1,2

9<=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. It is likely that a(12)=15; this would follow from Dickson's conjecture.

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.

Chris Caldwell, The Prime Glossary: Dickson's conjecture

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.

Examples of longest known sequences with 2n divisors, for 1<=n<=8, begin with 2, 33, 10093613546512321, 171893, 7939375, 1411925978538681821, 76571890623, 17476613, respectively.

Cf. A000005, A072507.

Cf. A006558.

Sequence in context: A079068 A133021 A131208 this_sequence A130008 A101809 A127203

Adjacent sequences: A119476 A119477 A119478 this_sequence A119480 A119481 A119482

hard,more,nonn

Frank Adams-Watters (FrankTAW(AT)Netscape.net), Jul 26 2006

Edited by Dean Hickerson (dean.hickerson(AT)yahoo.com), Aug 01 2006