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 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 R. J. Mathar, Entries proved or conjectured 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)*(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. 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

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Last modified September 19 02:21 EDT 2019. Contains 327186 sequences. (Running on oeis4.)