A119479 -id:A119479 - cp's OEIS frontend

A072507 Smallest start of n consecutive integers with n divisors, or 0 if no such number exists.

1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 120402988681658048433948, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Amarnath Murthy, Jul 22 2002

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

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

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

Cf. A000005 (number of divisors of n).
Cf. A006558 (start of first run of n consecutive integers with same number of divisors).
Cf. A119479.

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

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

A319045 Length of longest run of consecutive odd numbers having exactly n divisors.

Original entry on oeis.org

1, 3, 1, 8, 1, 8, 1, 17, 1

Offset: 1

Jon E. Schoenfield, Dec 22 2018

a(n)=1 for n odd, since every number with an odd number of divisors is a square, and no two squares are consecutive odd numbers.
The start of the first run of exactly k consecutive odd numbers having exactly n divisors is A319046(n,k).
From David Wasserman, May 04 2019: (Start)
7 <= a(10) <= 8.
14 <= a(12) <= 59. Dickson's conjecture implies a(12) >= 39. Schinzel's Hypothesis H implies a(12) >= 41. (End)

			From _David Wasserman_, May 04 2019: (Start)
A run of 17 consecutive odd numbers with 8 divisors begins at 237805775327, so a(8) >= 17; a run of 18 or more consecutive odd numbers would include at least two that are multiples of 9, and every multiple of 9 having 8 divisors is also a multiple of 27, but the two multiples of 9 cannot both be multiples of 27, so a(8) = 17.
A run of 5 consecutive odd numbers with 14 divisors begins at 10943266106145622193005970311, so a(14) >= 5. A run of 6 consecutive odd numbers with 14 divisors would include at least two that are multiples of 3, and these two would differ by 6. These must be 3^13, 3^6*p for p prime > 3, or 3*p^6 for p prime > 3. But 3*p^6 = 3 (mod 27), while 3^13 and 3^6*p = 0 (mod 27), so no two of these can differ by 6. Therefore no such run exists, and a(14) = 5. (End)

Cf. A119479 (analog for consecutive integers), A319046.

a(6)-a(9) from David Wasserman, Feb 07 2019

A325116 Length of longest run of consecutive even integers having exactly n divisors.

Original entry on oeis.org

0, 1, 1, 3, 1, 2, 1, 3, 1, 1, 1, 7, 1, 1, 1, 3, 1, 2, 1, 3, 1, 1, 1

Offset: 1

David Wasserman, Mar 27 2019

The start of the first run of exactly k consecutive even integers having exactly n divisors is A325117(n,k).
If m does not divide n, then a(n) < 2^(m-1). Proof: 2^(m-1) consecutive even integers must include one congruent to 2^(m-1) mod 2^m, and the number of divisors of this one is a multiple of m.
For m > 2, if 2*(m-1) does not divide n, then a(n) < 2^(m-1). Proof: 2^(m-1) consecutive even integers must include two congruent to 2^(m-2) mod 2^(m-1). These are 2^(m-2)*x and 2^(m-2)*(x+2) for some odd x, and x and x+2 each have n/(m-1) divisors. If 2*(m-1) does not divide n, then n/(m-1) is odd. Any number with an odd number of divisors is a square, so x and x+2 are squares, but squares cannot differ by 2.
14 <= a(24) <= 15. Dickson's conjecture implies a(24)=15.

Cf. A119479 (with consecutive integers), A319045 (with consecutive odd integers).

Cf. A325117.

A323743 Table read by rows: row n lists the numbers k for which there exist only finitely many runs of n consecutive integers whose number-of-divisors function sums to k.

Original entry on oeis.org

1, 3, 4, 5, 5, 7, 8, 9, 8, 9, 11, 12, 13, 14, 15, 10, 13, 15, 17, 18, 19, 14, 15, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27, 16, 19, 20, 21, 22, 23, 25, 26, 27, 29, 30, 31, 20, 22, 24, 27, 29, 30, 31, 33, 34, 35, 36, 37, 38, 39

Offset: 1

Jon E. Schoenfield, Apr 02 2019

Row n lists the numbers k such that
  0 < |{m : Sum_j={m..m+n-1} tau(j) = k}| < infinity
where tau(j) = A000005(j) is the number of divisors of j.

			There is only one number with exactly 1 divisor (namely, k=1), but there are infinitely many numbers with j divisors for every j >= 2, so row 1 consists only of the single term 1.
The sequence of values tau(k) for k >= 1 is A000005, which begins 1, 2, 2, 3, 2, 4, 2, 4, 3, 4, ..., from which the sums of two consecutive terms are 1+2=3, 2+2=4, 2+3=5, 3+2=5, 2+4=6, 4+2=6, 2+4=6, 4+3=7, 3+4=7, ...; no number j < 3 appears as such a sum, every j >= 6 appears infinitely many times as such a sum, and each j in {3,4,5} appears as such a sum only finitely many times, so row 2 is {3, 4, 5}.
Row 3 does not contain 6 as a term because there exists no run of 3 consecutive numbers whose sum of tau values is exactly 6.
The first six rows of the table are as follows:
  row 1: {1};
  row 2: {3, 4, 5};
  row 3: {5, 7, 8, 9};
  row 4: {8, 9, 11, 12, 13, 14, 15};
  row 5: {10, 13, 15, 17, 18, 19};
  row 6: {14, 15, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27}.

Cf. A000005, A005237, A006558, A048892, A072507, A100366, A119479, A141621, A284596, A284597, A292580, A319037, A319045, A319046.

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A072507 Smallest start of n consecutive integers with n divisors, or 0 if no such number exists.

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A319045 Length of longest run of consecutive odd numbers having exactly n divisors.

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A325116 Length of longest run of consecutive even integers having exactly n divisors.

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A323743 Table read by rows: row n lists the numbers k for which there exist only finitely many runs of n consecutive integers whose number-of-divisors function sums to k.

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