
EXAMPLE

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 m2 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 m2=2(r1)(r+1). r1 and r+1 are even and one of them is divisible by 4, so 2^4 divides m2. r1 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 m2 is divisible by 2^4*s*t and has at least 20 divisors, contrary to hypothesis. Therefore either r1 or r+1 is a power of 2; call it 2^j. Then the exponent of 2 in m2 is j+2, so j+3 divides 18, so j is 3 or 6. This leaves 4 possibilities for m2: 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, ?, ?, ?;
