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 1-3 of 3 results.

A051918 Start with n, apply k->2k+1 until reach new record prime; sequence gives number of steps needed.

Original entry on oeis.org

2, 2, 2, 3, 4, 5, 5, 10, 13, 13, 26, 32, 287, 18380, 21727, 23205, 24828, 35646, 48819, 51476
Offset: 0

Views

Author

N. J. A. Sloane, Dec 18 1999

Keywords

Comments

a(20) > 60000. [From Donovan Johnson, May 23 2010]

Examples

			0->1->3, new record prime 3 in 2 steps; 1->3->7, new record prime 7 in 2 steps; 2->5->11, new record prime 11 in 2 steps; 3->7->15->31, new record prime 31 in 3 steps.
For the next few terms we have: 4->9->19->39->79; 5->11->23->47->95->191; 6->13->27->55->111->223; 7->15->31->63->127->255->511->1023->2047->4095->8191; etc.
		

Crossrefs

Programs

  • Mathematica
    f[0] = {start=0, k=3, steps=2}; f[n_] := f[n] = (k=start=f[n-1][[1]]+1; steps=0; While[!PrimeQ[k] || k <= f[n-1][[2]], k=2k+1; steps++]; {start, k, steps}); A051918 = Table[Print[f[n] // Last]; f[n], {n, 0, 13}][[All, 3]] (* Jean-François Alcover, Dec 10 2014 *)

Extensions

More terms from Naohiro Nomoto, May 21 2001
a(13)-a(19) from Donovan Johnson, May 23 2010

A189657 Start with n, apply k->2k+1 until a semiprime is reached; sequence gives the semiprimes.

Original entry on oeis.org

15, 95, 15, 9, 95, 55, 15, 35, 39, 21, 95, 25, 55, 119, 511, 33, 35, 303, 39, 335, 87, 91, 95, 49, 51, 215, 55, 57, 119, 123, 511, 65, 543, 69, 143, 295, 303, 77, 159, 327, 335, 85, 87, 5759, 91, 93, 95, 391, 799, 203, 415, 54271, 215, 219, 111, 3647, 115
Offset: 0

Views

Author

Jonathan Vos Post, Apr 25 2011

Keywords

Comments

This is to semiprimes A001358 as A051919 is to primes A000040. Is this sequence defined for all n?

Examples

			a(0) = 15 in 4 steps because 2*(2*(2*((2*0)+1)+1)+1)+1 = 15 = 3*5 is semiprime.
a(1) = 15 in 3 steps because 2*(2*((2*1) + 1)+1)+1 = 15 = 3*5
a(2) = 95 in 5 steps because 2*(2*(2*(2*(2*2 + 1)+1)+1)+1)+1 = 95 = 5*19.
		

Crossrefs

Programs

  • Mathematica
    semiPrimeQ[n_] := Total[FactorInteger[n]][[2]]==2; Table[k = n; While[k = 2 k + 1; ! semiPrimeQ[k]]; k, {n, 100}] (* T. D. Noe, Apr 29 2011 *)

Extensions

Extended by T. D. Noe, Apr 29 2011

A287766 Numbers k such that (10^k*113 + 19)/3 is prime.

Original entry on oeis.org

1, 9, 13, 33, 81, 1113, 1609, 5697, 13577, 49949
Offset: 1

Views

Author

Mikk Heidemaa, May 31 2017

Keywords

Comments

For k > 2, numbers such that '37'||...'6'...||'73' ('6' concatenated k-2 times and prefixed by '37', and suffixed by '73') in decimal form is prime.

Examples

			13 is in the sequence because (10^13*113 + 19)/3 = 376666666666673 is prime.
		

Crossrefs

Programs

  • Mathematica
    ParallelTable[ If[ PrimeQ[ (10^n*113+19)/3], n, Nothing], {n, 2000}]
  • PARI
    isok(n) = isprime((10^n*113 + 19)/3); \\ Michel Marcus, Jul 21 2017
Showing 1-3 of 3 results.