A001656 -id:A001656 - 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

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

A015441 Generalized Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, 7, 13, 55, 133, 463, 1261, 4039, 11605, 35839, 105469, 320503, 953317, 2876335, 8596237, 25854247, 77431669, 232557151, 697147165, 2092490071, 6275373061, 18830313487, 56482551853, 169464432775, 508359743893, 1525146340543, 4575304803901, 13726182847159

Offset: 0

Olivier Gérard

a(n) is the coefficient of x^(n-1) in the bivariate Fibonacci polynomials F(n)(x,y) = xF(n-1)(x,y) + yF(n-2)(x,y), F(0)(x,y)=0, F(1)(x,y)=1, when y=6x^2. - Mario Catalani (mario.catalani(AT)unito.it), Dec 06 2002
For n>=1: number of length-(n-1) words with letters {0,1,2,3,4,5,6,7} where no two consecutive letters are nonzero, see fxtbook link below. - Joerg Arndt, Apr 08 2011
Starting with offset 1 and convolved with (1, 3, 3, 3, ...) = A003462: (1, 4, 13, 40, ...). - Gary W. Adamson, May 28 2009
a(n) is identical to its inverse binomial transform signed. Differences: A102901. - Paul Curtz, Feb 23 2010
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n>=2, 7*a(n-2) equals the number of 7-colored compositions of n with all parts >=2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011
Pisano period lengths: 1, 1, 1, 2, 20, 1, 6, 2, 3, 20, 5, 2, 12, 6, 20, 4, 16, 3, 18, 20, ... - R. J. Mathar, Aug 10 2012
A015441 and A015518 are the only integer sequences (from the family of homogeneous linear recurrence relation of order 2 with positive integer coefficients with initial values a(0)=0 and a(1)=1) whose ratio a(n+1)/a(n) converges to 3 as n approaches infinity. - Felix P. Muga II, Mar 14 2014
This is an autosequence of the first kind: the array of successive differences shows a main diagonal of zeros and the inverse binomial transform is identical to the sequence (with alternating signs). - Pointed out by Paul Curtz, Dec 05 2016
First two upper diagonals: A000400(n).
This is a variation on the Starhex honeycomb configuration A332243, see illustration in links. It is an alternating pattern of the 2nd iteration of the centered hexagonal numbers A003215 and centered 12-gonal 'Star' numbers A003154. - John Elias, Oct 06 2021

			G.f. = x + x^2 + 7*x^3 + 13*x^4 + 55*x^5 + 133*x^6 + 463*x^7 + 1261*x^8 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
Joerg Arndt, Matters Computational (The Fxtbook), p. 317-318
A. Abdurrahman, CM Method and Expansion of Numbers, arXiv:1909.10889 [math.NT], 2019.
John Elias, Illustration: A Starhex honeycomb variation
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
H. Li and T. MacHenry, Permanents and Determinants, Weighted Isobaric Polynomials, and Integer Sequences, J. Int. Seq. 16 (2013) #13.3.5, example 48.
F. P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, March 2014; Preprint on ResearchGate.
A. G. Shannon and J. V. Leyendekkers, The Golden Ratio family and the Binet equation, Notes on Number Theory and Discrete Mathematics, Vol. 21, No. 2, (2015), 35-42.
Index entries for linear recurrences with constant coefficients, signature (1,6).

Cf. A000079, A003462, A016153, A109466, A122117. Cf. A001656, A087451. Cf. A000400.

I:=[0,1]; [n le 2 select I[n] else Self(n-1) + 6*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018

A015441:=n->(1/5)*((3^n)-((-2)^n)); seq(A015441(n), n=0..30); # Wesley Ivan Hurt, Mar 14 2014

a[n_]:=(MatrixPower[{{1,4},{1,-2}},n].{{1},{1}})[[2,1]]; Table[Abs[a[n]], {n,-1,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
LinearRecurrence[{1,6},{0,1},30] (* Harvey P. Dale, Apr 26 2011 *)
CoefficientList[Series[x/((1 + 2 x) (1 - 3 x)), {x, 0, 29}], x] (* Michael De Vlieger, Dec 05 2016 *)

PARI
```
{a(n) = (3^n - (-2)^n) / 5};
```

[lucas_number1(n,1,-6) for n in range(0, 27)] # Zerinvary Lajos, Apr 22 2009

G.f.: x/((1+2*x)*(1-3*x)).
a(n) = a(n-1) + 6*a(n-2).
a(n) = (1/5)*((3^n)-((-2)^n)). - henryk.wicke(AT)stud.uni-hannover.de
E.g.f.: (exp(3*x) - exp(-2*x))/5. - Paul Barry, Apr 20 2003
a(n+1) = Sum_{k=0..ceiling(n/2)} 6^k*binomial(n-k, k). - Benoit Cloitre, Mar 06 2004
a(n) = (A000244(n) - A001045(n+1)(-1)^n - A001045(n)(-1)^n)/5. - Paul Barry, Apr 27 2004
The binomial transform of [1,1,7,13,55,133,463,...] is A122117. - Philippe Deléham, Oct 19 2006
a(n+1) = Sum_{k=0..n} A109466(n,k)*(-6)^(n-k). - Philippe Deléham, Oct 26 2008
a(n) = 3a(n-1) + (-1)^(n+1)*A000079(n-1). - Paul Curtz, Feb 23 2010
G.f.: Q(0) -1, where Q(k) = 1 + 6*x^2 + (k+2)*x - x*(k+1 + 6*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 06 2013
a(n) = (Sum_{1<=k<=n, k odd} binomial(n,k)*5^(k-1))/2^(n-1). - Vladimir Shevelev, Feb 05 2014
a(-n) = -(-1)^n * a(n) / 6^n for all n in Z. - Michael Somos, Mar 18 2014
From Peter Bala, Apr 01 2015: (Start)
Sum_{n >= 0} a(n+1)*x^n = exp( Sum_{n >= 1} A087451(n)*x^n/n ).
For k = 0, 1, 2, ... and for n >= 1, (5^k)*a(n) | a((5^k)*n).
The expansion of exp( Sum_{n >= 1} a(5*n)/(5*a(n))*x^n/n ) has integral coefficients. Cf. A001656. (End)
From  Peter Bala, Jun 27 2025: (Start)
Sum_{n >= 1} (-6)^n/(a(n)*a(n+1)) = -2, since (-6)^n/(a(n)*a(n+1)) = (-2)^n/a(n) - (-2)^(n+1)/a(n+1) for n >= 1.
The following are examples of telescoping infinite products:
Product_{n >= 0} (1 + 6^n/a(2*n+2)) = 6, since (1 + 6^(2*n-1)/a(4*n))*(1 + 6^(2*n)/a(4*n+2)) = (6 - 4^(n+1)/b(n)) / (6 - 4^n/b(n-1)), where b(n) = (2*4^n + 3*9^n)/5 = A096951(n). Similarly,
Product_{n >= 1} (1 - 6^n/a(2*n+2)) = 3/13.
Product_{n >= 0} (1 + (-6)^n/a(2*n+2)) = 6/5.
Product_{n >= 1} (1 - (-6)^n/a(2*n+2)) = 15/13.
exp( Sum_{n >= 1} a(2*n)/a(n)*x^n/n ) = Sum_{n >= 0} a(n+1)*x^n. (End)

A055870 Signed Fibonomial triangle.

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

Wolfdieter Lang, Jul 10 2000

Row n+1 (n >= 1) of the signed triangle lists the coefficients of the recursion relation for the n-th power of Fibonacci numbers A000045: Sum_{m=0..n+1} T(n+1,m)*(Fibonacci(k-m))^n = 0, k >= n+1; inputs: (Fibonacci(k))^n, k=0..n.
The inverse of the row polynomial p(n,x) := Sum_{m=0..n} T(n,m)*x^m is the g.f. for the column m=n-1 of the Fibonomial triangle A010048.
The row polynomials p(n,x) factorize according to p(n,x) = G(n-1)*p(n-2,-x), with inputs p(0,x)= 1, p(1,x)= 1-x and G(n):= 1 - A000032(n)*x + (-1)^n*x^2. (Derived from Riordan's result and Knuth's exercise).
The row polynomials are the characteristic polynomials of product of the binomial matrix binomial(i,j) and the exchange matrix J_n (matrix with 1's on the antidiagonal, 0 elsewhere). - Paul Barry, Oct 05 2004

			Row polynomial for n=4: p(4,x) = 1-3*x-6*x^2+3*x^3+x^4 = (1+x-x^2)*(1-4*x-x^2). 1/p(4,x) is G.f. for A010048(n+3,3), n >= 0: {1,3,15,60,...} = A001655(n).
For n=3: 1*(Fibonacci(k))^3 - 3*(Fibonacci(k-1))^3 - 6*(Fibonacci(k-2))^3 + 3*(Fibonacci(k-3))^3 + 1*(Fibonacci(k-4))^3 = 0, k >= 4; inputs: (Fibonacci(k))^3, k=0..3.
The triangle begins:
  n\m 0   1     2    3     4      5     6    7   8   9
  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
  ... [_Wolfdieter Lang_, Aug 06 2012; a(7,1) corrected, Oct 10 2012]

D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1969, Vol. 1, pp. 84-5 and 492.

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Katharine A. Ahrens, Combinatorial Applications of the k-Fibonacci Numbers: A Cryptographically Motivated Analysis, Ph. D. thesis, North Carolina State University (2020).
A. T. Benjamin, S. S. Plott, A combinatorial approach to fibonomial coefficients, Fib. Quart. 46/47 (1) (2008/9) 7-9.
A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
H. W. Gould, Extensions of the Hermite g.c.d. theorems for binomial coefficients, Fib Quart. 33 (1995) 386.
E. Kilic, The generalized Fibonomial matrix, Eur. J. Combinat. 31 (1) (2010) 193-209.
Ron Knott, The Fibonomials
Ewa Krot, An introduction to finite fibonomial calculus, Centr. Eur. J. Math. 2 (5) (2004) 754.
A. K. Kwasniewski, Fibonomial cumulative connection constants, arXiv:math/0406006 [math.CO], 2004-2009.
Phakhinkon Phunphayap, Various Problems Concerning Factorials, Binomial Coefficients, Fibonomial Coefficients, and Palindromes, Ph. D. Thesis, Silpakorn University (Thailand 2021).
Phakhinkon Phunphayap, Prapanpong Pongsriiam, Explicit Formulas for the p-adic Valuations of Fibonomial Coefficients, J. Int. Seq. 21 (2018), #18.3.1.
J. Riordan, Generating functions for powers of Fibonacci numbers, Duke. Math. J. 29 (1962) 5-12.
J. Seibert and P. Trojovsky, On some identities for the Fibonomial coefficients, Math. Slov. 55 (2005) 9-19.
P. Trojovsky, On some identities for the Fibonomial coefficients..., Discr. Appl. Math. 155 (15) (2007) 2017.

Cf. A000032, A000045, A001654, A001655, A001656, A001657, A001658, A010048, A051159, A056565, A056566, A056567.
Sums include: A055871 (signed row), A056569 (row).
Central column: A003268.
Cf. A383715.

Fibonomial:= func< n,k | k eq 0 select 1 else (&*[Fibonacci(n-j+1)/Fibonacci(j): j in [1..k]]) >;
[(-1)^Floor((k+1)/2)*Fibonomial(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 20 2024

A055870 := proc(n,k)
    (-1)^floor((k+1)/2)*A010048(n,k) ;
end proc: # R. J. Mathar, Jun 14 2015

T[n_, m_]:= {1,-1,-1,1}[[Mod[m,4] + 1]] * Product[ Fibonacci[n-j+1]/Fibonacci[j], {j, m}];
Table[T[n, m], {n, 0, 10}, {m, 0, n}]//Flatten (* Jean-François Alcover, Jul 05 2013 *)

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([[(-1)^((k+1)//2)*fibonomial(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 20 2024

T(n, m) = (-1)^floor((m+1)/2)*A010048(n, m), where A010048(n, m) := fibonomial(n, m).
G.f. for column m: (-1)^floor((m+1)/2)*x^m/p(m+1, x) with the row polynomial of the (signed) triangle: p(n, x) := Sum_{m=0..n} T(n, m)*x^m.
Sum_{k=0..n} T(n,k) * x^k = exp( -Sum_{k>=1} Fibonacci(n*k)/Fibonacci(k) * x^k/k ). - Seiichi Manyama, May 07 2025

A084175 Jacobsthal oblong numbers.

Original entry on oeis.org

0, 1, 3, 15, 55, 231, 903, 3655, 14535, 58311, 232903, 932295, 3727815, 14913991, 59650503, 238612935, 954429895, 3817763271, 15270965703, 61084037575, 244335800775, 977343902151, 3909374210503, 15637499638215, 62549992960455

Offset: 0

Paul Barry, May 18 2003

Inverse binomial transform is A001019 doubled up.
Binomial transform is A084177.
Partial sums of A003683.

Vincenzo Librandi, Table of n, a(n) for n = 0..500
N. Heninger, E. M. Rains and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, arXiv:math/0509316 [math.NT], 2005-2006.
N. 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.
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.
Index entries for linear recurrences with constant coefficients, signature (3,6,-8).

Except for initial terms, same as A015249 and A084152.

Cf. A001654, A001656, A084158, A084159, A099930.

List([0..30], n-> (2^(2*n+1) -(-2)^n -1)/9); # G. C. Greubel, Sep 21 2019

[(2*4^n-(-2)^n-1)/9: n in [0..30]]; // Vincenzo Librandi, Jun 04 2011

for n from 1 to 25 do print(round(2^n/3)*round(2^(n+1)/3)) od; # Gary Detlefs, Feb 10 2010

Table[(2*4^n -(-2)^n -1)/9, {n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Feb 05 2011, modified by G. C. Greubel, Sep 21 2019 *)
LinearRecurrence[{3,6,-8}, {0,1,3}, 25] (* Jean-François Alcover, Sep 21 2017 *)

a(n)=(2*4^n-(-2)^n-1)/9 \\ Charles R Greathouse IV, Sep 24 2015

def A084175(n): return ((m:=1<Chai Wah Wu, Apr 25 2025

[gaussian_binomial(n, 2, -2) for n in range(1, 26)] # Zerinvary Lajos, May 28 2009

a(n) = A001045(n)*A001045(n+1).
a(n) = (2*4^n - (-2)^n - 1)/9;
a(n) = 3*a(n-1) + 6*a(n-2) - 8*a(n-3), a(0)=0, a(1)=1, a(2)=3.
G.f.: x/((1+2*x)*(1-x)*(1-4*x)).
E.g.f.: (2*exp(4*x) - exp(x) - exp(-2*x))/9.
a(n+1) - 4*a(n) = 1, -1, 3, -5, 11, ... = A001045(n+1) signed. - Paul Curtz, May 19 2008
a(n) = round(2^n/3) * round(2^(n+1)/3). - Gary Detlefs, Feb 10 2010
From Peter Bala, Mar 30 2015: (Start)
The shifted o.g.f. A(x) := 1/( (1 + 2*x)*(1 - x)*(1 - 4*x) ) = 1/(1 - 3*x - 6*x^2 + 8*x^3). Hence A(x) == 1/(1 - 3*x + 3*x^2 - x^3) (mod 9) == 1/(1 - x)^3 (mod 9). It follows by Theorem 1 of Heninger et al. that (A(x))^(1/3) = 1 + x + 4*x^2 + 10*x^3 + ... has integral coefficients.
Sum_{n >= 0} a(n+1)*x^n = exp( Sum_{n >= 1} J(3*n)/J(n)*x^n/n ), where J(n) = A001045(n) are the Jacobsthal numbers. Cf. A001656, A099930. (End)

A001658 Fibonomial coefficients.

Original entry on oeis.org

1, 13, 273, 4641, 85085, 1514513, 27261234, 488605194, 8771626578, 157373300370, 2824135408458, 50675778059634, 909348684070099, 16317540120588343, 292806787575013635, 5254201798026392211, 94282845030238533383, 1691836875411111866723, 30358781826262552258596

Offset: 0

N. J. A. Sloane

It appears that 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) for n > 6. - John W. Layman, Apr 14 2000
Layman's formula is correct. - Wolfdieter Lang, Jul 13 2000
Layman's formula is a consequence of formula 2.8 (p. 116) of Lind (1971). - Dale Gerdemann, May 08 2016

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.
D. A. Lind, A Determinant Involving Binomial Coefficients, Part 1, Part 2, Fibonacci Quarterly 9.2, 1971.
Index entries for linear recurrences with constant coefficients, signature (13, 104, -260, -260, 104, 13, -1).

Cf. A001654, A001656, A001657.

f[n_] := Times @@ Fibonacci[Range[n+1, n+6]]/240; Table[f[n], {n, 0, 20}] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2010 *)
LinearRecurrence[{13,104,-260,-260,104,13,-1},{1,13,273,4641,85085,1514513,27261234},20] (* Harvey P. Dale, Aug 24 2014 *)

b(n,k)=prod(j=1,k,fibonacci(n+j)/fibonacci(j));
vector(20,n,b(n-1,6))  \\ Joerg Arndt, May 08 2016

From Wolfdieter Lang, Jul 13 2000: (Start)
G.f.: 1/(1-13*x-104*x^2+260*x^3+260*x^4-104*x^5-13*x^6+x^7) = 1/((1+x)*(1-3*x+x^2)*(1+7*x+x^2)*(1-18*x+x^2)) (see Comments to A055870).
a(n) = 5*a(n-1)+F(n-5)*Fibonomial(n+5, 5), n >= 1, a(0) = 1; F(n) = A000045(n) (Fibonacci). a(n) = 18*a(n-1)-a(n-2)+((-1)^n)*Fibonomial(n+4, 4), n >= 2; a(0) = 1, a(1) = 13; Fibonomial(n+4, 4) = A001656(n). (End)
From Gary Detlefs, Dec 03 2012: (Start)
a(n) = F(n+1)*F(n+2)*F(n+3)*F(n+4)*F(n+5)*F(n+6)/240.
a(n) = (F(n+5)^2 - F(n+4)^2)*(F(n+3)^4 - 1)/240, where F(n) = A000045(n). (End)
Conjecture: a(n) = F(7)^(n-6) + Sum_{i=3..n-5} F(i-2)F(6)^{i-1}F(7)^{n-i-5} + Sum_{j=3..i} F(i-2)F(j-2)F(5)^{j-1}F(6)^{i-j}F(7)^{n-i-5} + Sum_{k=3..j} F(i-2)F(j-2)F(k-2)F(4)^{k-1}F(5)^{j-k}F(6)^{i-j}F(7)^{n-i-5} + Sum_{l=3..k} F(i-2)F(j-2)F(k-2)F(l-2)F(3)^{l-1}F(4)^{k-l}F(5)^{j-k}F(6)^{i-j}F(7)^{n-i-5} + Sum_{m=3..l} F(i-2)F(j-2)F(k-2)F(l-2)F(m-2)F(m)F(3)^{l-m}F(4)^{k-l}F(5)^{j-k}F(6)^{i-j}F(7)^{n-i-5}, where F(n)=A000045(n). - Dale Gerdemann, May 08 2016
G.f.: exp( Sum_{k>=1} F(7*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

More terms from Wolfdieter Lang, Jul 13 2000

A099930 a(n) = Pell(n) * Pell(n-1) * Pell(n-2) / 10.

Original entry on oeis.org

1, 12, 174, 2436, 34307, 482664, 6791772, 95567064, 1344731653, 18921807828, 266250046986, 3746422451772, 52716164405255, 741772724044560, 10437534301224120, 146867252940711408, 2066579075472320521, 29078974309550454492, 409172219409185308518, 5757490046038128779316

Offset: 3

Ralf Stephan, Nov 03 2004

Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
N. Heninger, E. M. Rains and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, arXiv:math/0509316 [math.NT], 2005-2006.
N. 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.
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.
Index entries for linear recurrences with constant coefficients, signature (12,30,-12,-1).

Cf. A000129. Third column of triangle A099927. Cf. A001656, A084175.

Drop[CoefficientList[Series[x^3/((1+2x-x^2)(1-14x-x^2)),{x,0,20}],x],3] (* or *) LinearRecurrence[{12,30,-12,-1},{1,12,174,2436},20] (* Harvey P. Dale, Feb 26 2012 *)

G.f.: x^3 / ((1+2*x-x^2)*(1-14*x-x^2)).
a(n) = 12*a(n-1) + 30*a(n-2) - 12*a(n-3) - a(n-4); a(3)=1, a(4)=12, a(5)=174, a(6)=2436. - Harvey P. Dale, Feb 26 2012
From Peter Bala, Mar 30 2015: (Start)
The following remarks assume an offset of 0.
The o.g.f. A(x) = 1/( (1 + 2*x - x^2)*(1 - 14*x - x^2) ). Hence A(x) (mod 4) = 1/(1 + 2*x - x^2)^2 (mod 4). It follows by Theorem 1 of Heninger et al. that sqrt(A(x)) = 1 + 6*x + 69*x^2 + 804*x^3 + ... has integral coefficients.
Sum_{n >= 0} a(n+3)*x^n = exp( Sum_{n >= 1} Pell(4*n)/Pell(n)*x^n/n ). Cf. A001656, A084175. (End)
a(n+1) = (1/2)*Sum_{k=1..n-1} ( A014445(k)*A110272(n-k) ) for n > 1. - Michael A. Allen, Jan 25 2023

A088545 Quotient Fibonacci(5n)/(5Fibonacci(n)), where Fibonacci(n) = A000045(n).

Original entry on oeis.org

1, 11, 61, 451, 3001, 20801, 141961, 974611, 6675901, 45768251, 313671601, 2150012161, 14736206161, 101003973851, 692290189501, 4745031073651, 32522917584361, 222915417520961, 1527884938291801, 10472279325329251, 71778069881360701, 491974211042344811, 3372041404278257761

Offset: 1

Lekraj Beedassy, Nov 17 2003

The sequences {Fibonacci(k*n)/(Fibonacci(k)*Fibonacci(n)): n >= 1} are integral in the three cases k = 1 (A000012), k = 2 (A000032) and k = 5 (the present sequence). See Young, Section 4. - Peter Bala, Jan 09 2023

Chatchawan Panraksa and Aram Tangboonduangjit, On Some Arithmetic Properties of a Sequence Related to the Quotient of Fibonacci Numbers, Fibonacci Quart. 55 (2017), no. 1, 21-28.
Paul Thomas Young, p-adic congruences for generalized Fibonacci sequences, The Fibonacci Quarterly, Vol. 32, No. 1, 1994.

Cf. A000032, A000045, A001656, A103326.

a(n)=fibonacci(5*n)/(5*fibonacci(n)); \\ Joerg Arndt, Jul 16 2013

a(n) = 5*Fib(n)^2*(Fib(n)^2 + (-1)^n) + 1 = 5*A007598(n)*A059929(n+1) + 1.
a(n) = A103326(n) / 5.
G.f.: -x*(x^4-4*x^3-9*x^2+6*x+1) / ((x-1)*(x^2-7*x+1)*(x^2+3*x+1)). - Colin Barker, Jul 16 2013
The expansion of exp(Sum_{n >= 1} a(n)*x^n/n) = 1 + x + 6*x^2 + 26*x^3 + 151*x^4 + 851*x^5 + 5101*x^6 + ... has integral coefficients and is equal to G(x)^(1/5), where G(x) is the o.g.f. of A001656. See Young, Theorem 3. - Peter Bala, Jan 09 2023

A103326 a(n) = Fibonacci(5n)/Fibonacci(n).

Original entry on oeis.org

5, 55, 305, 2255, 15005, 104005, 709805, 4873055, 33379505, 228841255, 1568358005, 10750060805, 73681030805, 505019869255, 3461450947505, 23725155368255, 162614587921805, 1114577087604805, 7639424691459005, 52361396626646255, 358890349406803505

Offset: 1

Ralf Stephan, Feb 03 2005

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Fourth row of array A028412.

Cf. A000032, A000045, A001656, A088545.

[Fibonacci(5*n)/Fibonacci(n): n in [1..50]]; // Vincenzo Librandi, Apr 20 2011

p:= (1+5^(1/2))/2: q:=(1-5^(1/2))/2:
seq(simplify(q^(4*n)+(p-2)^n+(q-2)^n+(3*p+2)^n+(-1)^(2*n)/4+3/4),n=1..19);

Vec(-5*x*(x^4-4*x^3-9*x^2+6*x+1)/((x-1)*(x^2-7*x+1)*(x^2+3*x+1)) + O(x^30)) \\ Colin Barker, Jun 03 2016

a(n) = L(4n) + (-1)^n*L(2n) + 1, where L(n) = A000032, the Lucas numbers.
a(n) = 1 + L(n)*L(3n). - Neven Juric, Jan 05 2009
a(n) = 25*(Fibonacci(n)^4 + (-1)^n*Fibonacci(n)^2) + 5. - Gary Detlefs, Dec 22 2012
G.f.: -5*x*(x^4 - 4*x^3 - 9*x^2 + 6*x + 1) /((x - 1)*(x^2 - 7*x + 1)*(x^2 + 3*x + 1)). - Colin Barker, Jul 16 2013
a(n) = 5*A088545(n). - Joerg Arndt, Jul 16 2013
exp(Sum_{n >= 1} a(n)*x^n/n) = Sum_{n >= 0} A001656(n)*x^n. - Peter Bala, Mar 30 2015
a(n) = 1 + (1/2*(7 - 3*sqrt(5)))^n + (1/2*(-3 - sqrt(5)))^n + (1/2*(-3 + sqrt(5)))^n + (1/2*(7 + 3*sqrt(5)))^n. - Colin Barker, Jun 03 2016

More terms from Colin Barker, Jul 16 2013

A177727 a(0)=1; a(n) = a(n-1) * Fibonacci(3+n) * Fibonacci(1+n) / (Fibonacci(n))^2, n > 1.

Original entry on oeis.org

1, 3, 30, 180, 1300, 8736, 60333, 412335, 2829310, 19384200, 132882696, 910735488, 6242420665, 42785803515, 293259265950, 2010026277756, 13776931957468, 94428478367520, 647222466507045, 4436128656563175, 30405678471399166, 208403619747957648, 1428419662108160400

Offset: 0

Roger L. Bagula, May 12 2010

Similar recurrences a(n) = a(n-1)*F(a0+n-1)*F(b0+n-1)/(F(n)*F(c0+n-1)) are:
{a0,b0,c0} = {3,2,1} in A066258.
{a0,b0,c0} = {3,1,1} in A001654.
{a0,b0,c0} = {4,1,1} in A001655 and next for 5,6 as well.

Harry Hochstadt, The Functions of Mathematical Physics, Dover, New York, 1986, p. 93.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1)

Cf. A066258, A001654, A001655, A001656, A001657.

I:=[1, 3, 30, 180, 1300]; [n le 5 select I[n] else 5*Self(n-1)+15*Self(n-2)-15*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..30]]; // Vincenzo Librandi, Nov 18 2011

with (combinat):
A177727 := proc(n)
   if n = 0 then
           1;
   else
           procname(n-1)*fibonacci(3+n)*fibonacci(1+n)/fibonacci(n)^2 ;
   end if;
end proc:
seq(A177727(n),n=0..10) ; # R. J. Mathar, Nov 17 2011

a0 = 4; b0 = 2; c0 = 1;
a[0] = 1;
a[n_] := a[n] = (Fibonacci[(a0 + n - 1)]*Fibonacci[( b0 + n - 1)]/(Fibonacci[n]*Fibonacci[(c0 + n - 1)]))*a[n - 1];
Table[a[n], {n, 0, 30}]
LinearRecurrence[{5,15,-15,-5,1},{1,3,30,180,1300},30] (* Vincenzo Librandi, Nov 18 2011 *)

G.f.: ( -1+2*x ) / ( (x-1)*(x^2+3*x+1)*(x^2-7*x+1) ). - R. J. Mathar, Nov 17 2011

a(n) = A001656(n) - 2*A001656(n-1). - R. J. Mathar, Nov 17 2011

Showing 1-10 of 10 results.

cp's OEIS Frontend

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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A015441 Generalized Fibonacci numbers.

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A055870 Signed Fibonomial triangle.

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A084175 Jacobsthal oblong numbers.

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A088545 Quotient Fibonacci(5n)/(5Fibonacci(n)), where Fibonacci(n) = A000045(n).