login
Numbers n such that tau(n) < tau(n+1) where tau(x)=A000005(x).
8

%I #15 Feb 16 2019 03:36:01

%S 1,3,5,7,9,11,13,15,17,19,23,25,27,29,31,35,37,39,41,43,47,49,51,53,

%T 55,59,61,62,63,65,67,69,71,73,74,77,79,83,87,89,91,95,97,99,101,103,

%U 107,109,111,113,115,119,121,123,125,127,129,131,134,137,139,143,146,149

%N Numbers n such that tau(n) < tau(n+1) where tau(x)=A000005(x).

%H Charles R Greathouse IV, <a href="/A074775/b074775.txt">Table of n, a(n) for n = 1..10000</a>

%H P. Erdős, <a href="http://www.renyi.hu/~p_erdos/1936-03.pdf">On a problem of Chowla and some related problems</a>, Proc. Cambridge Philos. Soc. 32 (1936), pp. 530-540.

%H Vaclav Kotesovec, <a href="/A074775/a074775.jpg">Plot of a(n)/n for n = 1..5000000</a>

%F a(n) seems to be asymptotic to c*n with c=2.23...

%F In fact, Erdős proved that a(n) ~ 2n. - _Charles R Greathouse IV_, Dec 05 2012

%t Select[Range[1, 150], DivisorSigma[0,#] < DivisorSigma[0,#+1]&] (* _Vaclav Kotesovec_, Feb 16 2019 *)

%o (PARI) is(n)=numdiv(n) < numdiv(n+1) \\ _Charles R Greathouse IV_, Dec 05 2012

%K nonn

%O 1,2

%A _Benoit Cloitre_, Sep 07 2002