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

Leroy Quet q1qq2qqq3qqqq at yahoo.com
Wed Jun 10 17:34:22 CEST 2009

```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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