:= (n, k)-> (<<1|1>, <1|0>>^n)[1, 2]^k:
seq(seq(A(n, 1+d-n), n=1..d), d=1..12);  # Alois P. Heinz, Jun 17 2014

T[n_, k_] := Fibonacci[k]^n; Table[T[n-k+1, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 16 2015 *)

PARI
```
T(n,k)=fibonacci(k)^n
```

T(n, k) = A000045(k)^n, n, k > 0.

T(n, k) = Sum[i_1>=0, Sum[i_2>=0, ... Sum[i_{k-1}>=0, C(n, i_1)*C(n-i_1, i_2)*C(n-i_2, i_3)*...*C(n-i_{k-2}, i_{k-1}) ] ... ]].

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

A264142 T(n,k)=Number of (n+1)X(k+1) arrays of permutations of 0..(n+1)*(k+1)-1 with each element having index change +-(.,.) 0,0 1,-2 or 1,1.

Original entry on oeis.org

2, 8, 4, 18, 27, 8, 45, 144, 125, 16, 125, 720, 1440, 512, 32, 320, 3600, 15488, 14400, 2197, 64, 832, 18000, 168948, 297920, 144000, 9261, 128, 2197, 90000, 1903336, 7001316, 5953600, 1440000, 39304, 256, 5733, 450000, 20768650, 163887724

Offset: 1

R. H. Hardin, Nov 04 2015

Table starts
...2.....8.......18.........45..........125.............320...............832
...4....27......144........720.........3600...........18000.............90000
...8...125.....1440......15488.......168948.........1903336..........20768650
..16...512....14400.....297920......7001316.......163887724........3798503424
..32..2197...144000....5953600....302693857.....15238898080......758116015189
..64..9261..1440000..117534144..12968882161...1376159333484...147813083354689
.128.39304.14400000.2330885440.557669432592.125699825697364.29086230347332416

			Some solutions for n=3 k=4
..0..1..2..6..4....6..1..2..3..4....6..1..8..9..4....6..7..5..9..4
..5..3..7.11.12...11..0.10..8..9....2..0..7.11..3....2..0..1..8..3
.10.17..9.16..8....7..5.12.13.17...10..5.12.16.17...10.11.15.13.17
.15.13.14.18.19...15.16.14.18.19...15.13.14.18.19...12.16.14.18.19

R. H. Hardin, Table of n, a(n) for n = 1..97

Column 1 is A000079.
Column 2 is A056570(n+2).
Row 1 is A264054.

Empirical for column k:
k=1: a(n) = 2*a(n-1)
k=2: a(n) = 3*a(n-1) +6*a(n-2) -3*a(n-3) -a(n-4)
k=3: a(n) = 10*a(n-1) for n>2
k=4: [order 8]
k=5: [order 32]
k=6: [order 95]
Empirical for row n:
n=1: a(n) = 3*a(n-1) -a(n-2) +3*a(n-3) -9*a(n-4) +3*a(n-5) -a(n-6) +3*a(n-7) -a(n-8)
n=2: a(n) = 5*a(n-1) for n>3
n=3: [order 67]
n=4: [order 95]

A056573 Sixth power of Fibonacci numbers A000045.

Original entry on oeis.org

0, 1, 1, 64, 729, 15625, 262144, 4826809, 85766121, 1544804416, 27680640625, 496981290961, 8916100448256, 160005726539569, 2871098559212689, 51520374361000000, 924491486192068809, 16589354847268067929

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..124
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 (13,104,-260,-260,104,13,-1).

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

Sixth row of array A103323.

[Fibonacci(n)^6: n in [0..20]]; // Vincenzo Librandi, Jun 04 2011

with(combinat): A056573:=n->fibonacci(n)^6: seq(A056573(n), n=0..30); # Wesley Ivan Hurt, Jun 29 2015

f[n_]:=Fibonacci[n]^6; lst={}; Do[AppendTo[lst,f[n]],{n,0,5!}]; lst (* Vladimir Joseph Stephan Orlovsky, Feb 12 2010 *)
Fibonacci[Range[0,20]]^6 (* Harvey P. Dale, Sep 21 2024 *)

a(n)=fibonacci(n)^6 \\ Charles R Greathouse IV, Jun 29 2015

a(n) = F(n)^6, where F(n) = A000045(n).
G.f.: x*p(6, x)/q(6, x) with p(6, x) := sum_{m=0..5} A056588(5, m)*x^m = (1-x)*(1 - 11*x - 64*x^2 - 11*x^3 + x^4) and q(6, x) := sum_{m=0..7} A055870(7, m)*x^m = (1+x)*(1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2) (denominator factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): sum_{m=0..7} A055870(7, m)*a(n-m) = 0, n >= 7; inputs: a(n), n=0..6. a(n) = 13*a(n-1) + 104*a(n-2) - 260*a(n-3) - 260*a(n-4) + 104*a(n-5) + 13*a(n-6) - a(n-7).
From Gary Detlefs, Jan 07 2013: (Start)
a(n) = (F(3*n)^2 - (-1)^n*6*F(n)*F(3*n) + 9*F(n)^2)/25.
a(n) = (10*F(n)^3*F(3*n) - F(3*n)^2 + 9*F(n)^2)/25. (End)
a(n+1) = 2*[2*F(n+1)^2-(-1)^n]^3+3*F(n)^2*F(n+1)^2*F(n+2)^2-[F(n)^6+F(n+2)^6] = {Sum(0 <= j <= [n/2]; binomial(n-j, j))}^6, for n (this is Theorem 2.2 (vi) of Azarian's second paper in the references for this sequence). - Mohammad K. Azarian, Jun 29 2015

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A007598 Squared Fibonacci numbers: a(n) = F(n)^2 where F = A000045.

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A056588 Coefficient triangle of certain polynomials.

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A005968 Sum of cubes of first n Fibonacci numbers.

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A056572 Fifth 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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