# [seqfan] Re: A181930

israel at math.ubc.ca israel at math.ubc.ca
Tue Apr 3 04:50:35 CEST 2012

It telescopes.  If A_n = product(1 - 1/p_j,j=1..n) with A_0 = 1 then
A_{n-1} - A_n = 1/p A_{n-1}
so sum_{n=1}^infty 1/p A_{n-1} = 1 - lim_{n \to \infty} A_n .
Now it is well known that lim_{n \to \infty} A_n = 0 (see Euler's proof
that there are infinitely many primes).

Robert Israel
University of British Columbia

On Apr 2 2012, Benoît Jubin wrote:

>It appears that
>
>sum( 1/p * product( (q-1)/q, q prime, q<p), p prime) = 1
>
>since each summand is the asymptotic probability that the second
>factor of a number be p. Since any number has a prime as second
>factor, these probabilities add to 1.
>
>Do you know a proof of this identity which does not use this
>probability interpretation?
>
>This sum converges very slowly (remainder ~ 1/ln(p) by the prime
>number theorem).
>
>Thanks,
>Benoit
>PS.1: An alternative term is "asymptotic density of the set of natural
>numbers with n-th factor equal to d" (see
>http://en.wikipedia.org/wiki/Asymptotic_density)
>PS.2: David, I added in A181930 the formula for T(d,tau(d)). Have you
>submitted the associated sequence of "records" ?
>
>
>
> On Mon, Apr 2, 2012 at 9:35 AM, David Wilson <davidwwilson at comcast.net>
> wrote:
>> I have just edited and submitted A181930, which has to do with the
>> probability that number d is the nth divisor of an integer.
>>
>> I added the "tabl" keyword, but I am not sure whether if it will format
>> the table the way I would like.
>>
>>
>>
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