Of Different Prime-Factorization Exponents

[franktaw at netscape.net]

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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