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.

Showing 1-5 of 5 results.

A279795 Numbers n such that F(n) and F(n-2) are both prime where F(n) = A000045(n).

Original entry on oeis.org

5, 7, 13, 433, 571
Offset: 1

Views

Author

Altug Alkan, Dec 18 2016

Keywords

Comments

a(6) > 2904353. - Daniel Suteu, Dec 23 2016
Terms n of A001605 such that n-2 is also a term of A001605. Surprisingly, the first 4 terms minus 2, { 3, 5, 11, 431 }, are the first four terms of A101315 which also relates to simultaneously prime { m+2, F(m) and F(m)+2 }, but where F is a different function, m -> (m-1)^2 + 1. - M. F. Hasler, Dec 24 2016
Larger primes of the Fibonacci prime pairs in A073340. - Bobby Jacobs, Jan 18 2017

Examples

			13 is a term because Fibonacci(13) = 233 and Fibonacci(11) = 89 are both prime.
		

Crossrefs

Programs

Formula

a(n) = A281087(n) + 2. - Bobby Jacobs, Jan 18 2017

A073340 Fibonacci prime pairs: the indices of each pair differ by two and the relevant Fibonacci numbers are both prime.

Original entry on oeis.org

3, 5, 5, 7, 11, 13, 431, 433, 569, 571
Offset: 1

Views

Author

Harvey P. Dale, Aug 25 2002

Keywords

Comments

There are no other Fibonacci prime pairs up to Fibonacci(104911). (See A001605.) Are there any larger terms?

Examples

			The 431st Fibonacci number and the 433rd Fibonacci number are both prime and their indices differ by 2.
		

References

  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, Rev. ed. 1997, p. 46.

Crossrefs

Programs

  • Mathematica
    Flatten[Select[Partition[Select[Range[3000], PrimeQ[Fibonacci[ # ]]&], 2, 1], #[[2]] - #[[1]] == 2 &]]
  • Python
    from sympy import isprime
    def afind(limit):
      i, fnm2, fnm1 = 1, 1, 1
      while i < limit:
        if isprime(fnm2) and isprime(fnm2 + fnm1):
          print(i, i+2, sep=", ", end=", ")
        i, fnm2, fnm1 = i+1, fnm1, fnm2 + fnm1
    afind(600) # Michael S. Branicky, Mar 05 2021

Extensions

Offset changed to 1 by Joerg Arndt, Jan 18 2017
a(1) and a(2) prepended by Bobby Jacobs, Jan 18 2017

A297624 Numbers k such that Fibonacci(2*k+1) and Fibonacci(2*k-1) are prime.

Original entry on oeis.org

2, 3, 6, 216, 285
Offset: 1

Views

Author

Vincenzo Librandi, Jan 08 2018

Keywords

Comments

a(6) > 1622184 if it exists (see A001605). - Chai Wah Wu, Jan 23 2018

Examples

			2 is in the sequence because F(3)=2 and F(5)=5 are prime.
6 is in the sequence because F(11)=89 and F(13)=233 are prime.
		

Crossrefs

Programs

  • GAP
    o := [];; for k in [1..500] do if IsPrime(Fibonacci(2*k+1)) and IsPrime(Fibonacci(2*k-1)) then Add(o,k); fi; od; A297624 := o; # Muniru A Asiru, Jan 25 2018
  • Magma
    [n: n in [0..700] | IsPrime(Fibonacci(2*n+1)) and  IsPrime(Fibonacci(2*n-1))];
    
  • Maple
    with(combinat, fibonacci): select(k -> isprime(fibonacci(2*k+1)) and isprime(fibonacci(2*k-1)), [$1..500]); # Muniru A Asiru, Jan 25 2018
  • Mathematica
    Select[Range[0, 3000], PrimeQ[Fibonacci[2 # + 1]] && PrimeQ[Fibonacci[2 # - 1]] &]
  • PARI
    isok(n) = isprime(fibonacci(2*n-1)) && isprime(fibonacci(2*n+1)); \\ Michel Marcus, Jan 08 2018
    
  • Python
    from sympy import isprime
    A297624_list, k, a, b, c, aflag = [], 1, 1, 1, 2, False
    while k < 1000:
        cflag = isprime(c)
        if aflag and cflag:
            A297624_list.append(k)
        k, a, b, c, aflag = k + 1, c, b + c, b + 2*c, cflag # Chai Wah Wu, Jan 23 2018
    

Formula

From Chai Wah Wu, Jan 23 2018: (Start)
a(n) = (A279795(n)-1)/2 = (A281087(n)+1)/2 = (A073340(2n-1)+1)/2.
For n > 1, a(n) == 0 mod 3 as otherwise Fibonacci(2*k+1) or Fibonacci(2*k-1) is even. (End)

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

Views

Author

Amiram Eldar, Oct 14 2019

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

Programs

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

A350707 Numbers m such that all prime factors of m^2+1 are Fibonacci numbers.

Original entry on oeis.org

0, 1, 2, 3, 5, 7, 8, 18, 34, 57, 144, 239, 322, 610, 1134903170
Offset: 1

Views

Author

Michel Lagneau, Mar 27 2022

Keywords

Comments

The Fibonacci numbers in the sequence include 1, 2, 3, 5, 8, 144, 610 and 1134903170.
The sequence includes terms of the form sqrt(f(n) - 1) and sqrt(5 * f(n) - 1), where f(n) = Fibonacci(A281087(n)) * Fibonacci(A281087(n)+2) = A140362(n). - Daniel Suteu, Mar 29 2022

Examples

			57 is in the sequence because 57^2+1 = 2*5^3*13 and 2, 5 and 13 are Fibonacci numbers;
1134903170 = Fibonacci(45) is in the sequence because 1134903170^2+1 = 433494437*2971215073 = Fibonacci(43)*Fibonacci(47).
		

Crossrefs

The sequence contains A281618 and A285282.

Programs

  • Maple
    with(numtheory):
    A005478:={2, 3, 5, 13, 89, 233, 1597, 28657, 514229, 433494437, 2971215073, 99194853094755497,1066340417491710595814572169, 19134702400093278081449423917}:
    for n from 0 to 11000 do:
       y:=factorset(n^2+1):n0:=nops(y):
       if A005478 intersect y = y
           then
           print(n):
           else
         fi:
    od:
  • PARI
    isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8));
    isok(m) = my(f=factor(m^2+1)); for (i=1, #f~, if (!isfib(f[i,1]), return(0))); return(1); \\ Michel Marcus, Mar 29 2022
Showing 1-5 of 5 results.