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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.

%I A325117 #26 Dec 24 2021 02:30:54
%S A325117 2,4,14,0,6,16,12,18,64,24,40,182,36,48,1024,60,198,348,9050,25180,
%T A325117 25658650,584558736346,4096,192,144,120,918,5430,65536,180,17298,
%U A325117 262144,240,6640,4413038,576,3072,4194304,360,3400,19548,134044,182644,7126044,359208340,16074693138,419893531348,214932235538
%N 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.
%C A325117 The number of terms in row n is A325116(n).
%C A325117 2.46*10^12 <= T(24,11) <= 299005907036986132.
%C A325117 T(24,14) <= 1010085195622895590495442.
%C A325117 T(30,3) <= 1359906389476760004389052496.
%C A325117 5.17*10^12 < T(36, 6) <= 13707985134823441146.
%C A325117 T(36, 7) <= 1678936725442128595619270138.
%e A325117 T(4,3) = 6 because 6, 8, and 10 each have 4 divisors.
%e A325117 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.
%e A325117 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.
%e A325117 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.
%e A325117 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.
%e A325117 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.
%e A325117 Table begins:
%e A325117    n  T(n,1), T(n,2), ...
%e A325117   ==  =======================================================
%e A325117    2  2;
%e A325117    3  4;
%e A325117    4  14, 0, 6;
%e A325117    5  16;
%e A325117    6  12, 18;
%e A325117    7  64;
%e A325117    8  24, 40, 182;
%e A325117    9  36;
%e A325117   10  48;
%e A325117   11  1024;
%e A325117   12  60, 198, 348, 9050, 25180, 25658650, 584558736346;
%e A325117   13  4096;
%e A325117   14  192;
%e A325117   15  144;
%e A325117   16  120, 918, 5430;
%e A325117   17  65536;
%e A325117   18  180, 17298;
%e A325117   19  262144;
%e A325117   20  240, 6640, 4413038;
%e A325117   21  576;
%e A325117   22  3072;
%e A325117   23  4194304;
%e A325117   24  360, 3400, 19548, 134044, 182644, 7126044, 359208340, 16074693138, 419893531348, 214932235538, F, D, D, F, D;
%e A325117   25  1296;
%e A325117   26  12288;
%e A325117   27  900;
%e A325117   28  960, 640062, 32858781246;
%e A325117   29  268435456;
%e A325117   30  720, 110796496, F;
%e A325117   31  1073741824;
%e A325117   32  840, 18088, 180726;
%e A325117   33  9216;
%e A325117   34  196608;
%e A325117   35  5184;
%e A325117   36  1260, 41650, 406780, 3237731546, 3651712573692, F, F, ?, ?, ?;
%Y A325117 Cf. A292580 (analog for consecutive integers), A319046 (analog for consecutive odd integers), A325116.
%K A325117 nonn,tabf,more
%O A325117 2,1
%A A325117 _David Wasserman_, Mar 27 2019