A065108 -id:A065108 - cp's OEIS frontend

A287821 a(n) is the index of the largest Fibonacci number A065108(n) is divisible by.

2, 3, 4, 3, 5, 4, 6, 4, 5, 4, 7, 5, 6, 4, 5, 8, 6, 5, 7, 4, 5, 6, 9, 4, 7, 6, 8, 5, 6, 5, 7, 4, 10, 5, 8, 6, 7, 9, 6, 5, 7, 6, 4, 8, 11, 5, 6, 5, 9, 7, 8, 4, 10, 7, 6, 5, 8, 6, 7, 5, 9, 12, 5, 7, 6, 4, 10, 8, 7, 9, 11, 5, 8, 6, 7, 6, 9, 7, 8, 6, 10, 5, 13, 7, 6

Offset: 1

David A. Corneth, Jun 01 2017

nonn

			A065108(7) = 8 is divisible by fibonacci(6) = 8 and by no larger Fibonacci number. Therefore, a(7) = 6.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A065108.

A287820 Least number of factors to express A065108(n) as a product of Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 3, 1, 2, 2, 2, 3, 3, 3, 1, 4, 2, 2, 2, 3, 3, 3, 3, 4, 1, 4, 2, 2, 2, 2, 3, 3, 3, 3, 4, 3, 1, 4, 4, 4, 2, 2, 2, 5, 2, 3, 3, 3, 3, 3, 3, 4, 3, 1, 4, 4, 4, 5, 2, 2, 2, 2, 2, 5, 3, 3, 3, 3, 3, 3, 3, 4, 3, 4, 1, 4, 4, 5, 4, 4

Offset: 1

David A. Corneth, Jun 01 2017

Some terms of A065108 are a product of Fibonacci numbers in more than one way. For example, 8 is a product of Fibonacci numbers in more than one way as 8 = 2 * 2 * 2 and both 8 and 2 are Fibonacci numbers. Therefore, 'at least' is used in the name.

			8 = 2 * 2 * 2 are all ways to write A065108(7) = 8 as a product of Fibonacci numbers. 8 has one factor, the least number of all such factorizations. Therefore, a(7) = 1.
81 = 3^4. 81 isn't a Fibonacci number. 3^4 is the only factorization of A065108(43) = 81 into Fibonacci numbers and has four factors 3. Therefore, a(43) = 4.
144 = 2 * 3 * 3 * 8 = 2 * 2 * 2 * 2 * 3 * 3 are all ways to write A065108(62) = 144 as a product of Fibonacci numbers. 144 has one factor, the least number of all such factorizations. Therefore, a(62) = 1.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A065108, A261769, A287821.

Name clarified by Chai Wah Wu, Jun 02 2017

A160009 Numbers that are the product of distinct Fibonacci numbers.

Original entry on oeis.org

0, 1, 2, 3, 5, 6, 8, 10, 13, 15, 16, 21, 24, 26, 30, 34, 39, 40, 42, 48, 55, 63, 65, 68, 78, 80, 89, 102, 104, 105, 110, 120, 126, 130, 144, 165, 168, 170, 178, 195, 204, 208, 210, 233, 240, 267, 272, 273, 275, 288, 312, 315, 330, 336, 340, 377, 390, 432, 440, 442, 445

Offset: 1

T. D. Noe, Apr 29 2009

nonn

Starts the same as A049862, the product of two distinct Fibonacci numbers. This sequence has an infinite number of consecutive terms that are consecutive numbers (such as 15 and 16) because fib(k)*fib(k+3) and fib(k+1)*fib(k+2) differ by one for all k >= 0.
It follows from Carmichael's theorem that if u and v are finite sets of Fibonacci numbers such that (product of all the numbers in u) = (product of all the numbers in v), then u = v.  The same holds for many other 2nd order linear recurrence sequences with constant coefficients.  In the following guide to related "distinct product sequences", W = Wythoff array, A035513:
base sequence              distinct-product sequence
A000045 (Fibonacci)             A160009
A000032 (Lucas, without 2)      A274280
A000032 (Lucas, with 2)         A274281
A000285 (1,4,5,...)             A274282
A022095 (1,5,6,...)             A274283
A006355 (2,4,6,...)             A274284
A013655 (2,5,7,...)             A274285
A022086 (3,6,9,...)             A274191
row 2 of W: (4,7,11,...)        A274286
row 3 of W: (6,10,16,...)       A274287
row 4 of W: (9,15,24,...)       A274288
- Clark Kimberling, Jun 17 2016

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

A059844, A065108

s={1}; nn=30; f=Fibonacci[2+Range[nn]]; Do[s=Union[s,Select[s*f[[i]],#<=f[[nn]]&]], {i,nn}]; s=Prepend[s,0]

A049997 Numbers of the form Fibonacci(i)*Fibonacci(j).

Original entry on oeis.org

0, 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, 610, 699, 712, 714, 715, 720, 754

Offset: 0

Clark Kimberling

It follows from Atanassov et al. that a(n) << sqrt(phi)^n, which matches the trivial a(n) >> sqrt(phi)^n up to a constant factor. - Charles R Greathouse IV, Feb 06 2013

Conjecture: Fibonacci(m)*Fibonacci(n) with 2 < m < n is a perfect power only for (m,n) = (3,6). This has been verified for 2 < m < n <= 900. - Zhi-Wei Sun, Jan 02 2025

			25 is in the sequence since it is the product of two, not necessarily distinct, Fibonacci numbers, 5 and 5.
26 is in the sequence since it is the product of two Fibonacci numbers, 2 and 13.
27 is not in the sequence because there is no way whatsoever to represent it as the product of exactly two Fibonacci numbers.

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
K. T. Atanassov, Ron Knott, Kiyota Ozeki, A. G. Shannon, and László Szalay, Inequalities among related pairs of Fibonacci numbers, Fibonacci Quarterly 41:1 (2003), pp. 20-22.
Clark Kimberling, Orderings of products of Fibonacci numbers, Fibonacci Quarterly 42:1 (2004), pp. 28-35.
Zhi-Wei Sun, Perfect powers as products of two Fibonacci or Lucas numbersQuestion 485037 at MathOverflow, December 30, 2024.

Subsequence of A065108; apart from the first term, subsequence of A094563. Complement is A228523.
See A049998 for further information about this sequence. Cf. A080097.
Intersection with A059389 (sums of two Fibonacci numbers) is A226857.
Cf. also A090206, A005478.

Take[ Union@Flatten@Table[ Fibonacci[i]Fibonacci[j], {i, 0, 16}, {j, 0, i}], 61] (* Robert G. Wilson v, Dec 14 2005 *)

list(lim)=my(phi=(1+sqrt(5))/2, v=vector(log(lim*sqrt(5))\log(phi), i, fibonacci(i+1)), u=List([0]),t); for(i=1,#v,for(j=i,#v,t=v[i]*v[j];if(t>lim,break,listput(u,t)))); vecsort(Vec(u),,8) \\ Charles R Greathouse IV, Feb 05 2013

A351219 Multiplicative with a(p^e) = Fibonacci(e+1).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 5, 1, 2, 1, 2, 1, 1, 1, 3, 2, 1, 3, 2, 1, 1, 1, 8, 1, 1, 1, 4, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 5, 2, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 1, 2, 13, 1, 1, 1, 2, 1, 1, 1, 6, 1, 1, 2, 2, 1, 1, 1, 5, 5, 1, 1, 2, 1, 1, 1

Offset: 1

Amiram Eldar, Feb 05 2022

These numbers were called Zetanacci numbers by Bruckman (1983).

The distinct values of the terms are in A065108.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Paul S. Bruckman, Problem H-359, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 21, No. 3 (1983), p. 238; Zetanacci, Solution to Problem H-359 by C. Georghiou, ibid., Vol. 23, No. 1 (1985), pp. 91-92.

Cf. A000045, A005117, A065108, A065488.

f[p_, e_] := Fibonacci[e + 1]; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

a(n) = my(f=factor(n)); for (k=1, #f~, f[k,1] = fibonacci(f[k,2]+1); f[k,2]=1); factorback(f); \\ Michel Marcus, Feb 05 2022

for(n=1, 100, print1(direuler(p=2, n, 1/(1 - X - X^2))[n], ", ")) \\ Vaclav Kotesovec, Feb 10 2022

from math import prod
from sympy import factorint, fibonacci
def a(n): return prod(fibonacci(e+1) for p, e in factorint(n).items())
print([a(n) for n in range(1, 88)]) # Michael S. Branicky, Feb 05 2022

Dirichlet g.f.: Product_{p prime} 1/(1 - p^(-s) - p^(-2*s)).
a(n) = 1 if and only if n is a squarefree number (A005117).
Sum_{k=1..n} a(k) ~ c * n, where c = A065488 = Product_{p primes} (1 + 1/(p^2 - p - 1)) = 2.67411272557... - Vaclav Kotesovec, Feb 10 2022

A178772 Fibonacci integers.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 38, 39, 40, 41, 42, 44, 45, 46, 47, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 61, 62, 63, 64, 65, 66, 68, 69, 70, 72, 75, 76, 77, 78, 80, 81

Offset: 1

T. D. Noe, Jun 11 2010

A Fibonacci integer is a number that can be written as the product and/or quotient of Fibonacci numbers (A000045). For example, 33 is a Fibonacci integer because Fib(10) * Fib(4) / Fib(5) = 33. Of the numbers up to 100, only 8 are not Fibonacci integers: 37, 43, 53, 59, 67, 71, 73, 74, 79, 83, 86, and 97. See A178762 for the prime numbers in this sequence.
Integers of the form A065108(n)/A065108(m) for some m and n. - Charles R Greathouse IV, Jul 18 2012
Let F(x) be the number of terms of this sequence less than or equal to x. Then exp(c*sqrt(log x) - (log x)^e) < F(x) < exp(c*sqrt(log x) + (log x)^(1/6 + e)) for any e > 0, where c is this constant. Luca, Pomerance, & Wagner conjecture that 1/6 can be replaced by 0, and note that it can be replaced by 1/8 on a strong form of the abc conjecture. - Charles R Greathouse IV, Aug 31 2016

T. D. Noe, Table of n, a(n) for n = 1..1000
Florian Luca, Carl Pomerance, Stephan Wagner, Fibonacci Integers, J. Number Theory 131 (2011) 440-457. (Preprint)

Cf. A065108, A000045.

(* This naive program being quite slow, some terms are precomputed. *) max = Fibonacci[m = 20]; Clear[f]; Do[ f[n] = True, {n, {23, 31, 46, 62, 69, 92, 93}}]; f[] = False; Do[ f[fn = Fibonacci[n]] = True; f[fk = Fibonacci[k]] = True; If[ fn*fk < max, f[fn*fk] = True]; If[ IntegerQ[fk/fn] && fk/fk < max, f[fk/fn] = True], {n, 2, m}, {k, n, m}]; fp[] := (cnt = 0; Do[ If [f[n] && f[k], If[ n*k < max, f[n*k] = True; cnt++]; If[ IntegerQ[k/n], f[k/n] = True; cnt++]], {n, 1, max}, {k, n+1, max}]; Print[cnt, " Fibonacci integers"]; cnt); FixedPoint[fp, 0]; Reap[ Do[ If[ f[n], Sow[n]], {n, 1, 100}]][[2, 1]] (* Jean-François Alcover, Mar 01 2013 *)

A235383 Fibonacci numbers that are the product of other Fibonacci numbers.

Original entry on oeis.org

8, 144

Offset: 1

Robert C. Lyons, Jan 08 2014

This sequence and A229037 and A235265 are winners in the contest held at the 2014 AMS/MAA Joint Mathematics Meetings. - T. D. Noe, Jan 20 2014
Carmichael's theorem implies that 8 and 144 are the only terms of this sequence.
First two terms of A061899, A111687, A172150, A212703, and A231851. - Omar E. Pol, Jan 21 2014
Saha and Karthik conjectured (without reference to Carmichael's theorem) that the only positive integers k for which A001175(k^2) = A001175(k) are 6 and 12. (A000045(6) = 8 and A000045(12) = 144.) - L. Edson Jeffery, Feb 13 2014
Y. Bugeaud, M. Mignotte, and S. Siksek proved that 8 and 144 are the only nontrivial perfect power Fibonacci numbers. - Robert C. Lyons, Dec 23 2015

			The Fibonacci number 8 is in the sequence because 8=2*2*2, and 2 is a Fibonacci number that is not equal to 8. The Fibonacci number 144 is in the sequence because 144=3*3*2*2*2*2, and both 2 and 3 are Fibonacci numbers that are not equal to 144.

Yann Bugeaud, Maurice Mignotte, and Samir Siksek, Classical and modular approaches to exponential Diophantine equations I. Fibonacci and Lucas perfect powers, Annals of Mathematics, 163 (2006), pp. 969-1018.
Arpan Saha and Karthik C S, A few equivalences of Wall-Sun-Sun prime conjecture, arXiv:1102.1636 [math.NT], 2011.
Wikipedia, Carmichael's theorem.

Cf. A000045, A061899, A065108, A227875.

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

A065885 a(n)-1, a(n) and a(n)+1 form three consecutive integers that can be factored into Fibonacci numbers.

Original entry on oeis.org

2, 3, 4, 5, 9, 25, 26, 64, 169, 441, 1156, 3025, 7921, 20736, 54289, 142129, 372100, 974169, 2550409, 6677056, 17480761, 45765225, 119814916, 313679521, 821223649, 2149991424, 5628750625, 14736260449, 38580030724, 101003831721, 264431464441, 692290561600, 1812440220361

Offset: 1

John W. Layman, Nov 28 2001

In general it can be shown that F(n-1)F(n+1), F(n)^2, F(n-2)F(n+2) form three consecutive increasing integers when n is odd and F(n-2)F(n+2), F(n)^2, F(n-1)F(n+1) for three consecutive increasing integers when n is even. Thus the sequence is infinite. [Corrected by Charles R Greathouse IV, Jul 17 2012]

			440 = 8*55, 441 = 21^2, 442 = 13*34, so 441 is a term of the sequence.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000045, A007598, A065108.

a(n)=if(n>7,fibonacci(n-2)^2,[2,3,4,5,9,25,26][n]) \\ Charles R Greathouse IV, Jul 17 2012

Vec(x*(2-x-6*x^2-7*x^3-6*x^4+x^5-37*x^6-29*x^7+14*x^8+x^9)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 30 2016

Except for n = 1, 2, 4 and 7, a(n) is the square of a Fibonacci number.
From Colin Barker, Sep 30 2016: (Start) (based on the signature given in the link)
a(n) = 2*a(n-1)+2*a(n-2)-a(n-3) for n>10.
G.f.: x*(2-x-6*x^2-7*x^3-6*x^4+x^5-37*x^6-29*x^7+14*x^8+x^9) / ((1+x)*(1-3*x+x^2)).
(End)
a(n) = 3*a(n-1) - a(n-2) - 2*(-1)^n for n >= 10. - Greg Dresden, May 18 2020

A065105 Numbers not expressible as a product of Fibonacci numbers.

Original entry on oeis.org

7, 11, 14, 17, 19, 22, 23, 28, 29, 31, 33, 35, 37, 38, 41, 43, 44, 46, 47, 49, 51, 53, 56, 57, 58, 59, 61, 62, 66, 67, 69, 70, 71, 73, 74, 76, 77, 79, 82, 83, 85, 86, 87, 88, 91, 92, 93, 94, 95, 97, 98, 99, 101, 103, 106, 107, 109, 111, 112, 113, 114, 115, 116, 118, 119, 121

Offset: 1

Joseph L. Pe, Nov 20 2001

nonn

I conjecture that for this sequence, a(n + 1) - a(n) <= 5 for all n; and a(n + 1) - a(n) <= 3 for n >= 8.

			28 = 2*14 = 4*7 is a term in the sequence because 28, 14, and 7 are not Fibonacci numbers. 63 = 3*21 is not a term in the sequence because 3 and 21 are Fibonacci numbers.

Cf. A000045. Complement of A065108.

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 11 do for n from A000045(i) to A000045(i+1)-1 do m:=mulfib(n,i): if m=0 then printf("%d, ",n) fi od od: # C. Ronaldo

f[lst_] := Take[ Union[ Flatten[ Table[ lst[[i]]lst[[j]], {i, Length[lst]}, {j, i}]]], 70]; Complement[ Range[189], Nest[f, Fibonacci[Range[2, 20]], 3]] (* Robert G. Wilson v, Feb 12 2005 *)

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

Example corrected by Eric S. Egge, Dec 06 2013

cp's OEIS Frontend

A287821 a(n) is the index of the largest Fibonacci number A065108(n) is divisible by.

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A287820 Least number of factors to express A065108(n) as a product of Fibonacci numbers.

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A160009 Numbers that are the product of distinct Fibonacci numbers.

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A049997 Numbers of the form Fibonacci(i)*Fibonacci(j).

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A351219 Multiplicative with a(p^e) = Fibonacci(e+1).

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A178772 Fibonacci integers.

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A235383 Fibonacci numbers that are the product of other Fibonacci numbers.

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

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A065885 a(n)-1, a(n) and a(n)+1 form three consecutive integers that can be factored into Fibonacci numbers.

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