A072507 -id:A072507 - cp's OEIS frontend

A119479 Length of longest run of consecutive integers having exactly n divisors.

1, 2, 1, 3, 1, 5, 1, 7, 1, 3, 1, 15, 1, 3, 1, 7, 1, 5, 1, 7, 1, 3, 1

Offset: 1

Franklin T. Adams-Watters, Jul 26 2006

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)

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)

Cf. A000005, A072507.

Cf. A006558, A292580.

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.

Edited by Dean Hickerson, Aug 01 2006

a(12)-a(23) added by Vladimir Letsko, Apr 07 2022

A292580 T(n,k) is the start of the first run of exactly k consecutive integers having exactly 2n divisors. Table read by rows.

Original entry on oeis.org

5, 2, 6, 14, 33, 12, 44, 603, 242, 10093613546512321, 24, 104, 230, 3655, 11605, 28374, 171893, 48, 2511, 7939375, 60, 735, 1274, 19940, 204323, 368431323, 155385466971, 18652995711772, 15724736975643, 2973879756088065948, 9887353188984012120346

Offset: 1

Jon E. Schoenfield, Sep 19 2017

The number of terms in row n is A119479(2n).
Düntsch and Eggleton (1989) has typos for T(3,5) and T(10,3) (called D(6,5) and D(20,3) in their notation). Letsko (2015) and Letsko (2017) both have a wrong value for T(7,3).
The first value required to extend the data is T(6,13) <= 586683019466361719763403545; the first unknown value that may exist is T(12,19). See the a-file for other known values and upper bounds up to T(50,7).

			T(1,1) = 5 because 5 is the start of the first "run" of exactly 1 integer having exactly 2*1=2 divisors (5 is the first prime p such that both p-1 and p+1 are nonprime);
T(1,2) = 2 because 2 is the start of the first run of exactly 2 consecutive integers having exactly 2*1=2 divisors (2 and 3 are the only consecutive integers that are prime);
T(3,4) = 242 because the first run of exactly 4 consecutive integers having exactly 2*3=6 divisors is 242 = 2*11^2, 243 = 3^5, 244 = 2^2*61, 245 = 5*7^2.
Table begins:
   n  T(n,1), T(n,2), ...
  ==  ========================================================
   1  5, 2;
   2  6, 14, 33;
   3  12, 44, 603, 242, 10093613546512321;
   4  24, 104, 230, 3655, 11605, 28374, 171893;
   5  48, 2511, 7939375;
   6  60, 735, 1274, 19940, 204323, 368431323, 155385466971, 18652995711772, 15724736975643, 2973879756088065948, 9887353188984012120346, 120402988681658048433948, T(6,13), ...;
   7  192, 29888, 76571890623;
   8  120, 2295, 8294, 153543, 178086, 5852870, 17476613;
   9  180, 6075, 959075, 66251139635486389922, T(9,5);
  10  240, 5264, 248750, 31805261872, 1428502133048749, 8384279951009420621, 189725682777797295066519373;
  11  3072, 2200933376, 104228508212890623;
  12  360, 5984, 72224, 2919123, 15537948, 973277147, 33815574876, 1043710445721, 2197379769820, 2642166652554075, 17707503256664346, T(12,12), ...;
  13  12288, 689278976, 1489106237081787109375;
  14  960, 156735, 23513890624, 4094170438109373, 55644509293039461218749, 4230767238315793911295500109374, 273404501868270838132985214432619890621;
  15  720, 180224, 145705879375, 10868740069638250502059754282498, T(15,5);
  16  840, 21735, 318680, 6800934, 57645182, 1194435205, 14492398389;
  ...

Hugo van der Sanden, Table of n, a(n) for n = 1..32
Ivo Düntsch and Roger B. Eggleton, Equidivisible consecutive integers, 1989.
Vladimir A. Letsko, Some new results on consecutive equidivisible integers, arXiv:1510.07081 [math.NT], 2015.
Vladimir A. Letsko, Table of a(n) for all even n such that exact value of a(n) is proved, 2017.
Carlos Rivera, Problem 20: k consecutive numbers with the same number of divisors, The Prime Puzzles and Problems Connection.
Hugo van der Sanden, calculation of T(6,11).
Hugo van der Sanden, calculation of T(6,12).
Hugo van der Sanden, A-file with known values and bounds up to T(50,7)

Cf. A000005, A006558, A072507, A119479.
Cf. A005237, A005238, A006601, A049051, A049052, A049053.
Cf. A003680, A075036, A075040.

T(n,2) = A075036(n). - Jon E. Schoenfield, Sep 23 2017

a(1)-a(25) from Düntsch and Eggleton (1989) with corrections by Jon E. Schoenfield, Sep 19 2017
a(26)-a(27) from Giovanni Resta, Sep 20 2017
a(28)-a(29) from Hugo van der Sanden, Jan 12 2022
a(30) from Hugo van der Sanden, Sep 03 2022
a(31) added by Hugo van der Sanden, Dec 05 2022; see "calculation of T(6,11)" link for a list of the people involved.
a(32) added by Hugo van der Sanden, Dec 18 2022; see "calculation of T(6,12)" link for a list of the people involved.

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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A119479 Length of longest run of consecutive integers having exactly n divisors.

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A292580 T(n,k) is the start of the first run of exactly k consecutive integers having exactly 2n divisors. Table read by rows.

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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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Author

Keywords

Comments

Examples

Crossrefs