A022307 -id:A022307 - cp's OEIS frontend

A114704 Smallest primes p such that the p-th Fibonacci number consists of n prime factors.

3, 19, 37, 97, 229, 503, 367, 971

Offset: 1

Shyam Sunder Gupta, Feb 18 2006

			a(3)=37 because 37 is the smallest prime such that 37th Fibonacci number (that is, 24157817) consists of 3 prime factors (24157817 = 73*149*2221).

Cf. A000045, A022307.

a(7) corrected by Amiram Eldar, Oct 14 2019

A115051 Number of distinct prime factors of F(n + L(n)) where F(n) is the Fibonacci number and L(n) is the Lucas number.

Original entry on oeis.org

0, 0, 1, 1, 1, 3, 4, 5, 4, 4, 6, 15, 4, 9, 3, 8, 22, 42, 61

Offset: 0

Parthasarathy Nambi, Feb 28 2006

Added a(13)=9 from F(534) and a(14)=3 from F(857) using Kelly's factorizations. a(15)>=5 via F(1379) and a(16)=22 via F(2223). - R. J. Mathar, Apr 23 2006

a(19) >= 24. - Amiram Eldar, Feb 12 2020

			The first three terms are 1 since:
F(2 + L(2)) = 5 (a prime)
F(3 + L(3)) = 13 (a prime)
F(4 + L(4)) = 89 (a prime)

Blair Kelley, Factorizations
FactorDB, Status of F(9368).

Subsequence of A022307.

Cf. A000045, A000032.

lucas := proc(n::integer) if n = 0 then RETURN(2) ; elif n = 1 then RETURN(1) ; else RETURN(combinat[fibonacci](n-1)+combinat[fibonacci](n+1)) ; fi ; end : for n from 2 to 100 do print(n+lucas(n),"...") ; tst := combinat[fibonacci](n+lucas(n)) ; an := nops(op(2,ifactors(tst))) ; print(an) ; od : # R. J. Mathar, Apr 23 2006

Table[PrimeNu[Fibonacci[n+LucasL[n]]],{n,0,15}] (* Harvey P. Dale, Nov 12 2016 *)

More terms from R. J. Mathar, Apr 23 2006
Offset corrected and a(15)-a(18) added from factordb.com by Amiram Eldar, Feb 12 2020
a(0)=a(1)=0 inserted by Max Alekseyev, Jun 15 2025

A117517 Numbers k such that F(2*k + 1) is prime where F(m) is a Fibonacci number.

Original entry on oeis.org

1, 2, 3, 5, 6, 8, 11, 14, 21, 23, 41, 65, 68, 179, 215, 216, 224, 254, 284, 285, 1485, 2361, 2693, 4655, 4838, 7215, 12780, 15378, 17999, 18755, 25416, 40919, 52455, 65010, 74045, 100553, 198689, 216890, 295020, 296844, 302355, 465758, 524948, 642803, 818003, 901529, 984360, 1452176

Offset: 1

Parthasarathy Nambi, Apr 26 2006

nonn

For F(k) to be prime, with k > 4, it is necessary but not sufficient for k to be prime. Hence after F(4) = 3, every prime F(m) is of the form F(2*k+1) for some k. Every prime divides some Fibonacci number. See also comment to A093062. - Jonathan Vos Post, Apr 29 2006

			If k=68 then F(2*k + 1) = 19134702400093278081449423917, a prime, so 68 is a term.

H. Dubner and W. Keller, New Fibonacci and Lucas Primes, Math. Comp. 68 (1999) 417-427.

Cf. A000045, A001605, A117595.

Cf. A001602, A022307, A030427, A051694, A075737, A083668, A099000.

[n: n in [0..1000] | IsPrime(Fibonacci(2*n+1))]; // Vincenzo Librandi, May 24 2016

Select[Range[0, 5000], PrimeQ[Fibonacci[2 # + 1]] &] (* Vincenzo Librandi, May 24 2016 *)

a(n) = (A083668(n)-1)/2. - R. J. Mathar, Jul 08 2009

a(n) = (A001605(n+1)-1)/2, n > 1. - Vincenzo Librandi, May 24 2016

More terms from Vincenzo Librandi, May 24 2016

A135981 Number of distinct prime factors of A135972(n).

Original entry on oeis.org

0, 2, 2, 3, 2, 3, 2, 4, 3, 3, 4, 4, 5, 3, 4, 2, 6, 3, 3, 3, 6, 3, 6, 5, 4, 3, 4, 8, 2, 3, 4, 7, 2, 6, 3, 7, 6, 4, 3, 9, 2, 7, 5, 7, 3, 6, 6, 8, 4, 6, 2, 11, 3, 6, 7, 3, 8, 2, 7, 4, 9, 3, 12, 3, 5, 7, 7, 4, 7, 3, 9, 6, 5, 2, 12, 3, 5, 6, 10, 11, 5, 9, 3, 6, 5, 12, 2, 5, 8, 12

Offset: 2

Artur Jasinski, Dec 09 2007

nonn

			A135972(3) = 15 = 3*5 which has a(3)=2 distinct prime factors.

Amiram Eldar, Table of n, a(n) for n = 2..1193

Cf. A046051, A046053, A022307, A135972.

k = {}; Do[If[ ! PrimeQ[2^n - 1], c = FactorInteger[2^n - 1]; d = Length[c]; AppendTo[k, d]], {n, 1, 100}]; k

a(n) = A001221(A135972(n)) .

Offset set to 2, definition shortened - R. J. Mathar, Oct 01 2009

A174291 Numbers n such that bigomega(Fibonacci(n)) is a perfect square.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 11, 13, 17, 20, 23, 24, 27, 28, 29, 32, 43, 47, 52, 55, 74, 77, 80, 83, 85, 87, 88, 93, 96, 97, 110, 112, 115, 123, 131, 137, 143, 146, 149, 157, 161, 163, 178, 184, 186, 187, 189, 196, 197, 209, 211, 214, 215, 221, 223, 225, 232, 239, 242, 243, 246

Offset: 1

Michel Lagneau, Mar 15 2010

nonn

Places n such that A001222(A000045(n)) is a perfect square.

			bigomega(Fibonacci(1))= 0.
bigomega(Fibonacci(2))= bigomega(Fibonacci(3))=bigomega(Fibonacci(5))=1.
bigomega(Fibonacci(20))= 4, bigomega(Fibonacci(336))= 25.
bigomega(Fibonacci(359))= 1 because Fibonacci(359) is prime.

Majorie Bicknell and Verner E Hoggatt, Fibonacci's Problem Book, Fibonacci Association, San Jose, Calif., 1974.

Amiram Eldar, Table of n, a(n) for n = 1..236
Blair Kelly, Fibonacci and Lucas Factorizations

Cf. A038575, A022307, A000045.

[k:k in [1..240]| IsSquare(#PrimeDivisors(Fibonacci(k)))]; // Marius A. Burtea, Oct 15 2019

A174291 := proc(n) if issqr( numtheory[bigomega](combinat[fibonacci](n)) ) then printf("%d,",n) ; fi ; return ; end proc:
seq(A174291(n),n=1..90) ; # R. J. Mathar, Jun 01 2011

Select[Range@ 250, IntegerQ@ Sqrt@ PrimeOmega@ Fibonacci@ # &] (* Michael De Vlieger, Oct 15 2019 *)

isok(n) = issquare(bigomega(fibonacci(n))); \\ Michel Marcus, Oct 15 2019

{n: A038575(n) in A000290}.

a(1)=0 removed by Amiram Eldar, Oct 15 2019

A174323 Numbers n such that omega(Fibonacci(n)) is a square.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 11, 13, 17, 20, 23, 24, 27, 28, 29, 32, 43, 47, 52, 55, 74, 77, 80, 83, 84, 85, 87, 88, 91, 93, 96, 97, 100, 108, 115, 123, 131, 132, 137, 138, 143, 146, 149, 156, 157, 161, 163, 178, 184, 187, 189, 196, 197, 209, 211, 214, 215, 221, 222, 223, 232

Offset: 1

Michel Lagneau, Mar 15 2010

nonn

Numbers n such that omega(A000045(n)) is a square, where omega(p) is the number of distinct prime factors of p (A001221). Remark: for the larger Fibonacci numbers F(n) (n > 300), the Maple program (below) is very slow. So we use a two-step process: factoring F(n) with the elliptic curve method, and then calculate the distinct prime factors.

			omega(Fibonacci(1)) = omega(Fibonacci(2)) = omega(1) = 0,
omega(Fibonacci(3)) = omega(2) = 1,
omega(Fibonacci(20)) = omega(6765) = 4,
omega(Fibonacci(80)) = omega(23416728348467685) = 9.

Majorie Bicknell and Verner E Hoggatt, Fibonacci's Problem Book, Fibonacci Association, San Jose, Calif., 1974.
Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, The Fibonacci Association, 1972, pages 1-8.

Amiram Eldar, Table of n, a(n) for n = 1..258 (terms 1..200 from Robert Israel, derived from b-file for A022307)
Blair Kelly, Fibonacci and Lucas Factorizations
Pieter Moree, Counting Divisors of Lucas Numbers, Pacific J. Math, Vol. 186, No. 2, 1998, pp. 267-284.
Eric Weisstein's World of Mathematics, Fibonacci Number
Wikipedia, Fibonacci number

Cf. A038575 (number of prime factors of n-th Fibonacci number, with multiplicity).
Cf. A000045, A000213, A000288, A000322, A000383, A001221, A060455, A030186, A039834, A020695, A020701, A071679.
Cf. A022307 (number of distinct prime factors of n-th Fibonacci number), A086597 (number of primitive prime factors).

[k:k in [1..240]| IsSquare(#PrimeDivisors(Fibonacci(k)))]; // Marius A. Burtea, Oct 15 2019

with(numtheory):u0:=0:u1:=1:for p from 2 to 400 do :s:=u0+u1:u0:=u1:u1:=s: s1:=nops( ifactors(s)[2]): w1:=sqrt(s1):w2:=floor(w1):if w1=w2 then print (p): else fi:od:
# alternative:
P[1]:= {}: count:= 1: res:= 1:
for i from 2 to 300 do
  pn:= map(t -> i/t, numtheory:-factorset(i));
  Cprimes:= `union`(seq(P[t],t=pn));
  f:= combinat:-fibonacci(i);
  for p in Cprimes do f:= f/p^padic:-ordp(f,p) od;
  P[i]:= Cprimes union numtheory:-factorset(f);
  if issqr(nops(P[i])) then
     count:= count+1;
     res:= res, i;
  fi;
od:
res; # Robert Israel, Oct 13 2016

Select[Range[200], IntegerQ[Sqrt[PrimeNu[Fibonacci[#]]]] &] (* G. C. Greubel, May 16 2017 *)

is(n)=issquare(omega(fibonacci(n))) \\ Charles R Greathouse IV, Oct 13 2016

A072556 Numbers n such that n and the n-th Fibonacci number have the same number of distinct prime factors.

Original entry on oeis.org

1, 3, 4, 5, 7, 10, 11, 12, 13, 14, 17, 22, 23, 26, 29, 34, 43, 47, 83, 94, 131, 137, 359, 431, 433, 449, 509, 569, 571

Offset: 1

Benoit Cloitre, Aug 06 2002

			a(7)=10 because 10 and 10th Fibonacci number(i.e. 55) have the same number of prime factors i.e. 2. - _Shyam Sunder Gupta_, Feb 05 2006

Cf. A001221, A022307.

with(combinat): with(numtheory): a:=proc(n) if nops(factorset(fibonacci(n)))=nops(factorset(n)) then n else fi end: seq(a(n),n=1..150); # Emeric Deutsch, Apr 02 2006

Insert[Select[Range[1, 50], Length[FactorInteger[ # ]] ==Length[FactorInteger[Fibonacci[ # ]]] &], 2, 2] (* Stefan Steinerberger, Mar 20 2006 *)
Select[Range[600],PrimeNu[#]==PrimeNu[Fibonacci[#]]&] (* Harvey P. Dale, Oct 14 2023 *)

More terms from Sascha Kurz, Jan 25 2003

Edited by R. J. Mathar, Aug 11 2008

A277213 Least k such that Fibonacci(k + 1) has n times as many distinct prime factors as Fibonacci(k), or 0 if no such k exists.

Original entry on oeis.org

3, 7, 17, 23, 29, 47, 167, 419, 83

Offset: 1

Altug Alkan, Oct 05 2016

Least k such that A001221(Fibonacci(k+1)) / A001221(Fibonacci(k)) = n, or 0 if no such k exists.

			a(2) = 7 because Fibonacci(8) = 21 = 3*7 (2 distinct prime factors) and Fibonacci(7) = 13 (1 prime factor), and 2/1 is 2.

Cf. A000045, A001221, A022307, A277207.

Flatten[Table[Position[Partition[PrimeNu[Fibonacci[Range[2,420]]],2,1],?(k #[[1]]==#[[2]]&),1,1,Heads->False],{k,9}]]+1 (* _Harvey P. Dale, Jan 31 2025 *)

a(8)-a(9) from Charles R Greathouse IV, Oct 05 2016

A290498 Numbers m such that the set of distinct prime divisors of the number of divisors of Fibonacci(m) is equal to the set of distinct prime divisors of m.

Original entry on oeis.org

1, 4, 8, 16, 24, 32, 60, 64, 72, 96, 128, 192, 256, 300, 336, 512, 576, 648, 900, 1008, 1024, 1080, 1250

Offset: 1

Altug Alkan, Aug 04 2017

Thanks to squarefree terms of A058635, numbers of the form 2^k appear in this sequence for k > 1. However it is not proven yet whether it is always true.
From Jon E. Schoenfield, Aug 05 2017: (Start)
The difficulty in extending this sequence is that it becomes hard to obtain the complete prime factorization of Fibonacci(m) as m increases. However, since every number having an odd number of divisors is a square, and the largest Fibonacci number that is also a square is Fibonacci(12) = 144, we can confine the search for terms > 12 to even numbers only.
Even for values of m for which we are unable to completely factorize Fibonacci(m), we can determine with a high degree of confidence whether m is in the sequence by considering only the multiplicities of the smaller primes in those factorizations, because multiplicities greater than 1 in the prime factorizations of Fibonacci numbers rarely occur among the larger prime factors. If, in place of the actual complete factorization of Fibonacci(m) for each examined value of m, we were to use only the multiplicities of the prime factors of Fibonacci(m) that are less than 10000 (which are quickly and easily counted using trial division), the terms we would obtain for this sequence would begin with 1, 4, 8, 16, 24, 32, 60, 64, 72, 96, 128, 192, 256, 300, 336, 512, 576, 648, 900, 1008, 1024, 1080, 1250, 1500, 1536, 1620, 1920, 2048, 2352, 2500, 2592, 2700, 4096, 4608, 5000, 5184, 5400, 5832, 7500, 8100, 8192, 8448, 8640, 9072, 9600, 10000, 13608, 15000, ...
Perhaps surprisingly, we would get the same terms (up through at least a(141) = 960000) if, instead of the multiplicities of prime factors <= 10000, we were to use the multiplicities of just the prime factors <= 13. (End)

			72 is a term because d(Fibonacci(2^3*3^2)) = 2^9*3.
300 is a term because d(Fibonacci(2^2*3*5^2)) = 2^15*3^2*5.

Blair Kelly, Fibonacci and Lucas Factorizations

Cf. A000005, A000045, A022307, A038575, A058635, A063375, A081381, A086597, A135939.

Select[Range[12]~Join~Range[14, 300, 2], Apply[SameQ, Map[FactorInteger[#][[All, 1]] &, {#, DivisorSigma[0, Fibonacci@ #]}]] &] (* Michael De Vlieger, Aug 07 2017 *)

is(n) = factor(numdiv(fibonacci(n)))[,1]==factor(n)[,1] \\ David A. Corneth, Aug 04 2017

a(16)-a(19) from David A. Corneth, Aug 04 2017
a(20)-a(22) from Jon E. Schoenfield, Aug 05 2017
a(23) from Amiram Eldar, Oct 14 2019

A328381 Lesser of twin primes pair p, such that F(p) and F(p+2) have the same number of prime factors, where F(n) is the n-th Fibonacci number.

Original entry on oeis.org

3, 5, 11, 59, 71, 107, 179, 191, 311, 431, 569, 599, 827, 881

Offset: 1

Amiram Eldar, Oct 14 2019

No more terms below 1427.
The corresponding number of prime factors is 1, 1, 1, 2, 2, 2, 3, 2, 4, 1, 1, 2, 5, ...
Assuming that Fibonacci numbers with prime index are always squarefree, the distinction between number of prime factors with multiplicity (A001222) and number of distinct prime factors (A001221) is inessential.

			3 is in the sequence since 3 and 5 are twin primes, and F(3) = 2 and F(5) = 5 are both primes, thus having the same number of prime factors.
71 is in the sequence since 71 and 73 are twin primes, and F(71) and F(73) both have 2 prime factors.

Cf. A000045, A001359, A001605, A005478, A022307, A038575, A319908.

Supersequence of A281087.

s={}; Do[If[PrimeQ[n] && PrimeQ[n+2] && PrimeOmega[Fibonacci[n]] == PrimeOmega[ Fibonacci[n+2]], AppendTo[s, n]], {n, 1, 200}]; s

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A114704 Smallest primes p such that the p-th Fibonacci number consists of n prime factors.

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A115051 Number of distinct prime factors of F(n + L(n)) where F(n) is the Fibonacci number and L(n) is the Lucas number.

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A117517 Numbers k such that F(2*k + 1) is prime where F(m) is a Fibonacci number.

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A135981 Number of distinct prime factors of A135972(n).

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A174291 Numbers n such that bigomega(Fibonacci(n)) is a perfect square.

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A174323 Numbers n such that omega(Fibonacci(n)) is a square.

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A072556 Numbers n such that n and the n-th Fibonacci number have the same number of distinct prime factors.

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