

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.


3



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

Are these values known to be correct, or are they just conjectures?  Leroy Quet, Jun 20 2009
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..46.
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


STATUS

approved



