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%I #33 Feb 09 2021 02:44:06
%S 1,2,3,4,8,15,16,24,35,48,63,64,80,99,288,528,575,624,728,960,1023,
%T 1024,1088,1295,2303,2400,4095,4096,5328,6399,6723,9408,9999,14640,
%U 15624,28223,36863,38415,46655,50175,50624,57121,59048,59049,65535,65536,83520
%N Positive integers k such that there is no m different from k where both d(k) = d(m) and d(k+1) = d(m+1), where d(k) is the number of positive divisors of k.
%C Are these values known to be correct, or are they just conjectures? - _Leroy Quet_, Jun 20 2009 [Answer: all the numbers listed in the Data are known to be correct with the exception of 50175 and 59049, which remain conjectural at this time; see the Mathar link. - _Jon E. Schoenfield_, Feb 08 2021]
%C Numbers k that are uniquely identified by the values of the ordered pair (d(k), d(k+1)). - _Jon E. Schoenfield_, Aug 11 2019
%C Conjecture: 2 is the only term that is neither a square nor 1 less than a square. - _Jon E. Schoenfield_, Aug 12 2019
%H R. J. Mathar, <a href="http://vixra.org/abs/1911.0287">Recurring pairs of consecutive entries in the number-of-divisors function</a>, vixra:1911.0287 (2019).
%e d(15) = 4, and d(15+1) = 5. Any positive integer m+1 with exactly 5 divisors must be 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 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 m=15. Therefore 15 is in this sequence.
%Y Cf. A000005, A164119.
%K nonn
%O 1,2
%A _Leroy Quet_, Jun 10 2009
%E Extended with J. Brennen's values of Jun 11 2009 by _R. J. Mathar_, Jun 16 2009
%E a(47) from _Jon E. Schoenfield_, Feb 08 2021
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