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A072507 Smallest start of n consecutive integers with n divisors, or 0 if no such number exists. 3
1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 120402988681658048433948, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 (list; graph; refs; listen; history; text; internal format)



a(3) = 0 because only squares of primes have three divisors.

From T. D. Noe, Dec 04 2004: (Start)

"Note that a(n)=0 for odd n > 1 because a number has an odd number of divisors only if it is a square and there are no consecutive positive squares. Also, a(4)=0 because one of four consecutive numbers would be a multiple of 4 and have 4 divisors only if it is 8.

"Similarly, a(6)=0 because one of six consecutive number would be a multiple of 6 and the only multiples of 6 having 6 divisors are 12 and 18. For a(8), one of the eight consecutive numbers must be an odd multiple of 4, which cannot have 8 divisors. Interestingly, the 7 consecutive numbers starting at 171893 have 8 divisors.

"Similarly, for a(10), one of the ten consecutive numbers must be an odd multiple of 4, which would have 3x divisors. It is also easy to verify that a(n)=0 for n=14,16,20,22,26,28,32,34,... It seems likely that a(n)=0 for n>2." (End)

This sequence is zero for all but finitely many n. If k = floor(log_2(n)), there must be at least one term exactly divisible by 2^j for any j < k; hence the number of divisors must be divisible by j+1, or more generally by lcm_{i<=k} i. The only values of n divisible by this lcm are 1,2,3,4,6,12,24,60 and 120. For example, for n=30, there must be an element divisible by exactly 8, so its number of divisors is divisible by 4. For n = 60, there must by two numbers 8k and 8(k+2) with k odd; then k and k+2 must each have 15 divisors, making them squares. Together with the comments from T. D. Noe, this leaves only 12, 24 and 120 as open questions. - Franklin T. Adams-Watters, Jul 14 2006

If a(120) = k > 0, then a) k+i cannot be 64 (mod 128) since 7 would divide tau(k+i); b) k+i cannot be 120 (mod 144) since then we'd need k+i = 24x^2 with x==2 (mod 3); c) k+i cannot be 168 (mod 288) since then we'd need k+i = 24x^2 with x==3 (mod 4). Hence no possibility (mod 288) exists, and a(120) = 0. - Hugo van der Sanden, Jan 12 2022

a(12) <= 247239052981730986799644. - Hugo van der Sanden, Apr 25 2022


R. K. Guy, Unsolved Problems in Theory of Numbers, Springer-Verlag, Third Edition, 2004, B12.


Table of n, a(n) for n=1..23.


a(2) = 2 as 2 and 3 are the first (by chance the only) set of two consecutive integers with two divisors.


Cf. A000005 (number of divisors of n).

Cf. A006558 (start of first run of n consecutive integers with same number of divisors).

Cf. A119479.

Sequence in context: A069843 A069849 A138045 * A004530 A355549 A347438

Adjacent sequences: A072504 A072505 A072506 * A072508 A072509 A072510




Amarnath Murthy, Jul 22 2002


More terms from T. D. Noe, Dec 04 2004

a(12) to a(23) from Hugo van der Sanden, Dec 18 2022



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Last modified March 31 14:55 EDT 2023. Contains 361666 sequences. (Running on oeis4.)