A038575 -id:A038575 - cp's OEIS frontend

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

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

A277207 Least k such that A001222(Fibonacci(k+1)) / A001222(Fibonacci(k)) = n, or 0 if no such k exists.

Original entry on oeis.org

3, 7, 5, 53, 17, 11, 29, 293, 23, 167, 137, 47, 83

Offset: 1

Altug Alkan, Oct 05 2016

a(16) = 571, a(22) = 509, a(23) = 449, a(26) = 569, a(29) = 359. - Vaclav Kotesovec, Oct 05 2016

			a(1) = 3 because Fibonacci(3) = 2 and Fibonacci(4) = 3.

Cf. A000045, A001222, A038575.

A328733 List of numbers k such that Fibonacci(k) and Fibonacci(k+1) have the same number of prime factors, counted with multiplicity.

Original entry on oeis.org

1, 3, 4, 8, 9, 15, 27, 37, 38, 44, 68, 104, 116, 124, 170, 201, 202, 205, 214, 291, 302, 361, 381, 387, 403, 428, 469, 474, 502, 507, 514, 565, 584, 602, 603, 622, 628, 663, 668, 669, 675, 698, 710, 745, 763, 766, 865, 872, 873, 898, 922, 968, 1006, 1015, 1018, 1035, 1075, 1146, 1153, 1182

Offset: 1

Tomás Roca Sánchez, Oct 26 2019

nonn

F(1) and F(2), both being 1, count as having zero prime factors each.
0 is not a term since all primes divide 0.
For the corresponding Fibonacci numbers, see A328734.

			F(8) = 21 = 3 * 7, and F(9) = 34 = 2 * 17 have 2 prime factors each, so 8 is a part of the sequence.

Blair Kelly, Fibonacci factorizations up to 1000 terms
Tomás Roca Sánchez, Python script that uses already factorized numbers of the sequence

Cf. A000045, A038575, A328734.

isok(k) = bigomega(fibonacci(k)) == bigomega(fibonacci(k+1)); \\ Michel Marcus, Nov 11 2019

Python
```
# See link
```

More terms from Amiram Eldar, Oct 26 2019

A328734 Fibonacci numbers F(k) for k in A328733.

Original entry on oeis.org

1, 2, 3, 21, 34, 610, 196418, 24157817, 39088169, 701408733, 72723460248141, 2427893228399975082453, 781774079430987230203437, 36726740705505779255899443, 150804340016807970735635273952047185, 453973694165307953197296969697410619233826

Offset: 1

Tomás Roca Sánchez, Oct 26 2019

nonn

			The Fibonacci numbers (A000045) start 0,1,1,2,3,5,8,13,21,..., so the first 1 is a term here, but the second 1 is not. (Note that 0 has infinitely many prime divisors.)
21 = 3 * 7 and 34 = 2 * 17 both have two prime factors, and because they are contiguous in the Fibonacci sequence, 21 is included.

Blair Kelly, Fibonacci factorizations up to 1000 terms

Cf. A000045, A038575, A328733 (indices).

Edited to avoid confusion caused by the pair of 1's in the Fibonacci sequence. - N. J. A. Sloane, Nov 11 2019

A076212 Numbers k such that k and Fibonacci(k) have the same number of prime factors, counted with multiplicity.

Original entry on oeis.org

1, 3, 5, 7, 9, 10, 11, 13, 14, 17, 22, 23, 26, 29, 34, 43, 47, 64, 83, 94, 121, 131, 137, 359, 431, 433, 449, 509, 569, 571

Offset: 1

Joseph L. Pe, Nov 03 2002

More precisely, numbers n such that Omega(n) = Omega(Fibonacci(n)), where Omega(n) (A001222) denotes the number of prime factors of n, counting multiplicity.

a(31) > 1422, if it exists. - Amiram Eldar, Sep 10 2024

			9 is a term because 9 and 9th Fibonacci number (i.e., 34) have the same number of prime factors, i.e., 2.

Cf. A000045, A001222, A038575.

with(numtheory): with(combinat): a:=proc(n) if bigomega(n)=bigomega(fibonacci(n)) then n else fi end: seq(a(n),n=1..150); # Emeric Deutsch, Feb 15 2006

Select[Range[150], PrimeOmega[#] == PrimeOmega[Fibonacci[#]] &]

is(k) = bigomega(k) == bigomega(fibonacci(k)); \\ Amiram Eldar, Sep 10 2024

a(24) from Harvey P. Dale, May 01 2008
Edited by R. J. Mathar, Aug 11 2008
More terms from D. S. McNeil, Dec 23 2010

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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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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A277207 Least k such that A001222(Fibonacci(k+1)) / A001222(Fibonacci(k)) = n, or 0 if no such k exists.

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A328733 List of numbers k such that Fibonacci(k) and Fibonacci(k+1) have the same number of prime factors, counted with multiplicity.

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A328734 Fibonacci numbers F(k) for k in A328733.

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A076212 Numbers k such that k and Fibonacci(k) have the same number of prime factors, counted with multiplicity.

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Maple

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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.

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