

A161460


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.


4



1, 2, 3, 4, 8, 15, 16, 24, 35, 48, 63, 64, 80, 99, 288, 528, 575, 624, 728, 960, 1023, 1024, 1088, 1295, 2303, 2400, 4095, 4096, 5328, 6399, 6723, 9408, 9999, 14640, 15624, 28223, 36863, 38415, 46655, 50175, 50624, 57121, 59048, 59049, 65535, 65536, 83520
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OFFSET

1,2


COMMENTS

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]
Numbers k that are uniquely identified by the values of the ordered pair (d(k), d(k+1)).  Jon E. Schoenfield, Aug 11 2019
Conjecture: 2 is the only term that is neither a square nor 1 less than a square.  Jon E. Schoenfield, Aug 12 2019


LINKS

Table of n, a(n) for n=1..47.
R. J. Mathar, Recurring pairs of consecutive entries in the numberofdivisors function, vixra:1911.0287 (2019).


EXAMPLE

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)*(p1). 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)*(p1) = q^3. And the only p where (p^2+1)*(p+1)*(p1) = 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.


CROSSREFS

Cf. A000005, A164119.
Sequence in context: A117395 A006755 A005853 * A097029 A122774 A274166
Adjacent sequences: A161457 A161458 A161459 * A161461 A161462 A161463


KEYWORD

nonn


AUTHOR

Leroy Quet, Jun 10 2009


EXTENSIONS

Extended with J. Brennen's values of Jun 11 2009 by R. J. Mathar, Jun 16 2009
a(47) from Jon E. Schoenfield, Feb 08 2021


STATUS

approved



