# [seqfan] Re: Pairs Occurring Only Once Among # Of Divisors

Leroy Quet q1qq2qqq3qqqq at yahoo.com
Thu Jun 11 15:13:18 CEST 2009

```I wonder if someone can run a program to determine possible values for my sequence.

For instance, the program would make sure a pair occurs only once among the number-of-divisors of all positive integers < some big number.

Then the values can be independently proved to be unique, which shouldn't be hard for most of them, hopefully.

Thanks,
Leroy Quet

I wrote:
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I just submitted this sequence:

%I A161460
%S A161460 1,2,3,4,8,15,16,24
%N A161460 Those positive integers n such that there is no m different than n where both d(n) = d(m) and d(n+1) = d(m+1), where d(n) is the number of positive divisors of n.
%e A161460 d(15) = 4, and d(15+1) = 5. Any positive integers m+1 with exactly 5 divisors must by of the form p^4, where p is prime. So m = p^4 -1 = (p^2+1)*(p+1)*(p-1). Now, in order for d(m) to have exactly 4 divisors, m must either be of the form q^3 or q*r, where q and r are distinct primes. But no p is such that (p^2+1)*(p+1)*(p-1) = q^3. And the only p where (p^2+1)*(p+1)*(p-1) = q*r is when p = 2 ( and so q=5, r =3). So, there is only one m where both d(m) = 4 and d(m+1) = 5, which is when m=15. Therefore, 15 is in this sequence.
%K A161460 more,nonn
%O A161460 1,2

First, did I make a mistake, and this correct sequence is already in the EIS?

Second, is this sequence infinite or finite? (I suspect that it would be easier to prove it is infinite, if so, rather than prove it is finite, if finite.... unless the truth is already known regarding this, of course.)

Thanks,
Leroy Quet

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