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%I #86 Apr 03 2023 10:36:11
%S 1,2,1,3,1,5,1,7,1,3,1,15,1,3,1,7,1,5,1,7,1,3,1
%N Length of longest run of consecutive integers having exactly n divisors.
%C a(12) = 15. If there were 16 such consecutive integers, two would be consecutive multiples of 8. One would have the form 32p and the other the form 8q^2 with odd primes p and q; this implies that 8q^2 is congruent to 24 or 40 (mod 64), which is impossible. On the other hand _Dmitry Petukhov_ found a run of 15 consecutive integers each having 12 divisors. It starts with 66387422053662391209161093722597723545. - _Vladimir Letsko_, Apr 07 2022
%C a(14) = 3. If there were 4, two would be consecutive even numbers. One would have the form 64p and the other the form 2q^6 with odd primes p and q. Since 2q^6 == 2 (mod 16), this implies that 2q^6 = 64p+2, so p = (q^3-1)(q^3+1)/32 is prime, which is impossible.
%C a(16) = 7. If there were 8, one would be congruent to 4 (mod 8), which is impossible.
%C Schinzel's conjecture H would imply that:
%C a(2p) = 3 for all prime p > 3;
%C a(2pq) = 3 for all primes p, q such that gcd(p-1,q-1) > 4;
%C a(6p) = 5 for all odd prime p;
%C a(n) = 7 for all n > 4 such that n is divisible by 4 and nondivisible by 3. - _Vladimir Letsko_, Jul 18 2016
%C From _Vladimir Letsko_, Apr 09 2022: (Start)
%C One of any 32 consecutive integers is divisible by 16 but not by 32. The number of divisors of such an integer is divisible by 5. Therefore a(24) <= 31 and a(48) <= 31.
%C 768369049267672356024049141254832375543516 starts a run of 17 consecutive integers each having 24 divisors. Hence 17 <= a(24) <= 31.
%C 17668887847524548413038893976018715843277693308027547 starts a run of 20 consecutive integers each having 48 divisors. Therefore 20 <= a(48) <= 31. (End)
%C From _Vladimir Letsko_, May 31 2022: (Start)
%C Using Dmitry Petukhov's programs, Eugene Zhilitsky found a chain of 13 consecutive numbers with 36 divisors each. It starts with 1041358820322424595598704771003665679363657167077976401029442221233039097. Hence 13 <= a(36) <= 15. (End)
%H Chris Caldwell, <a href="https://t5k.org/glossary/page.php?sort=DicksonsConjecture">The Prime Glossary: Dickson's conjecture</a>
%H Vasilii A. Dziubenko and Vladimir A. Letsko, <a href="https://arxiv.org/abs/1811.05127">Consecutive positive integers with the same number of divisors</a>, arXiv:1811.05127 [math.NT], 2018.
%H Vladimir A. Letsko, <a href="http://www-old.fizmat.vspu.ru/lib/exe/fetch.php?media=marathon:consec_table_27-06-17.pdf">Table of a(n) for all even n such that exact value of a(n) is proved</a>
%H Vladimir A. Letsko, <a href="http://arxiv.org/abs/1510.07081">Some new results on consecutive equidivisible integers</a>, arXiv:1510.07081 [math.NT], 2015.
%H Vladimir A. Letsko and Vasilii Dziubenko <a href="http://grani.vspu.ru/files/publics/1464684416.pdf">On consecutive equidivisible integers</a> (in Russian)
%F a(2n+1) = 1, since numbers with an odd number of divisors must be squares. If n is not divisible by 3, a(2n) <= 7.
%Y Cf. A000005, A072507.
%Y Cf. A006558, A292580.
%K nonn,hard,more
%O 1,2
%A _Franklin T. Adams-Watters_, Jul 26 2006
%E Edited by _Dean Hickerson_, Aug 01 2006
%E a(12)-a(23) added by _Vladimir Letsko_, Apr 07 2022
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