|
%I #31 Aug 29 2020 02:18:29
%S 4,8,12,16,18,20,24,27,32,36,40,48,54,56,64,72,80,81,88,96,104,108,
%T 112,128,144,160,162,176,192,208,216,224,243,256,272,288,304,320,324,
%U 352,368,384,416,432,448,464,486,512,544,576,608,640,704,729,736,768,832
%N Numbers n such that n*tau(n) < sigma(n)*bigomega(n) where tau(n)=A000005(n), sigma(n)=A000203(n), and bigomega(n)=A001222(n).
%C This sequence contains numbers of the form p^u, u >= p, p prime; numbers of the form 2^k*p, p odd prime and k >= floor(log(p)/log(2)). Example: the first number of the form 2^k*67 in the sequence is 2^6*67=4288 because floor(log(67)/log(2))=6.
%C [Are we to understand that these are all the numbers in the sequence, or just some of them? - _N. J. A. Sloane_, Dec 27 2018]
%C If n=p^(p-1) with p prime, n*tau(n) - sigma(n)*bigomega(n) = 1.
%C From _David A. Corneth_, Aug 24 2020: (Start)
%C If n > 1 then each of n, tau(n), sigma(n) and bigomega(n) are > 0. We can then write the inequality as tau(n)/bigomega(n) < sigma(n)/n.
%C Note that tau(n) and bigomega(n) only depend on the prime signature of n.
%C Suppose we choose some k from A025487. If k is not a term then no number with that prime signature is not a term. k is the number with the value for sigma(m)/m for numbers m with the same prime signature as k. (End)
%H David A. Corneth, <a href="/A068306/b068306.txt">Table of n, a(n) for n = 1..10000</a> (first 4000 terms from Robert Israel)
%p filter:= proc(n) local L,t,j;
%p L:= ifactors(n)[2];
%p n * mul(t[2]+1,t=L) < mul(add(t[1]^j,j=0..t[2]),t=L)*add(t[2],t=L)
%p end proc:
%p select(filter, [$1..1000]); # _Robert Israel_, Dec 26 2018
%t filterQ[n_] := n DivisorSigma[0, n] < DivisorSigma[1, n] PrimeOmega[n];
%t Select[Range[1000], filterQ] (* _Jean-François Alcover_, Aug 24 2020 *)
%o (PARI) isok(n) = n*numdiv(n) < sigma(n)*bigomega(n); \\ _Michel Marcus_, Dec 27 2018
%Y Cf. A000005, A000203, A001222.
%K nonn
%O 1,1
%A _Benoit Cloitre_, Feb 24 2002
%E Corrected by _Robert Israel_, Dec 26 2018
|
