%I #13 Dec 16 2014 07:08:10
%S 4,5,7,11,13,17,19,23,25,26,29,31,34,35,37,41,43,49,51,53,55,56,59,61,
%T 65,67,71,76,79,81,83,89,91,92,97,101,109,111,113,125,127,129,131,139,
%U 149,151,155,161,169,179,181,187,191,197,199,209,223,235,239,241,251
%N 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.
%C 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.
%e 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.
%t 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_ *)
%Y Cf. A175607, A175901-A175904, A181447-A181471.
%K nonn
%O 1,1
%A _Artur Jasinski_, Oct 12 2010, Oct 21 2010
%E Name improved by _T. D. Noe_, Nov 15 2010