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 1-2 of 2 results.

A175901 Numbers n such that there exists a smaller number k such that k^2-1 has exactly the same set of distinct prime divisors as n^2-1.

Original entry on oeis.org

7, 17, 19, 26, 31, 41, 49, 53, 55, 65, 71, 76, 97, 109, 127, 129, 161, 191, 197, 199, 209, 239, 241, 251, 271, 289, 295, 351, 391, 401, 433, 449, 485, 511, 575, 577, 626, 647, 649, 685, 701, 703, 721, 727, 799, 811, 881, 883, 901, 967, 989, 1025, 1055, 1079
Offset: 1

Views

Author

Artur Jasinski, Oct 11 2010

Keywords

Crossrefs

Cf. A175902 (for corresponding k).

Programs

  • Mathematica
    aa = {}; bb = {}; Do[k = n^2 - 1; c = FactorInteger[k]; b = {}; Do[AppendTo[b, c[[m]][[1]]], {m, 1, Length[c]}]; If[Position[aa, b] != {}, AppendTo[bb, n], AppendTo[aa, b]], {n, 2, 10000}]; bb (*Artur Jasinski*)
  • PARI
    isok(n) = {pfs = factor(n^2-1)[,1]; for (k = 2, n-1, if (factor(k^2-1)[,1] == pfs, return (1));); return (0);} \\ Michel Marcus, Nov 04 2013

Formula

a(1)=7 because set of prime divisors of 7^2-1 is the same as for 5^2-1. The set is {2,3}.

Extensions

Edited by N. J. A. Sloane, Oct 14 2010

A175903 Numbers n such that there is another number k such that n^2-1 and k^2-1 have the same set of prime factors.

Original entry on oeis.org

4, 5, 7, 11, 13, 17, 19, 23, 25, 26, 29, 31, 34, 35, 37, 41, 43, 49, 51, 53, 55, 56, 59, 61, 65, 67, 71, 76, 79, 81, 83, 89, 91, 92, 97, 101, 109, 111, 113, 125, 127, 129, 131, 139, 149, 151, 155, 161, 169, 179, 181, 187, 191, 197, 199, 209, 223, 235, 239, 241, 251
Offset: 1

Views

Author

Artur Jasinski, Oct 12 2010, Oct 21 2010

Keywords

Comments

The difference from A175901 is that k may also be larger than n. So we obtain the sequence by building the union of the sets A175901 and A175902, and sorting.

Examples

			a(2)=5 because set of prime divisors of 5^2-1 =2^3*3 is {2,3}, the same as for example for 7^2-1 = 2^4*3.
		

Crossrefs

Programs

  • Mathematica
    aa = {}; bb = {}; cc = {}; ff = {}; Do[k = n^2 - 1; kk = FactorInteger[k]; b = {}; Do[AppendTo[b, kk[[m]][[1]]], {m, 1, Length[kk]}]; dd = Position[aa, b]; If[dd == {}, AppendTo[cc, n]; AppendTo[aa, b], AppendTo[ff, n]; AppendTo[bb, cc[[dd[[1]][[1]]]]]], {n, 2, 1000000}]; Take[Union[bb,ff],100] (* Artur Jasinski *)

Extensions

Name improved by T. D. Noe, Nov 15 2010
Showing 1-2 of 2 results.