A003266 -id:A003266 - cp's OEIS frontend

A372949 a(n) = 2f(2n)/(f(n)*f(n+2)) where f = A003266.

1, 2, 8, 91, 2618, 199716, 39690618, 20689636692, 28215085220016, 100763710906257557, 942012688139052139766, 23056957423045790791793932, 1477460537993359748548214768630, 247860656992078740305125996374953260, 108861324945456389643061592667638024842480

Offset: 1

Kendra Killpatrick, May 17 2024

nonn

Fibonacci analog of the super ballot numbers.

a(n) is also the generalized FiboCatalan number for r=1. Proof that the formula always gives a positive integer can be found in a recent paper of K. Killpatrick. The sequence is the Fibonacci analog of the super ballot numbers given by Gessel (A007054). The sequence is also the Fibonacci analog of the generalized Catalan numbers, J_r*(2n)!/(n!*(n+r+1)!) where J_r=(2r+1)!/r!, for r=1. Gessel defined the generalized Catalan numbers and proved they are integers.

			a(5) = 2*f(10)/(f(5)*f(7)) = 2*122522400/(30*3120) = 2618, where f=A003266.

Ira M. Gessel, Super ballot numbers, J. Symb. Comput. 14 (1992), 179-194.
Kendra Killpatrick, Super FiboCatalan Numbers and their Lucas Analogues, arXiv:2308.13457 [math.CO], (2023).

Cf. A003266, A007054.

a[n_]:=2Fibonorial[2n]/(Fibonorial[n]Fibonorial[n+2]); Array[a,15] (* Stefano Spezia, May 23 2024 *)

a(n) ~ 10 * phi^((n-3)*(n+1)) / A062073, where phi = A001622 is the golden ratio. - Vaclav Kotesovec, May 29 2024

A070826 One half of product of first n primes A000040.

Original entry on oeis.org

1, 3, 15, 105, 1155, 15015, 255255, 4849845, 111546435, 3234846615, 100280245065, 3710369067405, 152125131763605, 6541380665835015, 307444891294245705, 16294579238595022365, 961380175077106319535, 58644190679703485491635, 3929160775540133527939545, 278970415063349480483707695

Offset: 1

Wolfdieter Lang, May 10 2002

Also, with offset 0, product of first n odd primes. - N. J. A. Sloane, Feb 26 2017
Identical to A002110(n)/2, n>=1.
a(n+1) is the least odd number with exactly n distinct prime divisors. - Labos Elemer, Mar 24 2003
Also, odd numbers n for which sigma(n)*phi(n)/n^2 reaches a new record low, monotonically decreasing to the lower bound 8/Pi^2. - M. F. Hasler, Jul 08 2025

Vincenzo Librandi, Table of n, a(n) for n = 1..200

Cf. A003266 (for Fibonacci), A070825 (for Lucas), A003046 (for Catalan).
Cf. also A002110, A024451, A060389, A091852, A276086, A203008 [= A003415(a(1+n))].
Range of A196529.

a:=n->mul(ithprime(j), j=2..n):seq(a(n), n=1..17); # Zerinvary Lajos, Aug 24 2008

Rest[ FoldList[ Times, 1, Prime[ Range[ 18]] ]]/2 (* Robert G. Wilson v, Feb 17 2004 *)
FoldList[Times, 1, Prime[Range[2, 18]]] (* Zak Seidov, Jan 26 2009 *)

a(n) = prod(k=2, n, prime(k)) \\ Michel Marcus, Mar 25 2017, simplified by M. F. Hasler, Jul 09 2025

from sympy import primorial
def A070826(n): return primorial(n)>>1 # Chai Wah Wu, Jul 21 2022

a(n) = A002110(n)/2.
From Antti Karttunen, Feb 06 2024: (Start)
a(1) = 1, and for n > 1, a(n) = A276086(A060389(n-1)).
a(n) = A024451(n) - 2*A203008(n-1).
(End)
a(n) = A000040(n)*a(n-1) for n > 1, a(1) = 1. - M. F. Hasler, Jul 09 2025

Formula corrected by Gary Detlefs, Dec 07 2011

A079586 Decimal expansion of Sum_{k>=1} 1/F(k) where F(k) is the k-th Fibonacci number A000045(k).

Original entry on oeis.org

3, 3, 5, 9, 8, 8, 5, 6, 6, 6, 2, 4, 3, 1, 7, 7, 5, 5, 3, 1, 7, 2, 0, 1, 1, 3, 0, 2, 9, 1, 8, 9, 2, 7, 1, 7, 9, 6, 8, 8, 9, 0, 5, 1, 3, 3, 7, 3, 1, 9, 6, 8, 4, 8, 6, 4, 9, 5, 5, 5, 3, 8, 1, 5, 3, 2, 5, 1, 3, 0, 3, 1, 8, 9, 9, 6, 6, 8, 3, 3, 8, 3, 6, 1, 5, 4, 1, 6, 2, 1, 6, 4, 5, 6, 7, 9, 0, 0, 8, 7, 2, 9, 7, 0, 4

Offset: 1

Benoit Cloitre, Jan 26 2003

André-Jeannin proved that this constant is irrational.

This constant does not belong to the quadratic number field Q(sqrt(5)) (Bundschuh and Väänänen, 1994). - Amiram Eldar, Oct 30 2020

			3.35988566624317755317201130291892717968890513373...

Daniel Duverney, Number Theory, World Scientific, 2010, 5.22, pp.75-76.
Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 358.

Kenny Lau, Table of n, a(n) for n = 1..10000 (First 1000 terms computed by Joerg Arndt)
Richard André-Jeannin, Irrationalité de la somme des inverses de certaines suites récurrentes, Comptes Rendus de l'Académie des Sciences - Series I - Mathematics 308:19 (1989), pp. 539-541.
Richard André-Jeannin, Problem H-450, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 29, No. 1 (1991), p. 89; Comparable, Solution to Problem H-450 by Paul S. Bruckman, ibid., Vol. 30, No. 2 (1992), p. 191-192.
Richard André-Jeannin, Sequences of Integers Satisfying Recurrence Relations, The Fibonacci Quarterly, Vol. 29, No. 3 (1991), pp. 205-208;
Joerg Arndt, On computing the generalized Lambert series, arXiv:1202.6525v3 [math.CA], (2012).
Paul S. Bruckman, Problem B-602, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 25, No. 3 (1987), p. 279; Fibonacci Infinite Series, Solution to Problem B-602 by C. Georghiou, ibid., Vol. 26, No. 3 (1988), pp. 281-282.
Peter Bundschuh and Keijo Väänänen, Arithmetical investigations of a certain infinite product, Compositio Mathematica, Vol. 91, No. 2 (1994), pp. 175-199.
Daniel Duverney, Irrationalité de la somme des inverses de la suite Fibonacci, Elemente der Mathematik, Vol. 52, No. 1 (1997), pp. 31-36.
William Gosper, Acceleration of Series, Artificial Intelligence Memo #304 (1974).
W. E. Greig, Sums of Fibonacci reciprocals, The Fibonacci Quarterly, Vol. 15, No. 1 (1977), pp. 46-48.
Peter Griffin, Acceleration of the Sum of Fibonacci Reciprocals, The Fibonacci Quarterly, Vol. 30, No. 2 (1992), pp. 179-181.
Sarah H. Holliday and Takao Komatsu, On the sum of reciprocal generalized Fibonacci numbers, Integers 11A (2011), Article 11. Alternate link.
A. F. Horadam, Elliptic functions and Lambert series in the summation of reciprocals in certain recurrence-generated sequences, The Fibonacci Quarterly, Vol. 26, No.2 (May-1988), pp. 98-114.
Fredrik Johansson, The reciprocal Fibonacci constant.
Paul Kinlaw, Michael Morris, and Samanthak Thiagarajan, Sums related to the Fibonacci sequence, Husson University (2021).
Tapani Matala-Aho and Marc Prévost, Quantitative irrationality for sums of reciprocals of Fibonacci and Lucas numbers, Ramanujan J., Vol. 11 (2006), pp. 249-261.
Z.W. M. Trzaska, Fibonacci Polynomials their Properties and Applications, Z. Anal. Anwend. 15 (1996), no. 3, pp. 729-746.
Eric Weisstein's World of Mathematics, Reciprocal Fibonacci Constant.
Wikipedia, Reciprocal Fibonacci constant.

Cf. A000045, A000796, A001622, A002163, A002390, A084119, A093540, A016628, A105199, A153386, A153387.

Cf. A350901, A350902, A350903, A350904.

Digits := 120: c := Pi/2 + I*arccsch(2):
Jeannin := n -> sqrt(5/4)*add(I^(1-j)/sin(j*c), j = 1..n):
evalf(Jeannin(1000)); # Peter Luschny, Nov 15 2023

digits = 105; Sqrt[5]*NSum[(-1)^n/(GoldenRatio^(2*n + 1) - (-1)^n), {n, 0, Infinity}, WorkingPrecision -> digits, NSumTerms -> digits] // RealDigits[#, 10, digits] & // First (* Jean-François Alcover, Apr 09 2013 *)
First@RealDigits[Sqrt[5]/4 ((Log[5] + 2 QPolyGamma[1, 1/GoldenRatio^4] - 4 QPolyGamma[1, 1/GoldenRatio^2])/(2 Log[GoldenRatio]) + EllipticTheta[2, 0, 1/GoldenRatio^2]^2), 10, 105] (* Vladimir Reshetnikov, Nov 18 2015 *)

/* Fast computation without splitting into even and odd indices, see the Arndt reference */
lambert2(x, a, S)=
{
/* Return G(x,a) = Sum_{n>=1} a*x^n/(1-a*x^n) (generalized Lambert series)
   computed as Sum_{n=1..S} x^(n^2)*a^n*( 1/(1-x^n) + a*x^n/(1-a*x^n) )
   As series in x correct up to order S^2.
   We also have G(x,a) = Sum_{n>=1} a^n*x^n/(1-x^n) */
    return( sum(n=1,S, x^(n^2)*a^n*( 1/(1-x^n) + a*x^n/(1-a*x^n) ) ) );
}
inv_fib_sum(p=1, q=1, S)=
{
/* Return Sum_{n>=1} 1/f(n) where f(0)=0, f(1)=1, f(n) = p*f(n-1) + q*f(n-1)
   computed using generalized Lambert series.
   Must have p^2+4*q > 0 */
    my(al,be);
    \\ Note: the q here is -q in the Horadam paper.
    \\ The following numerical examples are for p=q=1:
    al=1/2*(p+sqrt(p^2+4*q));  \\ == +1.6180339887498...
    be=1/2*(p-sqrt(p^2+4*q));  \\ == -0.6180339887498...
    return( (al-be)*( 1/(al-1) + lambert2(be/al, 1/al, S) ) ); \\ == 3.3598856...
}
default(realprecision,100);
S = 1000; /* (be/al)^S == -0.381966^S == -1.05856*10^418 << 10^-100 */
inv_fib_sum(1,1,S) /* 3.3598856... */ /* Joerg Arndt, Jan 30 2011 */

suminf(k=1, 1/(fibonacci(k))) \\ Michel Marcus, Feb 19 2019

m=120; numerical_approx(sum(1/fibonacci(k) for k in (1..10*m)), digits=m) # G. C. Greubel, Feb 20 2019

Alternating series representation: 3 + Sum_{k >= 1} (-1)^(k+1)/(F(k)*F(k+1)*F(k+2)). - Peter Bala, Nov 30 2013
From Amiram Eldar, Oct 04 2020: (Start)
Equals sqrt(5) * Sum_{k>=0} (1/(phi^(2*k+1) - 1) - 2*phi^(2*k+1)/(phi^(4*(2*k+1)) - 1)), where phi is the golden ratio (A001622) (Greig, 1977).
Equals sqrt(5) * Sum_{k>=0} (-1)^k/(phi^(2*k+1) - (-1)^k) (Griffin, 1992).
Equals A153386 + A153387. (End)
From Gleb Koloskov, Sep 14 2021: (Start)
Equals 1 + c1*(c2 + 32*Integral_{x=0..infinity} f(x) dx),
where c1 = sqrt(5)/(8*log(phi)) = A002163/(8*A002390),
      c2 = 2*arctan(2)+log(5) = 2*A105199+A016628,
      phi = (1+sqrt(5))/2 = A001622,
f(x) = sin(x)*(4+cos(2*x))/((exp(Pi*x/log(phi))-1)*(2*cos(2*x)+3)*(7-2*cos(2*x))) (End)
From Amiram Eldar, Jan 27 2022: (Start)
Equals 3 + 2 * Sum_{k>=1} 1/(F(2*k-1)*F(2*k+1)*F(2*k+2)) (Bruckman, 1987).
Equals 2 + Sum_{k>=1} 1/A350901(k) (André-Jeannin, Problem H-450, 1991).
Equals lim_{n->oo} A350903(n)/(A350904(n)*A350902(n)) (André-Jeannin, 1991). (End)
Equals sqrt(5/4)*Sum_{j>=1} i^(1-j)/sin(j*c) where c = Pi/2 + i*arccsch(2). - Peter Luschny, Nov 15 2023
Equals lim_{n->oo} A203006(n)/A003266(n) (Z.W. M. Trzaska, 1996). - Raul Prisacariu, Sep 04 2024

A062073 Decimal expansion of Fibonacci factorial constant.

Original entry on oeis.org

1, 2, 2, 6, 7, 4, 2, 0, 1, 0, 7, 2, 0, 3, 5, 3, 2, 4, 4, 4, 1, 7, 6, 3, 0, 2, 3, 0, 4, 5, 5, 3, 6, 1, 6, 5, 5, 8, 7, 1, 4, 0, 9, 6, 9, 0, 4, 4, 0, 2, 5, 0, 4, 1, 9, 6, 4, 3, 2, 9, 7, 3, 0, 1, 2, 1, 4, 0, 2, 2, 1, 3, 8, 3, 1, 5, 3, 1, 2, 1, 6, 8, 4, 5, 2, 6, 2, 1, 5, 6, 2, 4, 9, 4, 7, 9, 7, 7, 4, 1, 2, 5, 9, 1, 3

Offset: 1

Jason Earls, Jun 27 2001

The Fibonacci factorial constant is associated with the Fibonacci factorial A003266.

Two closely related constants are A194159 and A194160. [Johannes W. Meijer, Aug 21 2011]

			1.226742010720353244417630230455361655871409690440250419643297301214...

Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.2.5.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison Wesley, 1990, pp. 478, 571.

Harry J. Smith, Table of n, a(n) for n=1..5000
M. Griffiths, Symmetric rational expressions in the Fibonacci numbers, Fib. Q., 46/47 (2008/2009), 262-267. [_N. J. A. Sloane_, Dec 05 2009]
Simon Plouffe, Fibonacci factorials
Eric Weisstein's World of Mathematics, Fibonacci Factorial Constant

Cf. A003266, A003267, A003268, A056569, A062072, A062381, A135407, A181926.

Cf. A218490, A253924, A256831, A259314, A259405.

RealDigits[N[QPochhammer[-1/GoldenRatio^2], 105]][[1]] (* Alonso del Arte, Dec 20 2010 *)
RealDigits[N[Re[(-1)^(1/24) * GoldenRatio^(1/12) / 2^(1/3) * EllipticThetaPrime[1,0,-I/GoldenRatio]^(1/3)], 120]][[1]] (* Vaclav Kotesovec, Jul 19 2015, after Eric W. Weisstein *)

\p 1300 a=-1/(1/2+sqrt(5)/2)^2; prod(n=1,17000,(1-a^n))

{ default(realprecision, 5080); p=-1/(1/2 + sqrt(5)/2)^2; x=prodinf(k=1, 1-p^k); for (n=1, 5000, d=floor(x); x=(x-d)*10; write("b062073.txt", n, " ", d)) } \\ Harry J. Smith, Jul 31 2009

C = (1-a)*(1-a^2)*(1-a^3)... 1.2267420... where a = -1/phi^2 and where phi is the Golden ratio = 1/2 + sqrt(5)/2.
C = QPochhammer[ -1/GoldenRatio^2]. [Eric W. Weisstein, Dec 01 2009]
C = A194159 * A194160. [Johannes W. Meijer, Aug 21 2011]
C = exp( Sum_{k>=1} 1/(k*(1-(-(3+sqrt(5))/2)^k)) ). - Vaclav Kotesovec, Jun 08 2013
C = Sum_{k = -inf .. inf} (-1)^((k-1)*k/2) / phi^((3*k-1)*k), where phi = (1 + sqrt(5))/2. - Vladimir Reshetnikov, Sep 20 2016

A035105 a(n) = LCM of Fibonacci sequence {F_1,...,F_n}.

Original entry on oeis.org

1, 1, 2, 6, 30, 120, 1560, 10920, 185640, 2042040, 181741560, 1090449360, 254074700880, 7368166325520, 449458145856720, 21124532855265840, 33735878969859546480, 640981700427331383120, 2679944489486672512824720, 109877724068953573025813520

Offset: 1

Fred Schwab (fschwab(AT)nrao.edu)

Alois P. Heinz, Table of n, a(n) for n = 1..124
Yuri V. Matiyasevich and Richard K. Guy, A new formula for Pi, The American Mathematical Monthly, Vol 93, No. 8 (1986), pp. 631-635.
Carlo Sanna, On the l.c.m. of shifted Fibonacci numbers, arXiv:2007.13330 [math.NT], 2020.
Index entries for sequences related to LCM's

Cf. A000045, A001622, A003266, A059248.

a:= proc(n) option remember; `if`(n=1, 1,
      ilcm(a(n-1), combinat[fibonacci](n)))
    end:
seq(a(n), n=1..25);  # Alois P. Heinz, Feb 12 2018

a[ n_ ] := LCM@@Table[ Fibonacci[ k ], {k, 1, n} ]
With[{fibs=Fibonacci[Range[20]]},Table[LCM@@Take[fibs,n],{n, Length[ fibs]}]] (* Harvey P. Dale, Apr 29 2019 *)

a(n)=lcm(apply(fibonacci,[1..n])) \\ Charles R Greathouse IV, Oct 07 2016

from math import lcm
from sympy import fibonacci
def A035105(n): return lcm(*(fibonacci(i) for i in range(1,n+1))) # Chai Wah Wu, Jul 17 2022

log(a(n)) ~ 3*n^2*log(phi)/Pi^2, where phi is the golden ratio, or equivalently lim_{n->oo} sqrt(6*log(A003266(n))/log(a(n))) = Pi. - Amiram Eldar, Jan 30 2019

A070825 One half of product of first n+1 Lucas numbers A000032.

Original entry on oeis.org

1, 1, 3, 12, 84, 924, 16632, 482328, 22669416, 1722875616, 211913700768, 42170826452832, 13579006117811904, 7074662187380001984, 5963940223961341672512, 8134814465483270041306368, 17953535525321576981163154176, 64112075360923351399733623562496, 370439571435415124387660876944101888

Offset: 0

Wolfdieter Lang, May 10 2002

Vincenzo Librandi, Table of n, a(n) for n = 0..90
R. Grünwald, E. Heidel, A. Strätz, M. Sünkel and R. Terbach, Induction on Number Series, Fakultät fur Wirtschaftsinformatik und Angewandte Informatik, Otto-Friedrich-Universität Bamberg, 2012. - _N. J. A. Sloane_, Feb 07 2013

Cf. A000032, A003266 (for Fibonacci), A003046 (for Catalan), A101690, A135407, A218490.

[1] cat [&*[Lucas(i+1): i in [0..n]]: n in [0..20]]; // Vincenzo Librandi, Sep 15 2016

c := arccsch(2) - I*Pi/2:
A070825 := n -> local j; 2^n*mul(I^j*cosh(c*j), j = 1..n):
seq(simplify(A070825(n)), n = 0..18);  # Peter Luschny, Jul 07 2025

FoldList[Times, LucasL[Range[0, 20]]]/2 (* or *)
Table[Round[GoldenRatio^(n(n+1)/2) QPochhammer[-1, GoldenRatio-2, n+1]]/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)

a(n) = prod(k=0, n, fibonacci(k+1)+fibonacci(k-1))/2; \\ Michel Marcus, Mar 18 2016

a(n) = (Product_{k=0..n} L(k))/2 with L = A000032.
Sum_{n>=0} 1/a(n) = 1 + A101690. - Amiram Eldar, Nov 09 2020
a(n) = 2^n*Product_{j=1..n} i^j*cosh(c*j), where c = arccsch(2) - i*Pi/2. - Peter Luschny, Jul 07 2025

A062381 Let A_n be the n X n matrix defined by A_n[i,j] = 1/F(i+j-1) for 1<=i,j<=n where F(k) is the k-th Fibonacci number (A000045). Then a_n = 1/det(A_n).

Original entry on oeis.org

1, -2, -360, 16848000, 1897448716800000, -3129723891582775706419200000, -541942196790147039091108680776954796441600000, 66373536294235576434745706427960099542896427384297349714149376000000

Offset: 1

Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 08 2001

In the reference it is proved that not only det(A_n) is a reciprocal of an integer but the inverse matrix (A_n)^(-1) is an integer matrix.

			a(3) = -360 because the matrix is / 1,1,1/2 / 1,1/2, 1/3 / 1/2, 1/3, 1/5 / with determinant -1/360.

Colin Barker, Table of n, a(n) for n = 1..19
T. M. Richardson, The Filbert Matrix, arXiv:math/9905079 [math.RA], 1999.

Cf. A000045, A010048.
Cf. A003266, A152686, A253268.
Cf. A062073, A253267, A253270.

Table[(-1)^Floor[n/2]*Product[Fibonacci[k]^(n-Abs[k-n]),{k,1,2*n-1}],{n,1,10}]/Table[Product[Product[Fibonacci[k],{k,1,m-1}],{m,1,n}],{n,1,10}]^2 (* Alexander Adamchuk, May 18 2006 *)

vector(8, n, 1/matdet(matrix(n, n, i, j, 1/fibonacci(i+j-1)))) \\ Colin Barker, May 01 2015

a(n) = s(n) * f(n) / h(n)^2, where s(n) = (-1)^Floor[n/2], f(n) = Product[Fibonacci[k]^(n-Abs[k-n]),{k,1,2*n-1}], h(n) = Product[Product[Fibonacci[k],{k,1,m-1}],{m,1,n}]. - Alexander Adamchuk, May 18 2006

a(n) ~ (-1)^floor(n/2) * A253270 * ((1+sqrt(5))/2)^(n*(2*n^2+1)/3) / (A253267^2 * 5^(n/2) * A062073^(2*n-2)). - Vaclav Kotesovec, May 01 2015

More terms from Vladeta Jovovic, Jul 11 2001

A067962 a(n) = F(n+2)*(Product_{i=1..n+1} F(i))^2 where F(i)=A000045(i) is the i-th Fibonacci number.

Original entry on oeis.org

1, 2, 12, 180, 7200, 748800, 204422400, 145957593600, 272940700032000, 1336044726656640000, 17122749216831498240000, 574502481723130428948480000, 50464872497041500009263431680000, 11605406728144633757130311383449600000

Offset: 0

R. H. Hardin, Feb 02 2002

Number of binary arrangements without adjacent 1's on n X n array connected nw-se.

Kitaev and Mansour give a general formula for the number of binary m X n matrices avoiding certain configurations.

			Neighbors for n=4 (dots represent spaces, circles represent grid points):
O..O..O..O
.\..\..\..
..\..\..\.
O..O..O..O
.\..\..\..
..\..\..\.
O..O..O..O
.\..\..\..
..\..\..\.
O..O..O..O

Reinhard Zumkeller, Table of n, a(n) for n = 0..68
Sergey Kitaev and Toufik Mansour, The problem of the pawns, arXiv:math/0305253 [math.CO], 2003; Annals of Combinatorics 8 (2004) 81-91.
Vaclav Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 69, 421.

Cf. circle A000204, line A000045, arrays: ne-sw nw-se A067965, e-w ne-sw nw-se A067963, n-s nw-se A067964, e-w n-s nw-se A066864, e-w ne-sw n-s nw-se A063443, n-s A067966, e-w n-s A006506, toruses: bare A002416, ne-sw nw-se A067960, ne-sw n-s nw-se A067959, e-w ne-sw n-s nw-se A067958, n-s A067961, e-w n-s A027683, e-w ne-sw n-s A066866.

Cf. A001654, A003266.

a067962 n = a067962_list !! n
a067962_list = 1 : zipWith (*) a067962_list (drop 2 a001654_list)
-- Reinhard Zumkeller, Sep 24 2015

a:= proc(n) option remember; `if`(n=0, 1, (F->
      F(n+1)*F(n+2)*a(n-1))(combinat[fibonacci]))
    end:
seq(a(n), n=0..14);  # Alois P. Heinz, May 20 2019

Rest[Table[With[{c=Fibonacci[Range[n]]},(Times@@Most[c])^2 Last[c]],{n,15}]] (* Harvey P. Dale, Dec 17 2013 *)

a(n)=fibonacci(n+2)*prod(i=0,n,fibonacci(i+1))^2

a(n) = (F(3) * F(4) * ... * F(n+1))^2 * F(n+2), where F(n) = A000045(n) is the n-th Fibonacci number.
a(n) is asymptotic to C^2*((1+sqrt(5))/2)^((n+2)^2)/(5^(n+3/2)) where C=1.226742010720353244... is the Fibonacci Factorial Constant, see A062073. - Vaclav Kotesovec, Oct 28 2011
a(n) = a(n-1) * A001654(n+1), n > 0. - Reinhard Zumkeller, Sep 24 2015

Edited by Dean Hickerson, Feb 15 2002

Revised by N. J. A. Sloane following comments from Benoit Cloitre, Nov 12 2003

A135407 Partial products of A000032 (Lucas numbers beginning at 2).

Original entry on oeis.org

2, 2, 6, 24, 168, 1848, 33264, 964656, 45338832, 3445751232, 423827401536, 84341652905664, 27158012235623808, 14149324374760003968, 11927880447922683345024, 16269628930966540082612736

Offset: 0

Jonathan Vos Post, Dec 09 2007

This is to A000032 as A003266 is to A000045. a(n) is asymptotic to C*phi^(n*(n+1)/2) where phi=(1+sqrt(5))/2 is the golden ratio and C = 1.3578784076121057013874397... (see A218490). - Corrected and extended by Vaclav Kotesovec, Oct 30 2012

			a(0) = L(0) = 2.
a(1) = L(0)*L(1) = 2*1 = 2.
a(2) = L(0)*L(1)*L(2) = 2*1*3 = 6.
a(3) = L(0)*L(1)*L(2)*L(3) = 2*1*3*4 = 24.

G. C. Greubel, Table of n, a(n) for n = 0..95

Cf. A000032, A000045, A000204, A003266, A070825, A218490.

Rest[FoldList[Times,1,LucasL[Range[0,20]]]] (* Harvey P. Dale, Aug 21 2013 *)
Table[Round[GoldenRatio^(n(n+1)/2) QPochhammer[-1, GoldenRatio-2, n+1]], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 14 2016 *)

a(n) = prod(k=0, n, fibonacci(k+1)+fibonacci(k-1)); \\ Michel Marcus, Oct 13 2016

a(n) = Product_{k=0..n} A000032(k).

C = exp( Sum_{k>=1} 1/(k*(((3-sqrt(5))/2)^k-(-1)^k)) ). - Vaclav Kotesovec, Jun 08 2013

cp's OEIS Frontend

A346045 Decimal expansion of Sum_{k>=0} 1/(k! * A003266(k)).

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A372949 a(n) = 2*f(2*n)/(f(n)*f(n+2)) where f = A003266.

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A070826 One half of product of first n primes A000040.

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A079586 Decimal expansion of Sum_{k>=1} 1/F(k) where F(k) is the k-th Fibonacci number A000045(k).

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A062073 Decimal expansion of Fibonacci factorial constant.

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A035105 a(n) = LCM of Fibonacci sequence {F_1,...,F_n}.

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A070825 One half of product of first n+1 Lucas numbers A000032.

A372949 a(n) = 2f(2n)/(f(n)*f(n+2)) where f = A003266.