D(12,k) calculation

Searching for minimal chains of 11 or more consecutive integers all with 12 divisors, we use the procedure described here. We rely on certain elementary mathematical proofs, which we also give below.

The logic of the calculation

Lemmas 1 to 7 below are all derived by the general logic built into pcoul. So in particular, any pattern pcoul attempts will include one of $32p$ and $8p^2$ but not both. Let us call those class A and class B patterns, respectively.

There is a proposed proof that no class B pattern can provide a complete chain of length $\ge 10$; however since these patterns fix a square they can be run very quickly, so it is as easy just to run them. I have done this for all class B patterns for $D(12,11)$ and $D(12,12)$.

For class A patterns, we refine things further following a scheme proposed by Dmitry Petukhov. Each element of a chain must have a factor of the form $p^2$, $p^5$ or $p^{11}$. Given a best known upper limit $n_x$, we select a threshold $Z > \sqrt[5]{n_x/2}$. Primes $p > Z$ we check globally for all patterns using the program sq12, to show that no element in a chain up to $n_x$ can be divisible by $p^2$. For the individual class A patterns we then run pcoul with the option -p<Z>, to have it try all combinations within the given pattern of qualifying prime powers $p^2, p^5, p^{11}: p \le Z$. Thus for each pattern all possible $p$ are considered.

For $p > Z$, the logic of sq12 is to generate all $q, r: q < r, p^2qr < n_x$ and $q: p^2q^3 < n_x$, and in each case select the pivot $h=32s$ as the nearest multiple of 32 to $p^2qr$ or $p^2q^3$. For every choice of $q$ and $r$, we then print $p^2qr$ (or $p^2q^3$) if all of $(h-3, h-2, ..., h+3)$ have 12 divisors; if any cases are printed, they are then fully checked manually. sq12 takes advantage of Lemmas 1 to 7 below to reject cases more efficiently.

Z

D(12,11)

For $D(12,11)$ we currently have $n_x = 9887353188984012120346$, and using the scheme described above we require $Z > \sqrt[5]{n_x/2} = 21817$.

I tested $Z > 13 \cdot 10^6$ by the time calculations for $D(12,11)$ were complete. However the introduction of the -W option with v20221201 meant that improving this value became less critical.

I currently recommend running pcoul with -p14e6.

Class B patterns have already been tested. (1260 patterns, total time 153.16s.)

D(12,12)

For $D(12,12)$ we currently have $n_x = 120402988681658048433948$, so we require $Z \ge \sqrt[5]{n_x/2} = 35968$.

I have currently tested $Z > 2 \cdot 10^8$.

Class B patterns have already been tested. (872 patterns, total time 271.93s.)

D(12,13)

For $D(12,13)$ we currently have $n_x = 586683019466361719763403545$, so we require $Z \ge \sqrt[5]{n_x/2} = 196550$.

D(12,14)

For $D(12,14)$ we currently have $n_x = 1966089440441196672524986345512345$, so we require $Z \ge \sqrt[5]{n_x/2} = 3967479$.

D(12,15)

For $D(12,15)$ we currently have $n_x = 80215613469168729088982885848674841$, so we require $Z \ge \sqrt[5]{n_x/2} = 8330014$.

Lemmas

Lemma 1: $\tau(32p) = 12$ implies $p$ is prime.

• the only other possibility is $p=64$, we show manually that this is not part of a chain.

Lemma 2: $n \equiv 16 \pmod{32} \implies \tau(n) \ne 12$.

• $\tau(n)$ must be divisible by 5.

Lemma 3: $n \equiv 24 \pmod{32} \implies \tau(n) \ne 12$.

• $n = 8m: m \equiv 3 \pmod{4}$, but then we must have $\tau(m) = 3$, so $m$ is the square of a prime, which contradicts $m \equiv 3 \pmod{4}$.

Lemma 4: Any chain of 9 or more consecutive integers with 12 divisors has at least one element that is either of the form $32p$ or of the form $32n + 8 = 8q^2$.

• This follows immediately from Lemmas 1 to 3.

Lemma 5: We cannot have both of $\tau(32p) = 12, \tau(32p + 8) = 12$.

• $\tau(32p + 8) = 12 \implies \tau(4p+1) = 3$. So $4p+1$ is the square of a prime, implying $4p \equiv 0 \pmod{24}$ which contradicts $p$ prime.
• the only other possibilities are when $p=2$ or $p=3$, we show manually that they are not parts of a chain.

Lemma 6: We cannot have both $\tau(32p) = 12$ and $\tau(32p - 2 = 6q^2) = 12$, and we cannot have both $\tau(32p) = 12$ and $\tau(32p + 2 = 6q^2) = 12$.

• In the first case we have $3q^2 = 16p + 1$, and in the second case $3q^2 = 16p - 1$. But with $q$ prime, $3q^2 \equiv 3 \pmod{8}$, a contradiction.

Lemma 7: If $(32p - 2, 32p, 32p + 2)$ are all elements of a chain, with $p > 3$, then either $32p - 2 = 18q$ or $32p + 2 = 18q$ for some prime $q$.

• This follows immediately from Lemma 6.