A292580 -id:A292580 - 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

A319046 Irregular table read by rows: T(n,k) is the start of the first run of exactly k consecutive odd numbers having exactly n divisors, or 0 if no such run exists.

Original entry on oeis.org

1, 23, 11, 3, 9, 15, 33, 91, 299, 213, 1383, 3091, 8129, 81, 45, 243, 3175, 2523, 3682662467, 164406964254894462023

Offset: 1

Jon E. Schoenfield, Dec 22 2018

The number of terms in row n is A319045(n).
For each odd n, row n contains only one term, i.e., T(n,1), since every number with an odd number of divisors is a square, and no two squares are consecutive odd numbers.
If n is prime, every number having n divisors is of the form p^(n-1) where p is an odd prime, so T(n,1) = 3^(n-1) if n is an odd prime.
Row 6 cannot cannot contain more than eight terms, because every number with six divisors is of the form p^5 or p^2 * q where p and q are distinct primes, and in any run of nine or more consecutive odd numbers, at least three would include be divisible by 3, of which at least two would not be divisible by 9 but would differ by at most 12; for any such pair of numbers (3*p1^2, 3*p2^2), p1^2 and p2^2 would differ by at most 4, and no such pair of primes (p1, p2) exists.
T(6,7) <= 7483570769727848971899774228580919;
T(6,7) > 3*10^22. - David Wasserman, May 04 2019
10^17 < T(6,8) <= 620228749187663825311276520397486295457519. - David Wasserman, Feb 05 2019
Row 7 consists of the single term T(7,1) = 3^6 = 729.
Row 8 cannot have more than 17 terms (see A319045); its first 15 terms are 105, 663, 6095, 10503, 35119, 58345, 195831, 247347, 1123281, 943607, 19235031, 148720547, 107473247, 1260718031, and 21470685.
T(8,17) = 237805775327. - David Wasserman, Feb 07 2019
T(10,7) <= 3*(7364195527360905184867386522361)^4 - 4 (approx. 8.8*10^123). - David Wasserman, May 04 2019
T(12,14) <= 1569073892509234696810905887582957. - David Wasserman, May 04 2019
1.7*10^14 < T(14,4) <= 4365641192113347078119. - David Wasserman, May 04 2019
T(14, 5) <= 10943266106145622193005970311. - David Wasserman, May 04 2019

			T(1,1) = 1 because 1 is the first (and only) number having 1 divisor.
T(2,1) = 23 because it is the first odd number having 2 divisors (i.e., the first prime) that is not part of a run of two or more consecutive odd numbers that are prime.
T(2,2) = 11 because it is the first odd prime that begins a run of exactly 2 consecutive odd numbers that are prime.
T(2,3) = 3 because it is the first (and only) number that begins a run of 3 consecutive odd numbers all of which are prime. (There exists no run of more than 3 consecutive odd numbers that are all prime, so T(2,3) is the last term in row 2.)
T(4,8) = 8129 because {8129 = 11*739, 8131 = 47*173, 8133 = 3*2711, 8135 = 5*1627, 8137 = 79*103, 8139 = 3*2713, 8141 = 7*1163, 8143 = 17*479} is the first run of 8 consecutive odd numbers with 4 divisors.
Table begins:
   n  T(n,1), T(n,2), ...
  ==  =======================================================
   1  1;
   2  23, 11, 3;
   3  9;
   4  15, 33, 91, 299, 213, 1383, 3091, 8129;
   5  81;
   6  45, 243, 3175, 2523, 3682662467, 164406964254894462023, ...;
   7  729;
   8  105, 663, 6095, 10503, 35119, 58345, 195831, 247347, 1123281, 943607, 19235031, 148720547, 107473247, 1260718031, 21470685, ...;
   9  225;
  10  405, 127251, 490219371, ...;
  11  59049;
  12  315, 2275, 22473, 1389683, 10753975, ...;
  13  531441;
  14  3645, 26890623, 136349453140621, ...;
  15  2025;
  16  945, 14875, 155701, 1343013, 4320561, 14906085, 88958433, 376675395, 957171679, ...;
  17  43046721;
  18  1575, 74725, 732665527, ...;
  19  387420489;
  20  2835, 244375, 608149373, ...;
  21  18225;
  22  295245, ...;
  23  31381059609;
  24  3465, 226525, 3720871, 39198573, ...;

Cf. A000005, A005408, A292580, A319045.

T(6,6) and table additions from David Wasserman, May 04 2019

A325117 Irregular table read by rows: T(n,k) is the start of the first run of exactly k consecutive even integers having exactly n divisors, or 0 if no such run exists.

Original entry on oeis.org

2, 4, 14, 0, 6, 16, 12, 18, 64, 24, 40, 182, 36, 48, 1024, 60, 198, 348, 9050, 25180, 25658650, 584558736346, 4096, 192, 144, 120, 918, 5430, 65536, 180, 17298, 262144, 240, 6640, 4413038, 576, 3072, 4194304, 360, 3400, 19548, 134044, 182644, 7126044, 359208340, 16074693138, 419893531348, 214932235538

Offset: 2

David Wasserman, Mar 27 2019

The number of terms in row n is A325116(n).
2.46*10^12 <= T(24,11) <= 299005907036986132.
T(24,14) <= 1010085195622895590495442.
T(30,3) <= 1359906389476760004389052496.
5.17*10^12 < T(36, 6) <= 13707985134823441146.
T(36, 7) <= 1678936725442128595619270138.

			T(4,3) = 6 because 6, 8, and 10 each have 4 divisors.
T(4,2) = 0. The runs 6, 8 and 8, 10 are excluded because they are part of a longer run, and there are no other consecutive even integers with 4 divisors.
T(18, 3) does not exist. This follows from the theorem: If m = 2 mod 4, and m has 18 divisors, then m-2 does not have 18 divisors.
Proof: Let d be the number of divisors function (A000005). Recall that it is multiplicative with d(p^i)=i+1. If m = 2 mod 4 and has 18 divisors, then m/2 is odd and has 9 divisors, so m=2*r^2 for some odd r. Then m-2=2(r-1)(r+1). r-1 and r+1 are even and one of them is divisible by 4, so 2^4 divides m-2. r-1 and r+1 have no prime factors in common except 2, so if they are both divisible by odd primes, call them s and t, then m-2 is divisible by 2^4*s*t and has at least 20 divisors, contrary to hypothesis. Therefore either r-1 or r+1 is a power of 2; call it 2^j. Then the exponent of 2 in m-2 is j+2, so j+3 divides 18, so j is 3 or 6. This leaves 4 possibilities for m-2: 2*6*8, 2*8*10, 2*62*64, or 2*64*66. Of these, only 2*62*64 has 18 divisors, and 2*62*64+2 does not have 18 divisors.
T(36, 11) does not exist. Proof: Suppose 11 consecutive even numbers with 36 divisors exist. Name them n_i where n_i = i (mod 32). n_16 and n_24 cannot have 36 divisors, so the 11 numbers are n_26 through n_14. Then n_8 is 8*x^2 for some odd x. Suppose 3 | x. Then 9 | n_8, so n_2 and n_14 are divisible by 3 but not 9, and by 2 but not 4. So n_2 = 6*y^2 and n_14 = 6*z^2 for some y and z, and z^2 = y^2 + 2, which is impossible. Therefore 3 doesn't divide x. Therefore x^2 = 1 (mod 3), and n_8 = 2 (mod 3). So 3 | n_6. Suppose n_6 = 0 (mod 9). Then n_26 = 6 (mod 9). So n_26 is divisible by 3 but not 9, and by 2 but not 4. So n_26 = 6*y^2. y^2 = 1 (mod 4), so n_26 = 6 (mod 8), but by definition n_26 = 2 (mod 8), a contradiction. Therefore n_6 != 0 (mod 9). Suppose n_6 = 3 (mod 9). Then n_6 is divisible by 3 but not 9, and by 2 but not 4. So n_6 = 6*y^2. y^2 = 1 (mod 3), so n_6 = 6 (mod 9), a contradiction. Therefore n_6 != 3 (mod 9), so n_6 = 6 (mod 9). Then n_26 = 3 (mod 9). So n_26 and n_6 are divisible by 3 but not 9, and by 2 but not 4. So n_26 = 6*y^2 and n_6 = 6*z^2 for some y and z, and z^2 = y^2 + 2, which is impossible.
In the table below, the following notation will be used for terms with unknown values: F: k consecutive even integers with n divisors have been found. D: Dickson's Conjecture implies the existence of k consecutive even integers with n divisors. H: Schinzel's Hypothesis H implies the existence of k consecutive even integers with n divisors. ?: It has not been proven that k consecutive even integers with n divisors do not exist. A semicolon indicates than no further terms exist.
Table begins:
   n  T(n,1), T(n,2), ...
  ==  =======================================================
   2  2;
   3  4;
   4  14, 0, 6;
   5  16;
   6  12, 18;
   7  64;
   8  24, 40, 182;
   9  36;
  10  48;
  11  1024;
  12  60, 198, 348, 9050, 25180, 25658650, 584558736346;
  13  4096;
  14  192;
  15  144;
  16  120, 918, 5430;
  17  65536;
  18  180, 17298;
  19  262144;
  20  240, 6640, 4413038;
  21  576;
  22  3072;
  23  4194304;
  24  360, 3400, 19548, 134044, 182644, 7126044, 359208340, 16074693138, 419893531348, 214932235538, F, D, D, F, D;
  25  1296;
  26  12288;
  27  900;
  28  960, 640062, 32858781246;
  29  268435456;
  30  720, 110796496, F;
  31  1073741824;
  32  840, 18088, 180726;
  33  9216;
  34  196608;
  35  5184;
  36  1260, 41650, 406780, 3237731546, 3651712573692, F, F, ?, ?, ?;

Cf. A292580 (analog for consecutive integers), A319046 (analog for consecutive odd integers), A325116.

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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A319046 Irregular table read by rows: T(n,k) is the start of the first run of exactly k consecutive odd numbers having exactly n divisors, or 0 if no such run exists.

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A325117 Irregular table read by rows: T(n,k) is the start of the first run of exactly k consecutive even integers having exactly n divisors, or 0 if no such run exists.

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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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