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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)
 OFFSET 1,2 COMMENTS 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 REFERENCES R. K. Guy, Unsolved Problems in Theory of Numbers, Springer-Verlag, Third Edition, 2004, B12. LINKS EXAMPLE a(2) = 2 as 2 and 3 are the first (by chance the only) set of two consecutive integers with two divisors. CROSSREFS 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 KEYWORD more,nonn AUTHOR Amarnath Murthy, Jul 22 2002 EXTENSIONS More terms from T. D. Noe, Dec 04 2004 a(12) to a(23) from Hugo van der Sanden, Dec 18 2022 STATUS approved

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Last modified March 31 05:21 EDT 2023. Contains 361634 sequences. (Running on oeis4.)