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
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)
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
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, ?, ?, ?;
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
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.
Showing 1-3 of 3 results.
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