OFFSET
1,2
COMMENTS
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
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.
a(16) = 7. If there were 8, one would be congruent to 4 (mod 8), which is impossible.
Schinzel's conjecture H would imply that:
a(2p) = 3 for all prime p > 3;
a(2pq) = 3 for all primes p, q such that gcd(p-1,q-1) > 4;
a(6p) = 5 for all odd prime p;
a(n) = 7 for all n > 4 such that n is divisible by 4 and nondivisible by 3. - Vladimir Letsko, Jul 18 2016
From Vladimir Letsko, Apr 09 2022: (Start)
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.
768369049267672356024049141254832375543516 starts a run of 17 consecutive integers each having 24 divisors. Hence 17 <= a(24) <= 31.
17668887847524548413038893976018715843277693308027547 starts a run of 20 consecutive integers each having 48 divisors. Therefore 20 <= a(48) <= 31. (End)
From Vladimir Letsko, May 31 2022: (Start)
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)
LINKS
Chris Caldwell, The Prime Glossary: Dickson's conjecture
Vasilii A. Dziubenko and Vladimir A. Letsko, Consecutive positive integers with the same number of divisors, arXiv:1811.05127 [math.NT], 2018.
Vladimir A. Letsko, Table of a(n) for all even n such that exact value of a(n) is proved
Vladimir A. Letsko, Some new results on consecutive equidivisible integers, arXiv:1510.07081 [math.NT], 2015.
Vladimir A. Letsko and Vasilii Dziubenko On consecutive equidivisible integers (in Russian)
FORMULA
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.
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Franklin T. Adams-Watters, Jul 26 2006
EXTENSIONS
Edited by Dean Hickerson, Aug 01 2006
a(12)-a(23) added by Vladimir Letsko, Apr 07 2022
STATUS
approved