cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A074691 Squarefree Fibonacci numbers with odd number of prime factors.

Original entry on oeis.org

2, 3, 5, 13, 89, 233, 610, 987, 1597, 10946, 28657, 514229, 3524578, 9227465, 24157817, 39088169, 63245986, 433494437, 701408733, 1134903170, 1836311903, 2971215073, 7778742049, 20365011074, 365435296162, 591286729879
Offset: 1

Views

Author

Felice Russo, Sep 03 2002

Keywords

Comments

Agrees for a long time with sequence of Fibonacci numbers whose number of divisors is a Fibonacci number.

Examples

			610 belongs to the sequence because it has 3 prime factors (2, 5, 61); it also has 8 divisors (1, 2, 5, 10, 61, 122, 305, 610).
		

Crossrefs

Subsequence of A061305 (squarefree Fibonacci numbers).
Cf. A000045.

Programs

  • Maple
    with(combinat, fibonacci): m2_fib := proc(n); if (numtheory[mobius](fibonacci(n))=-1) then RETURN(fibonacci(n)); fi; end: seq(m2_fib(i), i=1..100);
  • Mathematica
    Select[Fibonacci[Range[80]], MoebiusMu[#] == -1 &] (* Harvey P. Dale, Aug 23 2011 *)

Formula

Fibonacci(n) such that mu(Fibonacci(n)) = -1, where mu(n) is the Moebius mu function (A008683).

Extensions

More terms from Jani Melik, Oct 07 2002

A080345 a(n) is the number of prime factors in Fibonacci(prime(n)); that is, in the Fibonacci number whose index is the n-th prime.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 3, 2, 1, 1, 2, 2, 2, 3, 2, 2, 2, 1, 2, 4, 2, 3, 2, 2, 2, 2, 1, 1, 3, 4, 2, 4, 4, 2, 2, 3, 3, 2, 2, 4, 2, 4, 4, 2, 5, 3, 4, 3, 2, 3, 3, 4, 2, 2, 3, 4, 2, 4, 4, 4, 3, 2, 3, 5, 4, 2, 1, 7, 5, 4, 3, 3, 2, 2, 4, 3, 4, 1, 1, 5, 5, 1, 3, 5, 3, 2, 3, 4, 3, 4, 6, 1, 3, 4, 3
Offset: 1

Views

Author

T. D. Noe, Feb 16 2003

Keywords

Comments

In all known examples, Fibonacci(prime(n)) is squarefree, in which case a(n) is well-defined, i.e., the number of distinct prime factors equals the total number of prime factors. But if for some n, Fibonacci(prime(n)) has a repeated prime factor, then a(n) is not well-defined. - Jonathan Sondow, Oct 22 2015

Examples

			a(12) = 3 because the 12th prime is 37 and Fibonacci(37) = 24157817 = 73 * 149 * 2221 has 3 prime factors. - clarified by _Jonathan Sondow_, Oct 21 2015
		

Crossrefs

Programs

  • Mathematica
    Table[Length[FactorInteger[Fibonacci[Prime[n]]]], {n, 60}]
    PrimeNu[Fibonacci[Prime[Range[100]]]] (* Harvey P. Dale, Mar 13 2016 *)
  • PARI
    a(n) = omega(fibonacci(prime(n))); \\ Michel Marcus, Oct 22 2015

Formula

a(n) = A001221(A000045(A000040(n))). - Michel Marcus, Oct 22 2015

A366769 Number of distinct prime divisors of A001045(n) (Jacobsthal numbers).

Original entry on oeis.org

0, 0, 1, 1, 1, 2, 1, 2, 2, 2, 1, 4, 1, 2, 3, 3, 1, 4, 1, 4, 3, 3, 1, 6, 3, 2, 3, 5, 2, 6, 1, 4, 4, 2, 4, 8, 2, 2, 3, 6, 2, 6, 1, 6, 5, 3, 2, 9, 2, 6, 5, 6, 2, 6, 4, 7, 4, 5, 3, 11, 1, 2, 5, 6, 5, 8, 2, 6, 4, 8, 2, 12, 2, 4, 6, 6, 5, 7, 1, 8, 6, 4, 5, 12, 3, 4
Offset: 1

Views

Author

Sean A. Irvine, Oct 21 2023

Keywords

Examples

			a(12) = 4 because Jacobsthal(12) = 1365 has prime factors {3, 5, 7, 13}.
		

Crossrefs

Formula

a(n) = omega(Jacobsthal(n)) = A001221(A001045(n)).

A366780 Number of distinct prime divisors of A000073(n) (tribonacci numbers).

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 3, 2, 3, 2, 3, 1, 3, 4, 3, 3, 2, 2, 2, 3, 3, 3, 5, 3, 4, 5, 3, 4, 3, 5, 4, 5, 3, 5, 3, 2, 4, 2, 4, 5, 4, 4, 6, 2, 5, 5, 6, 3, 5, 7, 5, 2, 3, 5, 4, 6, 5, 4, 7, 3, 2, 4, 4, 3, 3, 4, 5, 2, 6, 6, 6, 5, 3, 6, 5, 4, 2, 6, 3, 6, 1, 7
Offset: 2

Views

Author

Sean A. Irvine, Oct 22 2023

Keywords

Examples

			a(8)=2 because the 8th tribonacci number 24 = 2^3*3 has 2 distinct prime factors.
		

Crossrefs

Programs

  • Mathematica
    PrimeNu[LinearRecurrence[{1, 1, 1}, {1, 1, 2}, 87]] (* Amiram Eldar, Oct 23 2023 *)

Formula

a(n) = A001221(A000073(n)).

A366781 Number of prime divisors of A000073(n) (tribonacci numbers) (counted with multiplicity).

Original entry on oeis.org

0, 0, 1, 2, 1, 1, 4, 3, 4, 1, 2, 6, 3, 3, 8, 5, 2, 3, 4, 4, 3, 6, 4, 3, 4, 4, 3, 8, 4, 5, 11, 6, 4, 4, 10, 5, 5, 5, 9, 4, 2, 4, 2, 6, 5, 4, 11, 11, 2, 6, 7, 9, 3, 5, 9, 6, 2, 3, 5, 8, 12, 5, 11, 12, 4, 2, 4, 6, 3, 3, 6, 6, 2, 10, 7, 8, 7, 5, 12, 7, 4, 2, 6, 4
Offset: 2

Views

Author

Sean A. Irvine, Oct 22 2023

Keywords

Examples

			a(8)=4 because the 8th tribonacci number 24 = 2^3*3 has 4 prime factors.
		

Crossrefs

Programs

  • Mathematica
    PrimeOmega[LinearRecurrence[{1, 1, 1}, {1, 1, 2}, 84]] (* Amiram Eldar, Oct 23 2023 *)

Formula

a(n) = A001222(A000073(n)).

A075735 Squarefree Fibonacci numbers with an even number of prime factors (mu(n)=1).

Original entry on oeis.org

1, 1, 21, 34, 55, 377, 4181, 6765, 17711, 121393, 196418, 317811, 1346269, 2178309, 5702887, 102334155, 165580141, 32951280099, 53316291173, 139583862445, 956722026041, 2504730781961, 10610209857723, 308061521170129
Offset: 1

Views

Author

Jani Melik, Oct 07 2002

Keywords

Examples

			21 is a Fibonacci number and 21=3*7, 34 is a Fibonacci numbers and 34=2*17, ...
		

Crossrefs

Subsequence of A061305 (squarefree Fibonacci numbers).

Programs

  • Maple
    with(combinat, fibonacci): m1_fib := proc(n); if (numtheory[mobius](fibonacci(n))=1) then RETURN(fibonacci(n)); fi; end: seq(m1_fib(i), i=1..100);

A114840 Indices of Fibonacci numbers with 5 distinct prime factors.

Original entry on oeis.org

30, 36, 42, 44, 45, 50, 57, 63, 66, 68, 69, 75, 76, 98, 111, 118, 124, 134, 141, 153, 169, 172, 183, 185, 201, 202, 203, 213, 218, 229, 247, 253, 267, 302, 303, 329, 335, 347, 363, 371, 373, 377, 381, 382, 386, 395, 398, 413, 415, 439, 443, 461
Offset: 1

Views

Author

Shyam Sunder Gupta, Feb 19 2006

Keywords

Examples

			a(1)=30 because 30th Fibonacci number consists of 5 distinct prime factors (i.e., 832040 = 2^3 * 5 * 11 * 31 * 61).
		

Crossrefs

Column k=5 of A303217.

Programs

  • PARI
    n=1;while(n<305,if(omega(fibonacci(n))==5,print1(n,", "));n++)

Formula

{n: A022307(n)=5}. - R. J. Mathar, Nov 29 2015

Extensions

More terms from Ryan Propper, Apr 26 2006
a(56)-a(106) from Max Alekseyev, Aug 18 2013

A335002 Integers m such that omega(F(m)) = omega(L(m)) where omega is the number of distinct prime divisors function, F(n) and L(n) are the n-th Fibonacci and n-th Lucas numbers.

Original entry on oeis.org

1, 3, 4, 5, 7, 9, 10, 11, 13, 14, 15, 17, 18, 21, 26, 33, 42, 46, 47, 55, 58, 66, 69, 73, 77, 85, 89, 93, 102, 103, 107, 111, 117, 121, 123, 132, 139, 167, 171, 177, 179, 181, 187, 201, 205, 207, 213, 219, 221, 233, 241, 246, 247, 253, 257, 262, 267, 269, 273, 279, 281, 282, 293, 295
Offset: 1

Views

Author

Michel Marcus, May 19 2020

Keywords

Comments

Numbers m such that A022307(m) = A086598(m).

Crossrefs

Programs

  • PARI
    lucas(n) = fibonacci(n+1)+fibonacci(n-1);
    isok(m) = omega(fibonacci(m))==omega(lucas(m));

A095224 Least squarefree Fibonacci number with exactly n prime divisors.

Original entry on oeis.org

1, 2, 21, 610, 6765, 701408733, 102334155, 190392490709135, 251728825683549488150424261, 23416728348467685, 13598018856492162040239554477268290, 81055900096023504197206408605
Offset: 0

Views

Author

Amarnath Murthy, Jun 10 2004

Keywords

Comments

Conjecture: The sequence is infinite.
Based on a table of the first 300 Fibonacci numbers, factorized.
Number of prime divisors counted with multiplicity. - Harvey P. Dale, Mar 06 2024

Examples

			a(5) = 701408733 = 3 * 43 * 89 * 199 * 307.
		

Crossrefs

Cf. A022307, A038575. - R. J. Mathar, Oct 14 2010

Programs

  • Maple
    From R. J. Mathar, Oct 14 2010: (Start)
    A001221 := proc(n) nops(numtheory[factorset](n)) ; end proc:
    A095224 := proc(n) for i from 1 do f := combinat[fibonacci](i) ; if A001221(f) =n and numtheory[bigomega](f) = n then return f ; fi; od ; end proc:
    for n from 1 do printf("%d,\n",A095224(n)) ; end do: (End)
  • Mathematica
    Table[SelectFirst[{#,PrimeOmega[#]}&/@Select[Fibonacci[Range[200]],SquareFreeQ],#[[2]] == n&],{n,0,11}][[;;,1]] (* Harvey P. Dale, Mar 06 2024 *)

Formula

min{A000045(i): A038575(i) = A022307(i) = n}. a(n) >= A060319(n). - R. J. Mathar, Oct 14 2010

Extensions

a(9) corrected and 3 terms added by R. J. Mathar, Oct 14 2010

A114305 Numbers k such that Fibonacci(k) has more distinct prime factors than k does.

Original entry on oeis.org

8, 9, 15, 16, 18, 19, 20, 21, 24, 25, 27, 28, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42, 44, 45, 46, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 84, 85, 86, 87, 88, 89, 90, 91
Offset: 1

Views

Author

Shyam Sunder Gupta, Feb 05 2006

Keywords

Crossrefs

Programs

  • Mathematica
    Select[Range[100],PrimeNu[Fibonacci[#]]>PrimeNu[#]&] (* Harvey P. Dale, Oct 25 2011 *)
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