A049997 -id:A049997 - cp's OEIS frontend

A049998 a(n) = b(n)-b(n-1), where b=A049997 are numbers of the form Fibonacci(i)*Fibonacci(j).

1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 5, 3, 1, 1, 8, 5, 1, 2, 13, 8, 1, 1, 3, 21, 13, 2, 1, 5, 34, 21, 3, 1, 1, 8, 55, 34, 5, 1, 2, 13, 89, 55, 8, 1, 1, 3, 21, 144, 89, 13, 2, 1, 5, 34, 233, 144, 21, 3, 1, 1, 8, 55, 377, 233, 34, 5, 1, 2, 13, 89, 610, 377, 55, 8, 1, 1, 3, 21, 144, 987, 610, 89

Offset: 1

Clark Kimberling

nonn

David W. Wilson conjectured (Dec 14 2005) that this sequence consists only of Fibonacci numbers. Proofs were found by Franklin T. Adams-Watters and Don Reble, Dec 14 2005. The following is Reble's proof:
Rearrange A049997, as suggested by Bernardo Boncompagni:
1
2
3 4
5 6
8 9 10
13 15 16
21 24 25 26
34 39 40 42
55 63 64 65 68
89 102 104 105 110
144 165 168 169 170 178
233 267 272 273 275 288
377 432 440 441 442 445 466
Then we know that
F(a+1) * F(a-1) - F(a) * F(a) = (-1)^a
F(a+1) * F(b-1) - F(a-1) * F(b+1)
= + (-1)^b F(a-b), if a>b
= - (-1)^a F(b-a), if aUse these to show that from F(x) to F(x+1), the representable numbers are
F(x) = F(x) * F(2)
< F(x-2) * F(4)
< F(x-4) * F(6)
< ...
< F(x-3) * F(5)
< F(x-1) * F(3)
< F(x+1) * F(1) = F(x+1)
(If x is even, the first identity is needed when the parity changes in the middle.)
Each Fibonacci-product is in one of those subsequences and the identities show that each difference is a Fibonacci number.

Clark Kimberling, Orderings of products of Fibonacci numbers, Fibonacci Quarterly 42:1 (2004), pp. 28-35. (Includes a proof of the conjecture proved in Comments.)

A049997 gives numbers of the form F(i)*F(j), when these Fibonacci-products are arranged in order without duplicates.

A049999(n) gives the smallest index k such that Fibonacci(k) equals a(n).

t = Take[ Union@Flatten@Table[ Fibonacci[i]Fibonacci[j], {i, 0, 20}, {j, 0, i}], 85]; Drop[t, 1] - Drop[t, -1] (* Robert G. Wilson v, Dec 14 2005 *)

More terms from Robert G. Wilson v, Dec 14 2005

Name edited by Michel Marcus, Mar 11 2016

A049999 a(n) = smallest index k such that Fibonacci(k) = d(n), where d = A049998 (sequence of first differences of ordered products of Fibonacci numbers, i.e., of A049997, with no duplicates).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 1, 1, 4, 3, 1, 5, 4, 1, 1, 6, 5, 1, 3, 7, 6, 1, 1, 4, 8, 7, 3, 1, 5, 9, 8, 4, 1, 1, 6, 10, 9, 5, 1, 3, 7, 11, 10, 6, 1, 1, 4, 8, 12, 11, 7, 3, 1, 5, 9, 13, 12, 8, 4, 1, 1, 6, 10, 14, 13, 9, 5, 1, 3, 7, 11, 15, 14, 10, 6, 1, 1, 4, 8, 12, 16, 15, 11

Offset: 1

Clark Kimberling

nonn

"David W. Wilson conjectured (Dec 14 2005) that" sequence A049998 "consists only of Fibonacci numbers. Proofs were found by Franklin T. Adams-Watters and Don Reble, Dec 14 2005." - Petros Hadjicostas, Nov 08 2019 [This comment was copied from A049998, which includes Don Reble's proof of the conjecture.]

			From _Petros Hadjicostas_, Nov 08 2019: (Start)
A049998(1) = 1 = Fibonacci(1) = Fibonacci(2), so a(1) = min(1,2) = 1.
A049998(7) = 2 = Fibonacci(3), so a(7) = 3.
A049998(10) = 3 = Fibonacci(4), so a(10) = 4.
A049998(13) = 5 = Fibonacci(5), so a(13) = 5.
A049998(17) = 8 = Fibonacci(6), so a(17) = 6. (End)

Clark Kimberling, Orderings of products of Fibonacci numbers, Fibonacci Quarterly 42:1 (2004), pp. 28-35. (Includes a proof of the conjecture proved in the Comments for sequence A049998.)

Cf. A000045, A049997, A049998, A094563, A226857, A271354.

A000045(a(n)) = A049998(n) = A049997(n) - A049997(n-1) for n >= 1. - Petros Hadjicostas, Nov 08 2019

Name edited by and more terms from Petros Hadjicostas, Nov 08 2019

A080097 a(n) = Fibonacci(n+2)^2 - 1.

Original entry on oeis.org

0, 3, 8, 24, 63, 168, 440, 1155, 3024, 7920, 20735, 54288, 142128, 372099, 974168, 2550408, 6677055, 17480760, 45765224, 119814915, 313679520, 821223648, 2149991423, 5628750624, 14736260448, 38580030723, 101003831720

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Jan 29 2003

a(n), a(n)+1 and a(n)+2 are consecutive members of A049997.

G. C. Greubel, Table of n, a(n) for n = 0..1000
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.
Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).

Equals A007598(n+2) - 1.

Cf. A000032, A000045, A001622, A049997, A059840, A064831, A080144.

List([0..30], n-> Fibonacci(n+2)^2 -1); # G. C. Greubel, Jul 23 2019

[Fibonacci(n+2)^2 -1: n in [0..30]]; // G. C. Greubel, Jul 23 2019

CoefficientList[Series[(3x+2x^2-x^3)/((1-x^2)(1-2x-2x^2+x^3)), {x, 0, 30}], x]
Table[Fibonacci[n+2]^2-1,{n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Apr 03 2011 *)
LinearRecurrence[{3,0,-3,1},{0,3,8,24},40] (* Harvey P. Dale, Nov 23 2024 *)

A080097(n):=fib(n+2)^2-1$ makelist(A080097(n),n,0,30); /* Martin Ettl, Nov 13 2012 */

a(n)=fibonacci(n+2)^2-1 \\ Charles R Greathouse IV, Feb 06 2013

[fibonacci(n+2)^2 -1 for n in (0..30)] # G. C. Greubel, Jul 23 2019

If n is odd, then a(n) = F(n+1)*F(n+3) = F(n)*F(n+4) - 2, else a(n) = F(n)*F(n+4) = F(n+1)*F(n+3) - 2, where F(n) = Fibonacci numbers (A000045).
a(n) = (Lucas(2*n+4) - 2*(-1)^n - 5)/5.
O.g.f.: x*(3-x)/((1-x^2)*(1-3*x+x^2)) (see a comment on A080144). - Wolfdieter Lang, Jul 30 2012
a(n) = Sum_{k=1..n} F(k+3)*F(k) = A027941(n) + 2*A001654(n), n>=0. - Wolfdieter Lang, Jul 27 2012
Sum_{n>=1} 1/a(n) = (43 - 15*sqrt(5))/18 = 29/9 - 5*phi/3, where phi is the golden ratio (A001622). - Amiram Eldar, Oct 20 2020
a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4). - Joerg Arndt, Nov 13 2023

Edited by Ralf Stephan, May 15 2005

A065108 Positive numbers expressible as a product of Fibonacci numbers.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 13, 15, 16, 18, 20, 21, 24, 25, 26, 27, 30, 32, 34, 36, 39, 40, 42, 45, 48, 50, 52, 54, 55, 60, 63, 64, 65, 68, 72, 75, 78, 80, 81, 84, 89, 90, 96, 100, 102, 104, 105, 108, 110, 117, 120, 125, 126, 128, 130, 135, 136, 144, 150, 156, 160, 162

Offset: 1

Joseph L. Pe, Nov 21 2001

nonn

There are infinitely many triples of consecutive terms of this sequence that are consecutive integers, see A065885. - John W. Layman, Nov 27 2001

Carmichael's theorem implies that 8 and 144 are the only Fibonacci numbers that are products of other Fibonacci numbers, cf. A235383. - Robert C. Lyons, Jan 13 2013

			52 = 2 * 2 * 13 is the product of Fibonacci numbers 2, 2 and 13.

T. D. Noe, Table of n, a(n) for n = 1..10000
Wawrzyniec Bieniawski, Piotr Masierak, Andrzej Tomski, and Szymon Łukaszyk, Assembly Theory - Formalizing Assembly Spaces and Discovering Patterns and Bounds, Preprints.org (2025).
Clemens Heuberger and Stephan Wagner, On the monoid generated by a Lucas sequence, arXiv:1606.02639 [math.NT], 2016.
David A. Corneth, Table of 10000*i, a(10000*i), log(a(10000*i))/log(10000*i) for i = 1..470

Cf. A000045, A065885. Complement of A065105.
Cf. A049997 and A094563: F(i)*F(j) and F(i)*F(j)*F(k) respectively.
Subsequence of A178772.

with(combinat): A000045:=proc(n) options remember: RETURN(fibonacci(n)): end: mulfib:=proc(m,i) local j,q,f: f:=0: for j from i by -1 to 3 while(f=0) do if(irem(m, A000045(j))=0) then q:=iquo(m, A000045(j)): if(q=1) then RETURN(1) else f:=mulfib(q,j) fi fi od: RETURN(f): end: for i from 3 to 12 do for n from A000045(i) to A000045(i+1)-1 do m:=mulfib(n,i): if m=1 then printf("%d, ",n) fi od od: # C. Ronaldo

nn = 1000; k = 1; fib = {}; While[k++; f = Fibonacci[k]; f <= nn, AppendTo[fib, f]]; s = fib; While[s2 = Select[Union[s, Flatten[Outer[Times, fib, s]]], # <= nn &]; Length[s2] > Length[s], s = s2]; s (* T. D. Noe, Jul 17 2012 *)

list(lim)=if(lim<7, return([1..lim\1])); my(v=List([1]), F=List([2,3]), curfib, t, idx, newidx); while((t=F[#F]+F[#F-1])<=lim, listput(F,t)); F=setminus(Set(F), [8,144]); for(i=1,#F, curfib=F[i]; idx=1; while(v[idx]*curfib<=lim, newidx=#v+1; for(j=idx,#v, t=curfib*v[j]; if(t<=lim, listput(v,t))); idx=newidx)); Set(v) \\ Charles R Greathouse IV, Jun 15 2017

As Charles R Greathouse IV recently remarked, it would be good to have an asymptotic formula for this sequence. - N. J. A. Sloane, Jul 22 2012

More terms from John W. Layman, Nov 27 2001

More terms from C. Ronaldo (aga_new_ac(AT)hotmail.com), Jan 02 2005

A271354 Products of two distinct Fibonacci numbers, both greater than 1.

Original entry on oeis.org

6, 10, 15, 16, 24, 26, 39, 40, 42, 63, 65, 68, 102, 104, 105, 110, 165, 168, 170, 178, 267, 272, 273, 275, 288, 432, 440, 442, 445, 466, 699, 712, 714, 715, 720, 754, 1131, 1152, 1155, 1157, 1165, 1220, 1830, 1864, 1869, 1870, 1872, 1885, 1974, 2961, 3016

Offset: 1

Clark Kimberling, May 02 2016

For n > 5, the numbers F(i)*F(j) satisfying F(n-1) <= F(i)*F(j) <= F(n) also satisfy F(n-1) < F(i)*F(j) < F(n). They are the numbers for which i + j = n + 1, where 2 < i < j, so that the number of such F(i)*F(j) is floor(n/2) - 2. The least is 3*F(n-3) and the greatest is 2*F(n-2).

			2*3 = 6, 2*5 = 10, 3*5 = 15, 2*8 = 16.

Clark Kimberling, Table of n, a(n) for n = 1..1000
Clark Kimberling, Orderings of products of Fibonacci numbers, Fibonacci Quarterly 42:1 (2004), pp. 28-35.

Cf. A000045, A004526, A094565, A271356 (difference sequence), subsequence of A049997.

z = 200; f[n_] := Fibonacci[n];
Take[Sort[Flatten[Table[f[m] f[n], {n, 3, z}, {m, 3, n - 1}]]], 100]
Times@@@Subsets[Fibonacci[Range[3,20]],{2}]//Union (* Harvey P. Dale, Jul 12 2025 *)

list(lim)=my(v=List,F=vector(A130233(lim\2),k,fibonacci(k)),t); for(i=2,#F, for(j=1,i-1, t=F[i]*F[j]; if(t>lim,break); listput(v,t))); Set(v) \\ Charles R Greathouse IV, Oct 07 2016

A004526(n) = number of numbers a(k) between F(n+3) and F(n+4), where F = A000045 (Fibonacci numbers).

A059389 Sums of two nonzero Fibonacci numbers.

Original entry on oeis.org

2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 18, 21, 22, 23, 24, 26, 29, 34, 35, 36, 37, 39, 42, 47, 55, 56, 57, 58, 60, 63, 68, 76, 89, 90, 91, 92, 94, 97, 102, 110, 123, 144, 145, 146, 147, 149, 152, 157, 165, 178, 199, 233, 234, 235, 236, 238, 241, 246, 254, 267

Offset: 1

Avi Peretz (njk(AT)netvision.net.il), Jan 29 2001

The sums of two distinct nonzero Fibonacci numbers is essentially the same sequence: 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 18, 21, ... (only 2 is missing), since F(i) + F(i) = F(i-2) + F(i+1). - Colm Mulcahy, Mar 02 2008

To elaborate on Mulcahy's comment above: all terms of A078642 are in this sequence; those are numbers with two distinct representations as the sum of two Fibonacci numbers, which are, as Alekseyev proved, numbers of the form 2*F(i) greater than 2. - Alonso del Arte, Jul 07 2013

			10 is in the sequence because 10 = 2 + 8.
11 is in the sequence because 11 = 3 + 8.
12 is not in the sequence because no pair of Fibonacci numbers adds up to 12.

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A000045, A059390 (complement). Similar in nature to A048645. Essentially the same as A084176. Intersection with A049997 is A226857.

N:= 1000: # to get all terms <= N
R:= NULL:
for j from 1 do
  r:= combinat:-fibonacci(j);
  if r > N then break fi;
  R:= R, r;
end:
R:= {R}:
select(`<=`, {seq(seq(r+s, s=R),r=R)},N);
# if using Maple 11 or earlier, uncomment the next line
# sort(convert(%,list)); # Robert Israel, Feb 15 2015

max = 13; Select[Union[Total/@Tuples[Fibonacci[Range[2, max]], {2}]], # <= Fibonacci[max] &] (* Harvey P. Dale, Mar 13 2011 *)

list(lim)=my(upper=log(lim*sqrt(5))\log((1+sqrt(5))/2)+1, t, tt, v=List([2])); if(fibonacci(t)>lim,t--); for(i=3,upper, t=fibonacci(i); for(j=2,i-1,tt=t+fibonacci(j); if(tt>lim, break, listput(v,tt)))); vecsort(Vec(v)) \\ Charles R Greathouse IV, Jul 24 2012

a(1) = 2 and for n >= 2 a(n) = F_(trinv(n-2)+2) + F_(n-((trinv(n-2)*(trinv(n-2)-1))/2)) where F_n is the n-th Fibonacci number, F_1 = 1 F_2 = 1 F_3 = 2 ... and the definition of trinv(n) is in A002262. - Noam Katz (noamkj(AT)hotmail.com), Feb 04 2001

log a(n) ~ sqrt(n log phi) where phi is the golden ratio A001622. There are (log x/log phi)^2 + O(log x) members of this sequence up to x. - Charles R Greathouse IV, Jul 24 2012

More terms from Larry Reeves (larryr(AT)acm.org), Jan 31 2001

A272909 Numbers that are the product of two Lucas numbers L(i), for i >= 1, using the Lucas numbers as defined in A000204.

Original entry on oeis.org

1, 3, 4, 7, 9, 11, 12, 16, 18, 21, 28, 29, 33, 44, 47, 49, 54, 72, 76, 77, 87, 116, 121, 123, 126, 141, 188, 198, 199, 203, 228, 304, 319, 322, 324, 329, 369, 492, 517, 521, 522, 532, 597, 796, 836, 841, 843, 846, 861, 966, 1288, 1353, 1363, 1364, 1368, 1393

Offset: 1

Clark Kimberling, May 10 2016

Conjecture: if c and d are consecutive terms, then d - c is a product of two Lucas numbers or a product of two Fibonacci numbers.

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A049997 (Fibonacci(i)*Fibonacci(j)), A000204.

Take[Union@Flatten@Table[LucasL[i] LucasL[j], {i, 0, 15}, {j, i}], 60] (* adapted by Vincenzo Librandi, Sep 04 2016 *)

A094563 Triple products of Fibonacci numbers: F(i)F(j)F(k), 2 <= i <= j <= k.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 13, 15, 16, 18, 20, 21, 24, 25, 26, 27, 30, 32, 34, 39, 40, 42, 45, 48, 50, 52, 55, 63, 64, 65, 68, 72, 75, 78, 80, 84, 89, 102, 104, 105, 110, 117, 120, 125, 126, 128, 130, 136, 144, 165, 168, 169, 170, 178, 189, 192

Offset: 1

Clark Kimberling, May 12 2004

nonn

This sequence contains A049997 as a subsequence (aside from its first term), so a(n) << sqrt(phi)^n. All Fibonacci factors must be at most the number, so a(n) >> (phi^(1/3))^n. - Charles R Greathouse IV, Feb 06 2013

			F(2)*F(2)*F(2) = 1 < F(2)*F(2)*F(3) = 2 < ...
< F(4)*F(4)*F(4) = 27 < F(3)*F(4)*F(5) = 30 < F(3)*F(3)*F(6) = 32 < ...

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
C. Kimberling, Orderings of products of Fibonacci numbers, Fibonacci Quart. 42:1 (2004), pp. 28-35.

Subsequence of A065108.

Cf. A000045, A094564.

Select[Union[Times@@@Tuples[Fibonacci[Range[12]],3]],#<200&] (* Harvey P. Dale, Dec 13 2011 *)

list(lim)=my(phi=(1+sqrt(5))/2, v=vector(log(lim*sqrt(5))\log(phi),i,fibonacci(i+1)), u=List(), t, t1); for(i=1,#v, for(j=i,#v, t1=v[i]*v[j];if(t1>lim,break); for(k=j, #v, t=t1*v[k]; if(t>lim,break,listput(u,t))))); vecsort(Vec(u),,8) \\ Charles R Greathouse IV, Feb 06 2013

A226857 Numbers that are both the sum of two Fibonacci numbers and the product of two Fibonacci numbers.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 8, 9, 10, 13, 15, 16, 21, 24, 26, 34, 39, 42, 55, 63, 68, 89, 102, 110, 144, 165, 178, 233, 267, 288, 377, 432, 466, 610, 699, 754, 987, 1131, 1220, 1597, 1830, 1974, 2584, 2961, 3194, 4181, 4791, 5168, 6765, 7752, 8362, 10946, 12543, 13530

Offset: 1

Alonso del Arte, Jun 19 2013

All Fibonacci numbers are in the sequence. The only prime numbers in this sequence are prime Fibonacci numbers.

			5 + 21 = 2 * 13 = 26, therefore 26 is in the sequence.
8 + 21 = 1 * 34 = 34, therefore 34 is in the sequence.
5 + 34 = 3 * 13 = 39, therefore 39 is in the sequence.

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A059389, A049997.

t = Fibonacci[Range[0, 25]]; t1 = Select[Union[Flatten[Table[a + b, {a, t}, {b, t}]]], # <= t[[-1]] &]; t2 = Select[Union[Flatten[Table[a*b, {a, t}, {b, t}]]], # <= t[[-1]] &]; Intersection[t1, t2] (* T. D. Noe, Jul 03 2013 *)

Conjecture: a(n) = a(n-3)+a(n-6) for n>12. - Colin Barker, Nov 09 2014

Empirical g.f.: -x^2*(x^10 +x^9 +x^8 +2*x^7 +3*x^6 +3*x^5 +3*x^4 +3*x^3 +3*x^2 +2*x +1) / (x^6 +x^3 -1). - Colin Barker, Nov 09 2014

A272900 Fibonacci-products fractal sequence.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling, May 10 2016

Let F = A000045, the Fibonacci numbers. Let s be the sequence of all products F(i)F(j), for 2 <= i < = j, arranged in increasing order; viz., (1,2,3,4,5,6,8,9,10,13,15,...) = (F(2)F(2), F(2)F(3), F(2)F(4), F(3)F(3), F(2)F(5), ... ), as at A049997. The sequence of first factors is (F(2), F(2), F(2), F(3), F(2),...), represented by indices (2,2,2,3,2,...). Subtracting 1 from each term leaves A272900, which is a fractal sequence; i.e., the removal of the first occurrence of each term in A272900 leaves A272900, so that the sequence contains itself infinitely many times.

Clark Kimberling, Table of n, a(n) for n = 1..1000
Clark Kimberling, Orderings of products of Fibonacci numbers, Fibonacci Quarterly 42:1 (2004), pp. 28-35.

Cf. A272904 (the associated interspersion), A000045, A049997, A272907 (Lucas-products fractal sequence).

z = 200; f[n_] := Fibonacci[n + 1]; u1 = Table[f[n], {n, 1, z}];
u2 = Sort[Flatten[Table[f[i]*f[j], {i, 1, z}, {j, i, z}]]];
Table[Select[Range[30], MemberQ[u1, u2[[i]]/f[#]] &][[1]], {i, 1, z}]

Showing 1-10 of 16 results. Next

cp's OEIS Frontend

A049998 a(n) = b(n)-b(n-1), where b=A049997 are numbers of the form Fibonacci(i)*Fibonacci(j).

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A049999 a(n) = smallest index k such that Fibonacci(k) = d(n), where d = A049998 (sequence of first differences of ordered products of Fibonacci numbers, i.e., of A049997, with no duplicates).

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A080097 a(n) = Fibonacci(n+2)^2 - 1.

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A065108 Positive numbers expressible as a product of Fibonacci numbers.

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A271354 Products of two distinct Fibonacci numbers, both greater than 1.

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A059389 Sums of two nonzero Fibonacci numbers.

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A272909 Numbers that are the product of two Lucas numbers L(i), for i >= 1, using the Lucas numbers as defined in A000204.

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A094563 Triple products of Fibonacci numbers: F(i)F(j)F(k), 2 <= i <= j <= k.