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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A085917 Least k such that k*2^n - 1 is a semiprime.

Original entry on oeis.org

5, 4, 2, 1, 3, 5, 4, 2, 1, 2, 1, 3, 6, 3, 3, 3, 3, 6, 3, 3, 3, 2, 1, 3, 6, 3, 6, 3, 6, 3, 3, 11, 16, 8, 4, 2, 1, 8, 4, 2, 1, 15, 13, 15, 16, 8, 4, 2, 1, 6, 3, 17, 15, 12, 6, 3, 4, 2, 1, 3, 6, 3, 3, 5, 3, 2, 1, 5, 6, 3, 48, 24, 12, 6, 3, 12, 6, 3, 10, 5, 4, 2, 1, 3, 3, 31, 21, 17, 15, 11, 13, 24, 12, 6, 3
Offset: 1

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Author

Jason Earls, Aug 16 2003

Keywords

Comments

The first few values of n such that 509203*2^n - 1 is a semiprime, where k = 509203 (the conjectured smallest Riesel number), are: 3,4,16,34,61,82,124,142,163,171,... Conjecture: there are infinitely many semiprimes of this form.

Examples

			a(33)=16 because k*2^33 - 1 is not a semiprime for k=1,2,...15, but 16*2^33 - 1 = 223 * 616318177 is.
		

Crossrefs

A239677 Least numbers k such that k*3^n-1 is prime.

Original entry on oeis.org

1, 2, 2, 8, 4, 6, 2, 2, 16, 8, 6, 2, 16, 10, 4, 14, 8, 18, 6, 2, 18, 6, 2, 20, 18, 6, 2, 38, 30, 10, 16, 20, 18, 6, 2, 60, 20, 10, 10, 40, 58, 48, 16, 12, 4, 32, 90, 30, 10, 8, 130, 62, 26, 10, 6, 2, 30, 10, 32, 18, 6, 2, 74, 28, 46, 18, 6, 2, 30, 10, 46, 80, 94, 52
Offset: 1

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Author

Derek Orr, Mar 23 2014

Keywords

Comments

All the numbers in this sequence, excluding a(1), are even.

Examples

			1*3^2-1 = 8 is not prime. 2*3^2-1 = 17 is prime. Thus, a(2) = 2.
1*3^5-1 = 242 is not prime. 2*3^5-1 = 485 is not prime. 3*3^5-1 = 728 is not prime. 4*3^5-1 = 971 is prime. Thus, a(5) = 4.
		

Crossrefs

Cf. A085427.

Programs

  • PARI
    for(n=1, 100, k=0; while(!isprime(k*3^n-1), k++); print1(k, ", ")) \\ Colin Barker, Mar 24 2014
  • Python
    import sympy
    from sympy import isprime
    def Pow_3(n):
      for k in range(10**4):
        if isprime(k*(3**n)-1):
          return n
    n = 1
    while n < 100:
      print(Pow_3(n))
      n += 1
    
Previous Showing 11-12 of 12 results.