[seqfan] Re: n < 2^tau(n)
Vladimir Shevelev
shevelev at bgu.ac.il
Sun May 30 18:21:27 CEST 2010
It follows from the Wiman-Ramanujan theorem that, for every eps>0 and n>n_0(eps), we have n>tau(n)^(lnln(n)/(ln2+eps)). Therefore, in particular, A034884 is finite.
On the other hand, for 0<eps<ln2, it is known that there exist infinitely many numbers for which n<tau(n)^(lnln(n)/(ln2-eps)), i.e. tau(n)>n^((ln2-eps)/lnln(n)) and 2^tau(n)>2^( n^((ln2-eps)/lnln(n)))>>n. Thus sequence A175495 is infinite.
Regards,
Vladimr
----- Original Message -----
From: zak seidov <zakseidov at yahoo.com>
Date: Sunday, May 30, 2010 16:31
Subject: [seqfan] Re: n < 2^tau(n)
To: seqfan at list.seqfan.eu
> --- On Sun, 5/30/10, Leroy Quet <q1qq2qqq3qqqq at yahoo.com> wrote:
>
> > %C A175495 After the initial 1 in this sequence, the first
> integer in this sequence but not in A034884 is 44.
>
> Me: All terms of A034884 are also in A175495.
> Zak
>
> --- On Sun, 5/30/10, Leroy Quet <q1qq2qqq3qqqq at yahoo.com> wrote:
>
> > From: Leroy Quet <q1qq2qqq3qqqq at yahoo.com>
> > Subject: [seqfan] n < 2^tau(n)
> > To: seqfan at seqfan.eu
> > Date: Sunday, May 30, 2010, 8:38 AM
> > I just submitted this sequence.
> >
> > %S A175495
> > 1,2,3,4,6,8,10,12,14,15,16,18,20,24,28,30,32,36,40,42,44
> > %N A175495 Those positive integers n where n < 2^d(n),
> > where d(n) = number of divisors of n.
> > %C A175495 After the initial 1 in this sequence, the first
> > integer in this sequence but not in A034884 is 44.
> > %C A175495 This sequence is those n where A175494(n) = 1.
> > %Y A175495 CF. A175494, A034884
> > %K A175495 more,nonn
> > %O A175495 1,2
> >
> > (By the way, A175494(n) = floor(n^(1/d(n))).)
> >
> >
> > Is this sequence (A175495) finite?
> >
> > Thanks,
> > Leroy Quet
> >
> > [ ( [ ([( [ ( ([[o0Oo0Ooo0Oo(0)oO0ooO0oO0o]]) ) ] )]) ] )
> > ]
> >
> >
> >
> >
> >
> > _______________________________________________
> >
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Shevelev Vladimir
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