Пентадекатлон мечты

[EUgeneUS]

UPD: отдельное спасибо нужно сказать DemIS'у.
Без его поддержки я бы на такой длительный расчет не взялся. Его мощности уменьшили время расчета примерно в три раза, то есть были около двух раз больше, чем мои.

[EUgeneUS]

Расчет группы Б (с ключом -p500 и до известной цепочки) окончен.

[Huz]

Hi guys, it has been quiet for a while. Is anyone still running my code? If so, I'll try to organise a new version - there are further speed improvements, and a couple of new features.

(Оффтоп)

I have also been looking at a somewhat similar sequence, A309981. Is anyone able to prove the given value $a(49) = 2$? I have similar problems with $a(324) ?= 2$ and $a(361) ?= 8$, all reliant on similar Pell equations. Please feel free to email me about this if it is too off-topic here.

[EUgeneUS]

Huz
Hi, Hugo!
AFAIK, after finding $U(6,14)$ by Demis no one is doing the calculations.
When we were searching for $U(6,14)$ we had 40 threads, and this has been going on for several months.
According to rough estimates, several thousand threads will be required to obtain some of the following results in an acceptable time
Both for searching for examples of chains where there are none, and for proving the minimality of chains where it has not been proven

As for me, I am ready to join the calculation in some team for the task in A292580 or A119579, if the estimated time is acceptable - no more than a few months. IMHO, the best or even the only option is to organize BOINC settlements with a team of several hundred participants. So that the total number of threads is several thousand.

[Huz]

A119579 doesn't seem likely to be what you meant, is it a typo?

It turns out that the way my code breaks down the A292580 problem into chunks is not a good fit for BOINC - they really want predictably same-sized chunks, so that participants know they will get an appropriate credit for doing any given amount of work. I had started talking with a BOINC expert who offered to work with me on adapting it, but he flaked out on me so I've got no further with that. I suspect it isn't helped by the fact that choosing an optimal set of parameters (mainly -g, -W, -f) is an art in itself, and can make a huge difference to the runtime - I'd love to have the code determine optimal settings for itself, but I don't currently have any idea how to achieve that.

In the meantime I've mainly been working on A309981 and a variant of A292580 (which seems to be much harder to calculate) where we look for a chain $a(n) = \max k: \exists d: \tau(d + in) = n: 0 \le i < k$ - those have been driving most of the code improvements in recent months. For the first, I've proven 920 values (with minimal examples) of the first 1000; for the second I haven't even managed to find enough values to make an oeis entry - I think it would need Vladimir's skills to show whether $a(6) = 10$ and $a(10) = 8$.

[EUgeneUS]

Huz в сообщении #1633577 писал(а):

A119579
doesn't seem likely to be what you meant, is it a typo?

Of course, I meant A119479

Huz в сообщении #1633577 писал(а):

It turns out that the way my code breaks down the A292580
problem into chunks is not a good fit for BOINC

It's a pity

Huz в сообщении #1633577 писал(а):

I suspect it isn't helped by the fact that choosing an optimal set of parameters (mainly -g, -W, -f) is an art in itself, and can make a huge difference to the runtime - I'd love to have the code determine optimal settings for itself, but I don't currently have any idea how to achieve that.

IMHO.
A rough estimate of the launch parameters can be made through trial runs. And submit it for settlement to BOINC with them. Accurate automatic parameter estimation seems so complicated that it is no longer necessary.

[Huz]

@VAL: this argument can be refined slightly, since both $n_2 = 2 \cdot x_2^2$ and $n_3 = 3 \cdot x_3^2$, the quadratic residues mean that any prime $p$ dividing $n_1$ must be of form $p \in \{1, -1\} \pmod{24} = \{23, 47, 71, 73, ...\}$.

For $n = 2pqr < 1000$, I find that this immediately gives $\max(v_0) + 1 < \min(v_1)$ for $n$ in (54, 90, 126, 150, 198, 210, 234, 250, 294, 306, 330, 342, 390, 414, 462, 522, 546, 558, 666, 738, 774, 846, 954), while there are still some values that need explicit testing for $n$ in (350, 490, 510, 550, 570, 650, 686, 690, 714, 726, 770, 798, 850, 858, 870, 910, 930, 950, 966). Always assuming that my code to check these is correct.

VAL в сообщении #1554809 писал(а):

Приведу доказательство того, что $M(90) \le 5$.
Возможно, это кому-то интересно. Но даже если не интересно, все равно, приведу, в надежде, что меня проверят. Ну и чтобы доказательство не потерялось (как у меня бывало с доказательствами на бумажке).

Я буду придерживаться фактов и обозначений из Теоремы 1 (в этой ситуации доказательство будет совсем коротким).
Согласно этой теореме, если числа $n_0, n_1, n_2$ и $n_3$ имеют по 90 делителей, то $n_0$ кратно 3 и 5. Кроме кого, оно кратно 8. Чтобы у него оказалось 90 делителей оно должно иметь вид $2^4\cdot3^2\cdot5^2\cdot p$, или $2^4\cdot3^2\cdot5\cdot p^2$ или $2^4\cdot3\cdot5^2\cdot p^2$ или вовсе не зависеть от параметра $p$.
В то же время, $n_2=2x^2$ для некоторого нечетного $x$. Тогда $2x^2-2=2^4\cdot3^2\cdot5^2\cdot p$ (или другому возможному варианту $n_0$).
Отсюда $(x-1)(x+1)=2^3\cdot3^2\cdot5^2\cdot p$. НОД чисел $x-1$ и $x+1$ равен 2. Отсюда $p \le 2\cdot3^2\cdot5^2+1$. И проверка свелась к короткому конечному перебору.

Остальные варианты (и остальные числа из $\{90, 126, 150, 162, 198, 210, 234, 294, 306, 330, 342, 390, 462, 510, 546, 726\}$) проверяются аналогично (включая 162, для которого на показатель двойки у $n_0$ не хватит одного простого делителя).

Надеюсь, что нигде не соврал. А если соврал, остается надеяться, что меня поймают за руку.

[EUgeneUS]

VAL
Вам выше Хуго написал.
(сделал корректное упоминание ника участника, чтобы нотификация пришла, если настроены нотификации на упоминания)