Of Different Prime-Factorization Exponents

[Dan Dima]

I was rushing so the previous post is dumb. It's a little bit more
complicated but it can be expressed involving all terms like [n/(p_k^{u_k})]
in the same manner. It's even more complicated to write it in plain text.

Regards,
Dan


On 3/19/06, Dan Dima <dimad72 at gmail.com> wrote:
>
>  There is a straightforward formula for this - as simple as for the
> Euler-phi or sigma functions.
> It is quite difficult to write it in plain text but I'll try :D :
>
> n=p^u:
> a(n) = p^u-1
>
> n=p_1^{u_1} * p_2^{u_2}:
> a(n) = p_1^{u_1} * p_2^{u_2} - p_1^{u_1} + 1 - p_2^{u_2} + 1 = (p_1^{u_1}
> - 1)*(p_2^{u_2} - 1) + 1
>
> n=p_1^{u_1} * p_2^{u_2} * p_3^{u_3}:
> a(n) = (p_1^{u_1} - 1)*(p_2^{u_2} - 1)*(p_3^{u_3} - 1) + 1
>
> The general formula for n=p_1^{u_1} * p_2^{u_2} * ... * p_k^{u_k} as:
> a(n) = (p_1^{u_1} - 1)*(p_2^{u_2} - 1)*...*(p_k^{u_k} - 1) + 1
>
> Just count for the inverse condition and pay attention for the multiple
> counting.
>
> Best regards,
>  Dan Dima
>
>
>
> On 3/18/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
> >
> > 1) Here's a few more values:
> >
> > 0,1,2,3,4,3,6,7,8,6,10,8,12,9,9,15,16,13,18,15,14,15,22,17,24,18,26,21,28
> > ,15
> >
> > 2) I think it would make more sense to have a(1) = 1.  You would have
> > to change the "m < n" to "m <= n" - which affects only this value.
> >
> > 3) It might be more productive to look at the inverse condition, the
> > number of m s.t. b(m,p) = b(n,p) for some p|n.  This is n - a(n):
> >
> > 0,1,1,1,1,3,1,1,1,4,1,4,1,5,6,1,1,5,1,5,7,7,1,7,1,8,1,7,1,15
> >
> > 4) Looking at the function in (3), we have a(p^i) = 1, and  if n =
> > p^i*q^j,
> >
> > a(n) = floor(n/p^i) - floor(n/p^{i+1}) + floor(n/q^j) -
> > floor(n/q^{j+1}) - 1.
> >
> > For numbers with 3 or more prime factors, we can get similar but
> > increasingly more complex formulae based on the inclusion/exclusion
> > principle.
> >
> > 5) I don't at this point see any sort of simple formula.
> >
> > Franklin T. Adams-Watters
> >
> > -----Original Message-----
> > From: Leroy Quet <qq-quet at mindspring.com >
> >
> > A couple of days ago I sent to the EIS this sequence:
> >
> > >%S A000001 0,1,2,3,4,3,6,7,8,6,10,8,12,9
> > >%N A000001 If n = product{p=primes,p|n} p^b(n,p), where each b(n,p) is
> > a
> > >positive integer, then a(n) is number of positive integers m, m < n,
> > >such that each b(m,p) does not equal b(n,p).
> > >%e A000001 12 = 2^2 * 3^1. Of the positive integers < 12, there are 8
> > >integers where no prime divides these integers the same number of
> > times
> > >the prime divides 12: 1, 2 = 2^1, 5 = 5^1, 7 = 7^1, 8 = 2^3, 9 = 3^2,
> > 10 =
> > >2^1 *5^1, and 11 = 11^1.
> > >So a(12) = 8.
> > >The other positive integers < 12 (3 = 3^1, 4 = 2^2, and 6 = 2^1
> > >* 3^1) each are divisible by at least one prime the same number of
> > times
> > >this prime divides 12.
> > >%O A000001 1
> > >%K A000001 ,more,nonn,
> >
> > (Unfortunately, this sequence has not yet appeared, given that Neil is
> > currently on vacation.)
> >
> > Is there a direct way (say, a sum involving the Moebius function,
> > perhaps) to calculate this sequence's terms?
> >
> > If someone finds a formula, please post it as a comment to this
> > sequence
> > when the sequence finally appears (as well as posting it to seq.fan).
> > I myself am not as good at math as I used to be, and cannot find a
> > formula for a(n).
> >
> > thanks,
> > Leroy Quet
> >
> >
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>
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