A055870 -id:A055870 - cp's OEIS frontend

A001654 Golden rectangle numbers: F(n) * F(n+1), where F(n) = A000045(n) (Fibonacci numbers).

0, 1, 2, 6, 15, 40, 104, 273, 714, 1870, 4895, 12816, 33552, 87841, 229970, 602070, 1576239, 4126648, 10803704, 28284465, 74049690, 193864606, 507544127, 1328767776, 3478759200, 9107509825, 23843770274, 62423800998, 163427632719, 427859097160, 1120149658760

Offset: 0

N. J. A. Sloane

a(n)/A007598(n) ~= golden ratio, especially for larger n. - Robert Happelberg (roberthappelberg(AT)yahoo.com), Jul 25 2005
Let phi be the golden ratio (cf. A001622). Then 1/phi = phi - 1 = Sum_{n>=1} (-1)^(n-1)/a(n), an alternating infinite series consisting solely of unit fractions. - Franz Vrabec, Sep 14 2005
a(n+2) is the Hankel transform of A005807 aerated. - Paul Barry, Nov 04 2008
A more exact name would be: Golden convergents to rectangle numbers. These rectangles are not actually golden (ratio of sides is not phi) but are golden convergents (sides are numerator and denominator of convergents in the continued fraction expansion of phi, whence ratio of sides converges to phi). - Daniel Forgues, Nov 29 2009
The Kn4 sums (see A180662 for definition) of the "Races with Ties" triangle A035317 lead to this sequence. - Johannes W. Meijer, Jul 20 2011
Numbers m such that m(5m+2)+1 or m(5m-2)+1 is a square. - Bruno Berselli, Oct 22 2012
In pairs, these numbers are important in finding binomial coefficients that appear in at least six places in Pascal's triangle. For instance, the pair (m,n) = (40, 104) finds the numbers binomial(n-1,m) = binomial(n,m-1). Two additional numbers are found on the other side of the triangle. The final two numbers appear in row binomial(n-1,m). See A003015. - T. D. Noe, Mar 13 2013
For n>1, a(n) is one-half the area of the trapezoid created by the four points (F(n),L(n)), (L(n),F(n)), (F(n+1), L(n+1)), (L(n+1), F(n+1)) where F(n) = A000045(n) and L(n) = A000032(n). - J. M. Bergot, May 14 2014
[Note on how to calculate: take the two points (a,b) and (c,d) with aa(n) = A067962(n-1) / A067962(n-2), n > 1. - Reinhard Zumkeller, Sep 24 2015
Can be obtained (up to signs) by setting x = F(n)/F(n+1) in g.f. for Fibonacci numbers - see Pongsriiam. - N. J. A. Sloane, Mar 23 2017

			G.f. = x + 2*x^2 + 6*x^3 + 15*x^4 + 40*x^5 + 104*x^6 + 273*x^7 + 714*x^8 + ...

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 9.
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).

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972. See p. 17.
Ömer Egecioglu, Elif Saygi, and Zülfükar Saygi, The Mostar index of Fibonacci and Lucas cubes, arXiv:2101.04740 [math.CO], 2021. Mentions this sequence.
Shalosh B. Ekhad and Doron Zeilberger, Automatic Counting of Tilings of Skinny Plane Regions, arXiv preprint arXiv:1206.4864 [math.CO], 2012.
Sergio Falcon, On the Sequences of Products of Two k-Fibonacci Numbers, American Review of Mathematics and Statistics, March 2014, Vol. 2, No. 1, pp. 111-120.
Dale Gerdemann, Golden Ratio Base Digit Patterns for Columns of the Fibonomial Triangle, "Another interesting pattern is for Golden Rectangle Numbers A001654. I made a short video illustrating this pattern, along with other columns of the Fibonomial Triangle A010048".
Jonny Griffiths and Martin Griffiths, Fibonacci-related sequences via iterated QRT maps, Fib. Q., 51 (2013), 218-227.
James P. Jones and Péter Kiss, Representation of integers as terms of a linear recurrence with maximal index, Acta Academiae Paedagogicae Agriensis, Sectio Mathematicae, 25. (1998) pp. 21-37. See Lemma 4.1 p. 34.
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. 11.
Ronald Orozco López, Simplicial d-Polytopic Numbers Defined on Generalized Fibonacci Polynomials, arXiv:2501.11490 [math.CO], 2025. See p. 6.
C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7.
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
Prapanpong Pongsriiam, Integral Values of the Generating Functions of Fibonacci and Lucas Numbers, College Math. J., 48 (No. 2 2017), pp 97ff.
M. Renault, Dissertation
Wikipedia, Illustration of 273 as a golden rectangle number.
Robert G. Wilson v, Letter to N. J. A. Sloane, circa 1993
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A001655, A001656, A001657, A001658, A010048, A067962.
Cf. A005968, A005969, A098531, A098532, A098533, A119283, A128697.
Cf. A000071, A079472, A080145, A239798, A290565.
Bisection of A006498, A070550, A080239.
First differences of A064831.
Partial sums of A007598.

a001654 n = a001654_list !! n
a001654_list = zipWith (*) (tail a000045_list) a000045_list
-- Reinhard Zumkeller, Jun 08 2013

I:=[0,1,2]; [n le 3 select I[n] else 2*Self(n-1) + 2*Self(n-2) - Self(n-3): n in [1..30]]; // G. C. Greubel, Jan 17 2018

with(combinat): A001654:=n->fibonacci(n)*fibonacci(n+1):
seq(A001654(n), n=0..28); # Zerinvary Lajos, Oct 07 2007

LinearRecurrence[{2,2,-1}, {0,1,2}, 100] (* Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *)
Times@@@Partition[Fibonacci[Range[0,30]],2,1] (* Harvey P. Dale, Aug 18 2011 *)
Accumulate[Fibonacci[Range[0, 30]]^2] (* Paolo Xausa, May 31 2024 *)

A001654(n)=fibonacci(n)*fibonacci(n+1);

b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
vector(30, n, b(n-1, 2))  \\ Joerg Arndt, May 08 2016

from sympy import fibonacci as F
def a(n): return F(n)*F(n + 1)
[a(n) for n in range(101)] # Indranil Ghosh, Aug 03 2017

from math import prod
from gmpy2 import fib2
def A001654(n): return prod(fib2(n+1)) # Chai Wah Wu, May 19 2022

a(n) = A010048(n+1, 2) = Fibonomial(n+1, 2).
a(n) = A006498(2*n-1).
a(n) = a(n - 1) + A007598(n) = a(n - 1) + A000045(n)^2 = Sum_{j <= n} Fibonacci(j)^2. - Henry Bottomley, Feb 09 2001 [corrected by Ridouane Oudra, Apr 12 2025]
For n > 0, 1 - 1/a(n+1) = Sum_{k=1..n} 1/(F(k)*F(k+2)) where F(k) is the k-th Fibonacci number. - Benoit Cloitre, Aug 31 2002.
G.f.: x/(1-2*x-2*x^2+x^3) = x/((1+x)*(1-3*x+x^2)). (Simon Plouffe in his 1992 dissertation; see Comments to A055870),
a(n) = 3*a(n-1) - a(n-2) - (-1)^n = -a(-1-n).
Let M = the 3 X 3 matrix [1 2 1 / 1 1 0 / 1 0 0]; then a(n) = the center term in M^n *[1 0 0]. E.g., a(5) = 40 since M^5 * [1 0 0] = [64 40 25]. - Gary W. Adamson, Oct 10 2004
a(n) = Sum{k=0..n} Fibonacci(k)^2. The proof is easy. Start from a square (1*1). On the right side, draw another square (1*1). On the above side draw a square ((1+1)*(1+1)). On the left side, draw a square ((1+2)*(1+2)) and so on. You get a rectangle (F(n)*F(1+n)) which contains all the squares of side F(1), F(2), ..., F(n). - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 19 2007
With phi = (1+sqrt(5))/2, a(n) = round((phi^(2*n+1))/5) = floor((1/2) + (phi^(2*n+1))/5), n >= 0. - Daniel Forgues, Nov 29 2009
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3), a(1)=1, a(2)=2, a(3)=6. - Sture Sjöstedt, Feb 06 2010
a(n) = (A002878(n) - (-1)^n)/5. - R. J. Mathar, Jul 22 2010
a(n) = 1/|F(n+1)/F(n) - F(n)/F(n-1)| where F(n) = Fibonacci numbers A000045. b(n) = F(n+1)/F(n) - F(n)/F(n-1): 1/1, -1/2, 1/6, -1/15, 1/40, -1/104, ...; c(n) = 1/b(n) = a(n)*(-1)^(n+1): 1, -2, 6, -15, 40, -104, ... (n=1,2,...). - Thomas Ordowski, Nov 04 2010
a(n) = (Fibonacci(n+2)^2 - Fibonacci(n-1)^2)/4. - Gary Detlefs, Dec 03 2010
Let d(n) = n mod 2, a(0)=0 and a(1)=1. For n > 1, a(n) = d(n) + 2*a(n-1) + Sum_{k=0..n-2} a(k). - L. Edson Jeffery, Mar 20 2011
From Tim Monahan, Jul 11 2011: (Start)
a(n+1) = ((2+sqrt(5))*((3+sqrt(5))/2)^n+(2-sqrt(5))*((3-sqrt(5))/2)^n+(-1)^n)/5.
a(n) = ((1+sqrt(5))*((3+sqrt(5))/2)^n+(1-sqrt(5))*((3-sqrt(5))/2)^n-2*(-1)^n)/10. (End)
From Wolfdieter Lang, Jul 21 2012: (Start)
a(n) = (2*A059840(n+2) - A027941(n))/3, n >= 0, with A059840(n+2) = Sum_{k=0..n} F(k)*F(k+2) and A027941(n) = A001519(n+1) - 1, n >= 0, where A001519(n+1) = F(2*n+1). (End)
a(n) = (-1)^n * Sum_{k=0..n} (-1)^k*F(2*k), n >= 0. - Wolfdieter Lang, Aug 11 2012
a(-1-n) = -a(n) for all n in Z. - Michael Somos, Sep 19 2014
0 = a(n)*(+a(n+1) - a(n+2)) + a(n+1)*(-2*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014
a(n) = (L(2*n+1) - (-1)^n)/5 with L(k) = A000032(k). - J. M. Bergot, Apr 15 2016
E.g.f.: ((3 + sqrt(5))*exp((5+sqrt(5))*x/2) - 2*exp((2*x)/(3+sqrt(5))+x) - 1 - sqrt(5))*exp(-x)/(5*(1 + sqrt(5))). - Ilya Gutkovskiy, Apr 15 2016
From Klaus Purath, Apr 24 2019: (Start)
a(n) = A061646(n) - Fibonacci(n-1)^2.
a(n) = (A061646(n+1) - A061646(n))/2. (End)
a(n) = A226205(n+1) + (-1)^(n+1). - Flávio V. Fernandes, Apr 23 2020
Sum_{n>=1} 1/a(n) = A290565. - Amiram Eldar, Oct 06 2020
Product_{n>=2} (1 + (-1)^n/a(n)) = phi^2/2 (A239798). - Amiram Eldar, Dec 02 2024
G.f.: x * exp( Sum_{k>=1} F(3*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

Extended by Wolfdieter Lang, Jun 27 2000

A015448 a(0) = 1, a(1) = 1, and a(n) = 4*a(n-1) + a(n-2) for n >= 2.

Original entry on oeis.org

1, 1, 5, 21, 89, 377, 1597, 6765, 28657, 121393, 514229, 2178309, 9227465, 39088169, 165580141, 701408733, 2971215073, 12586269025, 53316291173, 225851433717, 956722026041, 4052739537881, 17167680177565, 72723460248141, 308061521170129, 1304969544928657, 5527939700884757

Offset: 0

Olivier Gérard

If one deletes the leading 0 in A084326, takes the inverse binomial transform, and adds a(0)=1 in front, one obtains this sequence here. - Al Hakanson (hawkuu(AT)gmail.com), May 02 2009
For n >= 1, row sums of triangle
  m |k=0     1     2     3     4     5     6     7
====+=============================================
  0 |  1
  1 |  1     4
  2 |  1     4    16
  3 |  1     8    16    64
  4 |  1     8    48    64   256
  5 |  1    12    48   256   256  1024
  6 |  1    12    96   256  1280  1024  4096
  7 |  1    16    96   640  1280  6144  4096 16384
which is triangle for numbers 4^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 12 2012
a(n) = a(n;-2) = 3^n*Sum_{k=0..n} binomial(n,k)*F(k+1)*(-2/3)^k, where a(n;d), n=0,1,...,d, denotes the delta-Fibonacci numbers defined in comments to A000045 (see also the papers of Witula et al.). We note that (see A033887) F(3n+1) = 3^n*a(n,2/3) = Sum_{k=0..n} binomial(n,k)*F(k-1)*(-2/3)^k, which implies F(3n+1) + 3^(-n)*a(n) = Sum_{k=0..n} binomial(n,k)*L(k)*(-2/3)^k, where L(k) denotes the k-th Lucas number. - Roman Witula, Jul 12 2012
a(n+1) is (for n >= 0) the number of length-n strings of 5 letters {0,1,2,3,4} with no two adjacent nonzero letters identical. The general case (strings of L letters) is the sequence with g.f. (1+x)/(1-(L-1)*x-x^2). - Joerg Arndt, Oct 11 2012
Starting with offset 1 the sequence is the INVERT transform of (1, 4, 4*3, 4*3^2, 4*3^3, ...); i.e., of A003946: (1, 4, 12, 36, 108, ...). - Gary W. Adamson, Aug 06 2016
a(n+1) equals the number of quinary sequences of length n such that no two consecutive terms differ by 3. - David Nacin, May 31 2017

T. D. Noe, Table of n, a(n) for n = 0..200
Joerg Arndt, Matters Computational (The Fxtbook), pp. 313-315.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876 (see Corollary 1 (vi)).
Gary Detlefs and Wolfdieter Lang, Improved Formula for the Multi-Section of the Linear Three-Term Recurrence Sequence, arXiv:2304.12937 [math.CO], 2023.
Amelia Gilson, Hadley Killen, Steven J. Miller, Nadia Razek, Joshua M. Siktar, and Liza Sulkin, Zeckendorf's Theorem Using Indices in an Arithmetic Progression, arXiv:2005.10396 [math.NT], 2020.
Edyta Hetmaniok, Bozena Piatek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Math. 15 (2017), 477-485.
I. M. Gessel and Ji Li, Compositions and Fibonacci identities, J. Int. Seq. 16 (2013) 13.4.5.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Bahar Kuloğlu, Engin Özkan, and Marin Marin, Fibonacci and Lucas Polynomials in n-gon, An. Şt. Univ. Ovidius Constanţa (Romania 2023) Vol. 31, No 2, 127-140.
Roman Witula, Binomials transformation formulae of scaled Lucas numbers, Demonstratio Math. 46 (2013), 15-27.
R. Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math 3 (2009) 310-329, MR2555042
Index entries for linear recurrences with constant coefficients, signature (4,1).

Cf. A001076, A147722 (INVERT transform), A109499 (INVERTi transform), A154626 (Binomial transform), A086344 (inverse binomial transform), A003946, A049310.

Cf. A000032, A000045, A014445, A014448, A033887, A046854, A055830, A055870, A084326, A167808.

[Fibonacci(3*n-1): n in [0..40]]; // Vincenzo Librandi, Jul 04 2015

with(combinat): a:=n->fibonacci(n,4)-3*fibonacci(n-1,4): seq(a(n), n=1..23); # Zerinvary Lajos, Apr 04 2008

Fibonacci/@(3*Range[0,30]-1) (* Vladimir Joseph Stephan Orlovsky, Mar 01 2010 *)
LinearRecurrence[{4,1},{1,1},30] (* Harvey P. Dale, May 15 2019 *)

a[0]:1$
a[1]:1$
a[n]:=4*a[n-1]+a[n-2]$
A015448(n):=a[n]$
makelist(A015448(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

a(n) = fibonacci(3*n-1); \\ Altug Alkan, Dec 10 2015

a(n) = Fibonacci(3n-1) = ( (1+sqrt(5))*(2-sqrt(5))^n - (1-sqrt(5))*(2+sqrt(5))^n )/ (2*sqrt(5)).
O.g.f.: (1-3*x)/(1-4*x-x^2). - Len Smiley, Dec 09 2001
a(n) = Sum_{k=0..n} 3^k*A055830(n,k). - Philippe Deléham, Oct 18 2006
a(n) = upper left term in the 2 X 2 matrix [1,2; 2,3]^n. - Gary W. Adamson, Mar 02 2008
[a(n), A001076(n)] = [1,4; 1,3]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = A167808(3*n-1) for n > 0. - Reinhard Zumkeller, Nov 12 2009
a(n) = Fibonacci(3n+1) mod Fibonacci(3n), n > 0.
a(n) = (A000032(3*n)-Fibonacci(3*n))/2 = (A014448(n)-A014445(n))/2.
For n >= 2, a(n) = F_n(4) + F_(n+1)(4), where F_n(x) is a Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} binomial(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012
a(n) = A001076(n+1) - 3*A001076(n). - R. J. Mathar, Jul 12 2012
From Gary Detlefs and Wolfdieter Lang, Aug 20 2012: (Start)
a(n) = (5*F(n)^3 + 5*F(n-1)^3 + 3*(-1)^n*F(n-2))/2,
a(n) = (F(n+1)^3 + 2*F(n)^3 - F(n-2)^3)/2, n >= 0, with F(-1) = 1 and F(-2) = -1. Second line from first one with 3*(-1)^n* F(n-2) = F(n-1)^3 - 4*F(n-2)^3 - F(n-3)^3 (in Koshy's book, p. 89, 32. (with a - sign) and 33. For the Koshy reference see A000045) and the F^3 recurrence (see row n=4 of A055870, or Koshy p. 87, 1.). First line from the preceding R. J. Mathar formula with F(3*n) = 5*F(n)^3 + 3*(-1)^n*F(n) (Koshy p. 89, 46.) and the above mentioned formula, Koshy's 32. and 33., with n -> n+2 in order to eliminate F(n+1)^3. (End)
For n > 0, a(n) = L(n-1)*L(n)*F(n) + F(n+1)*(-1)^n with L(n)=A000032(n). - J. M. Bergot, Dec 10 2015
For n > 1, a(n)^2 is the denominator of continued fraction [4,4,...,4, 6, 4,4,...4], which has n-1 4's before, and n-1 4's after, the middle 6. - Greg Dresden, Sep 18 2019
From Gary Detlefs and Wolfdieter Lang, Mar 06 2023: (Start)
a(n) = A001076(n) + A001076(n-1), with A001076(-1) = 1. See the R. J. Mathar formula above.
a(n+1) = i^n*(S(n-1,-4*i) - i*S(n-2,-4*i)), for n >= 0, with i = sqrt(-1), and the Chebyshev S-polynomials (see A049310) with S(n, -1) = 0. From the simplified Fibonacci trisection formula for {F(3*n+2)}_{n>=0}. (End)
a(n) = Sum_{k=0..n} A046854(n-1,k)*4^k. - R. J. Mathar, Feb 10 2024
E.g.f.: exp(2*x)*(5*cosh(sqrt(5)*x) - sqrt(5)*sinh(sqrt(5)*x))/5. - Stefano Spezia, Jun 03 2024

A010048 Triangle of Fibonomial coefficients, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 3, 6, 3, 1, 1, 5, 15, 15, 5, 1, 1, 8, 40, 60, 40, 8, 1, 1, 13, 104, 260, 260, 104, 13, 1, 1, 21, 273, 1092, 1820, 1092, 273, 21, 1, 1, 34, 714, 4641, 12376, 12376, 4641, 714, 34, 1, 1, 55, 1870, 19635, 85085, 136136, 85085, 19635, 1870, 55, 1

Offset: 0

N. J. A. Sloane

Conjecture: polynomials with (positive) Fibonomial coefficients are reducible iff n odd > 1. - Ralf Stephan, Oct 29 2004

			First few rows of the triangle T(n, k) are:
  n\k 0   1    2     3     4      5      6      7     8   9  10
   0: 1
   1: 1   1
   2: 1   1    1
   3: 1   2    2     1
   4: 1   3    6     3     1
   5: 1   5   15    15     5      1
   6: 1   8   40    60    40      8      1
   7: 1  13  104   260   260    104     13      1
   8: 1  21  273  1092  1820   1092    273     21     1
   9: 1  34  714  4641 12376  12376   4641    714    34   1
  10: 1  55 1870 19635 85085 136136  85085  19635  1870  55   1
... - Table extended and reformatted by _Wolfdieter Lang_, Oct 10 2012
For n=7 and k=3, n - k + 1 = 7 - 3 + 1 = 5, so T(7,3) = F(7)*F(6)*F(5)/( F(3)*F(2)*F(1)) = 13*8*5/(2*1*1) = 520/2 = 260. - _Michael B. Porter_, Sep 26 2016

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 15.
D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, p. 84 and 492.

T. D. Noe, Rows n = 0..50 of triangle, flattened
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
A. T. Benjamin and S. S. Plott, A combinatorial approach to fibonomial coefficients, Fib. Quart. 46/47 (1) (2008/9) 7-9.
A. Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972.
Johann Cigler, Pascal triangle, Hoggatt matrices, and analogous constructions, arXiv:2103.01652 [math.CO], 2021.
M. Dziemianczuk, Cobweb Sequences Map, See sequence (4).2. [Dead link]
Tom Edgar and Michael Z. Spivey, Multiplicative functions, generalized binomial coefficients, and generalized Catalan numbers, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.6.
P. F. F. Espinosa, J. F. González, J. P. Herrán, A. M. Cañadas, and J. L. Ramírez, On some relationships between snake graphs and Brauer configuration algebras, Algebra Disc. Math. (2022) Vol. 33, No. 2, 29-59.
Sergio 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.
Dale Gerdemann, Golden Ratio Base Digit Patterns for Columns of the Fibonomial Triangle, "Another interesting pattern is for Golden Rectangle Numbers A001654. I made a short video illustrating this pattern, along with other columns of the Fibonomial Triangle A010048".
Dale K. Hathaway and Stephen L. Brown, Fibonacci Powers and a Fascinating Triangle, The College Mathematics Journal, 28 (No. 2, 1997), 124-128. See Fig. 1.
Ron Knott, The Fibonomials.
E. Krot, An introduction to finite Fibonomial calculus, arXiv:math/0503210 [math.CO], 2005.
E. Krot, Further developments in Fibonomial calculus, arXiv:math/0410550 [math.CO], 2004.
D. Marques and P. Trojovsky, On Divisibility of Fibonomial Coefficients by 3, J. Int. Seq. 15 (2012) #12.6.4.
D. Marques and P. Trojovsky, The p-adic order of some fibonomial coefficients, J. Int. Seq. 18 (2015) # 15.3.1.
Romeo Mestrovic, Lucas' theorem: its generalizations, extensions and applications (1878--2014), arXiv preprint arXiv:1409.3820 [math.NT], 2014.
Phakhinkon Phunphayap, Various Problems Concerning Factorials, Binomial Coefficients, Fibonomial Coefficients, and Palindromes, Ph. D. Thesis, Silpakorn University (Thailand 2021).
Phakhinkon Phunphayap and Prapanpong Pongsriiam, Explicit Formulas for the p-adic Valuations of Fibonomial Coefficients, J. Int. Seq. 21 (2018), #18.3.1.
C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7.
C. J. Pita Ruiz Velasco, Sums of Products of s-Fibonacci Polynomial Sequences, J. Int. Seq. 14 (2011) # 11.7.6.
T. M. Richardson, The Filbert Matrix, arXiv:math/9905079 [math.RA], 1992.
Bruce Sagan, Two Binomial Coefficient Analogues, Slides, 2013.
Jeremiah Southwick, A Conjecture concerning the Fibonomial Triangle, arXiv:1604.04775 [math.NT], 2016.
Ralf Stephan, A recurrence for the fibonomials.
Eric Weisstein's World of Mathematics, Fibonacci Coefficient, q-Binomial Coefficient.

Cf. A055870 (signed version of triangle).
Columns include: A000045, A001654, A001655, A001656, A001657, A001658, A056565, A056566, A056567.
Sums include: A056569 (row), A181926 (antidiagonal), A181927 (row square-sums).
Cf. A003267 and A003268 (central Fibonomial coefficients), A003150 (Fibonomial Catalan numbers), A144712, A099927, A385732/A385733 (Lucas).

Fibonomial:= func< n,k | k eq 0 select 1 else (&*[Fibonacci(n-j+1)/Fibonacci(j): j in [1..k]]) >;
[Fibonomial(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 20 2024

A010048 := proc(n,k)
    mul(combinat[fibonacci](i),i=n-k+1..n)/mul(combinat[fibonacci](i),i=1..k) ;
end proc:
seq(seq(A010048(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Feb 05 2015

f[n_, k_] := Product[ Fibonacci[n - j + 1]/Fibonacci[j], {j, k}]; Table[ f[n, i], {n, 0, 10}, {i, 0, n}] (* Robert G. Wilson v, Dec 04 2009 *)
Column[Round@Table[GoldenRatio^(k(n-k)) QBinomial[n, k, -1/GoldenRatio^2], {n, 0, 10}, {k, 0, n}], Center] (* Round is equivalent to FullSimplify here, but is much faster - Vladimir Reshetnikov, Sep 25 2016 *)
T[n_, k_] := With[{c = ArcCsch[2] - I Pi/2}, Product[I^j Sinh[c j], {j, k + 1, n}] / Product[I^j Sinh[c j], {j, 1, n - k}]]; Table[Simplify[T[n, k]], {n, 0, 10}, {k, 0, n}] // Flatten  (* Peter Luschny, Jul 08 2025 *)

ffib(n):=prod(fib(k),k,1,n);
fibonomial(n,k):=ffib(n)/(ffib(k)*ffib(n-k));
create_list(fibonomial(n,k),n,0,20,k,0,n); /* Emanuele Munarini, Apr 02 2012 */

T(n, k) = prod(j=0, k-1, fibonacci(n-j))/prod(j=1, k, fibonacci(j));
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print); \\ Michel Marcus, Jul 20 2018

def fibonomial(n,k): return 1 if k==0 else product(fibonacci(n-j+1)/fibonacci(j) for j in range(1,k+1))
flatten([[fibonomial(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 20 2024

T(n, k) = ((n, k)) = (F(n)*F(n-1)*...*F(n-k+1))/(F(k)*F(k-1)*...*F(1)), F(i) = Fibonacci numbers A000045.
T(n, k) = Fibonacci(n-k-1)*T(n-1, k-1) + Fibonacci(k+1)*T(n-1, k).
T(n, k) = phi^(k*(n-k)) * C(n, k)A001622%20is%20the%20golden%20ratio,%20and%20C(n,%20k)_q%20is%20the%20q-binomial%20coefficient.%20-%20_Vladimir%20Reshetnikov">{-1/phi^2}, where phi = (1+sqrt(5))/2 = A001622 is the golden ratio, and C(n, k)_q is the q-binomial coefficient. - _Vladimir Reshetnikov, Sep 26 2016
G.f. of column k: x^k * exp( Sum_{j>=1} Fibonacci((k+1)*j)/Fibonacci(j) * x^j/j ). - Seiichi Manyama, May 07 2025
T(n, k) = Product_{j=k+1..n} i^j*sinh(c*j) / Product_{j=1..n-k} i^j*sinh(c*j) where c = arccsch(2) - i*Pi/2 and i is the imaginary unit. If you substitute sinh by cosh you get the Lucas triangle A385732/A385733, which is a rational triangle. - Peter Luschny, Jul 08 2025

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

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

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

A001656 Fibonomial coefficients.

Original entry on oeis.org

1, 5, 40, 260, 1820, 12376, 85085, 582505, 3994320, 27372840, 187628376, 1285992240, 8814405145, 60414613805, 414088493560, 2838203264876, 19453338487220, 133335155341960, 913892777190965, 6263914210945105

Offset: 0

N. J. A. Sloane

			G.f. = 1 + 5*x + 40*x^2 + 260*x^3 + 1820*x^4 + 12376*x^5 + 85085*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).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Alfred Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972. See p. 17.
Nadia Heninger, E. M. Rains, and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.
Thomas Koshy, Infinite Sums Involving Jacobsthal Polynomial Products Revisited, The Fibonacci Quarterly, Vol. 60, No. 1 (2022), pp. 3-14.
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.
Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Cf. A001622, A001654, A001655, A001657, A001658, A084175, A099930.

with (combinat): a:=n->1/6*fibonacci(n)*fibonacci(n+1)*fibonacci(n+2)*fibonacci(n+3): seq(a(n), n=1..18); # Zerinvary Lajos, Oct 07 2007
A001656:=-1/(z-1)/(z**2-7*z+1)/(z**2+3*z+1); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation

Table[(Fibonacci[n+3]*Fibonacci[n+2]*Fibonacci[n+1]*Fibonacci[n])/6,{n,0,50}] (* Vladimir Joseph Stephan Orlovsky, Nov 23 2009 *)
LinearRecurrence[{5,15,-15,-5,1},{1,5,40,260,1820},20] (* Vincenzo Librandi, Aug 02 2012 *)
Times@@@Partition[Fibonacci[Range[30]],4,1]/6 (* Harvey P. Dale, Oct 13 2016 *)

b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
vector(20, n, b(n-1, 4))  \\ Joerg Arndt, May 08 2016

a(n) = ((4+n, 4)) (see A010048), or fibonomial(4+n, 4).
G.f.: 1/(1-5*x-15*x^2+15*x^3+5*x^4-x^5) = 1/((1-x)*(1+3*x+x^2)*(1-7*x+x^2)) (see Comments to A055870). a(n)= 7*a(n-1)-a(n-2)+((-1)^n)*fibonomial(n+2, 2), n >= 2; a(0)=1, a(1)=5; fibonomial(n+2, 2)= A001654(n+1).
a(n) = Product_{k=1..n} Fibonacci(k+4)/Fibonacci(k). - Gary Detlefs, Feb 06 2011
a(n) = (F(n+3)^2-F(n+2)^2)*F(n+3)*F(n+2)/6, where F(n) is the n-th Fibonacci number. - Gary Detlefs, Oct 12 2011
a(n) = a(-5-n) for all n in Z. - Michael Somos, Sep 19 2014
0 = a(n)*(+a(n+1) - 2*a(n+2)) + a(n+1)*(-5*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014
From Peter Bala, Mar 30 2015: (Start)
The o.g.f. A(x) = 1/(1 - 5*x - 15*x^2 + 15*x^3 + 5*x^4 - x^5). Hence A(x) (mod 25) = 1/(1 - 5*x + 10*x^2 - 10^x^3 + 5*x^4 - x^5) (mod 25) = 1/(1 - x)^5 (mod 25). It follows by Theorem 1 of Heninger et al. that A(x)^(1/5) = 1 + x + 6*x^2 + 26*x^3 + ... has integral coefficients.
Sum_{n >= 0} a(n)*x^n = exp( Sum_{n >= 1} Fibonacci(5*n)/Fibonacci(n)*x^n/n ). Cf. A084175, A099930. (End)
Sum_{n>=0} 1/a(n) = 51/2 - 15*phi, where phi is the golden ratio (A001622) (Koshy, 2022, section 3.3, p. 9). - Amiram Eldar, Jan 23 2025

Corrected and extended by Wolfdieter Lang, Jun 27 2000

More terms from Vladimir Joseph Stephan Orlovsky, Nov 23 2009

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A055871 Row sums of signed Fibonomial triangle A055870.

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A001654 Golden rectangle numbers: F(n) * F(n+1), where F(n) = A000045(n) (Fibonacci numbers).

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A010048 Triangle of Fibonomial coefficients, read by rows.

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A056588 Coefficient triangle of certain polynomials.

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