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.
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
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
Comments