Пентадекатлон мечты

[Yadryara]

Пока насчёт замеченных ошибок по датам напишу.

На 1-й странице.
D(48, 23) найдена в январе 26-го.

В Марафоне
D(96, 19) найдена в марте 26-го.

[Yadryara]

EUgeneUS в сообщении #1719831 писал(а):

У меня считалось задание для улучшения D(24,20) (кстати, найденная Демисом цепочка на первой страницы темы ещё не отражена) примерно в 15 раз.
По всем 400 паттернам, со всеми перестановками простых до $100$.
Досчиталось, улучшение не найдено.

Приплыли. Опять считали одно и то же? И вместо предварительного согласования, об этом сообщается уже пост-фактум: "досчиталось".

Почему думаю, что одно и то же считали. Да потому что различных подходящих серий паттернов для D(24,20) гораздо меньше. И речь, видимо, про те самые 400 комплектов, которые в серии 2-12-6-8!

EUgeneUS в сообщении #1719831 писал(а):

По всем 400 паттернам, со всеми перестановками простых до $100$.

Вроде переставлять простые для 24 делителей это совсем глупо. Возможно, участник имел в виду всё-таки перестановки квадратов простых.

И выполнять именно все перестановки квадратов простых до $100$ весьма странно. Разве не очевидно что надо смотреть именно по величине шага, чтобы было больше попыток.

Вот первые 80 перестановок.

(Оффтоп)

pors8[1] = [23, 29, 31, 37, 41, 43, 47, 53];  \\ 0
pors8[2] = [23, 29, 31, 37, 41, 43, 47, 59];  \\ 23
pors8[3] = [23, 29, 31, 37, 41, 43, 47, 61];  \\ 32
pors8[4] = [23, 29, 31, 37, 41, 43, 53, 59];  \\ 57
pors8[5] = [23, 29, 31, 37, 41, 43, 47, 67];  \\ 59
pors8[6] = [23, 29, 31, 37, 41, 43, 53, 61];  \\ 68
pors8[7] = [23, 29, 31, 37, 41, 43, 47, 71];  \\ 79
pors8[8] = [23, 29, 31, 37, 41, 47, 53, 59];  \\ 88
pors8[9] = [23, 29, 31, 37, 41, 43, 47, 73];  \\ 89
pors8[10] = [23, 29, 31, 37, 41, 47, 53, 61];  \\ 101
pors8[11] = [23, 29, 31, 37, 41, 43, 53, 67];  \\ 103
pors8[12] = [23, 29, 31, 37, 43, 47, 53, 59];  \\ 107
pors8[13] = [23, 29, 31, 37, 41, 43, 59, 61];  \\ 108
pors8[14] = [23, 29, 31, 37, 43, 47, 53, 61];  \\ 121
pors8[15] = [23, 29, 31, 37, 41, 43, 47, 79];  \\ 122
pors8[16] = [23, 29, 31, 37, 41, 43, 53, 71];  \\ 128
pors8[17] = [23, 29, 31, 37, 41, 43, 53, 73];  \\ 141
pors8[18] = [23, 29, 31, 37, 41, 47, 53, 67];  \\ 142
pors8[19] = [23, 29, 31, 37, 41, 43, 47, 83];  \\ 145
pors8[20] = [23, 29, 31, 37, 41, 47, 59, 61];  \\ 149
pors8[21] = [23, 29, 31, 37, 41, 43, 59, 67];  \\ 151
pors8[22] = [23, 29, 31, 41, 43, 47, 53, 59];  \\ 154
pors8[23] = [23, 29, 31, 37, 43, 47, 53, 67];  \\ 167
pors8[24] = [23, 29, 31, 37, 41, 43, 61, 67];  \\ 169
pors8[25] = [23, 29, 31, 41, 43, 47, 53, 61];  \\ 171
pors8[26] = [23, 29, 31, 37, 41, 47, 53, 71];  \\ 172
pors8[27] = [23, 29, 31, 37, 43, 47, 59, 61];  \\ 174
pors8[28] = [23, 29, 31, 37, 41, 43, 47, 89];  \\ 181
pors8[29] = [23, 29, 31, 37, 41, 43, 53, 79];  \\ 182
pors8[30] = [23, 29, 31, 37, 41, 43, 59, 71];  \\ 182
pors8[31] = [23, 29, 31, 37, 41, 47, 53, 73];  \\ 188
pors8[32] = [23, 29, 31, 37, 41, 43, 59, 73];  \\ 198
pors8[33] = [23, 29, 31, 37, 43, 47, 53, 71];  \\ 199
pors8[34] = [23, 29, 31, 37, 41, 47, 59, 67];  \\ 200
pors8[35] = [23, 29, 31, 37, 41, 43, 61, 71];  \\ 202
pors8[36] = [23, 29, 31, 37, 41, 43, 53, 83];  \\ 211
pors8[37] = [23, 29, 31, 37, 43, 47, 53, 73];  \\ 217
pors8[38] = [23, 29, 31, 37, 41, 53, 59, 61];  \\ 217
pors8[39] = [23, 29, 31, 37, 41, 43, 61, 73];  \\ 219
pors8[40] = [23, 29, 31, 37, 41, 47, 61, 67];  \\ 221
pors8[41] = [23, 29, 31, 41, 43, 47, 53, 67];  \\ 227
pors8[42] = [23, 29, 31, 37, 43, 47, 59, 67];  \\ 230
pors8[43] = [23, 29, 31, 37, 41, 43, 47, 97];  \\ 234
pors8[44] = [23, 29, 31, 41, 43, 47, 59, 61];  \\ 236
pors8[45] = [23, 29, 31, 37, 41, 47, 59, 71];  \\ 237
pors8[46] = [23, 29, 31, 37, 41, 47, 53, 79];  \\ 237
pors8[47] = [23, 29, 31, 37, 43, 53, 59, 61];  \\ 248
pors8[48] = [23, 29, 31, 37, 41, 43, 59, 79];  \\ 250
pors8[49] = [23, 29, 31, 37, 43, 47, 61, 67];  \\ 253
pors8[50] = [23, 29, 31, 37, 41, 47, 59, 73];  \\ 257
pors8[51] = [23, 29, 31, 37, 41, 43, 53, 89];  \\ 258
pors8[52] = [23, 29, 31, 37, 41, 47, 61, 71];  \\ 261
pors8[53] = [23, 29, 37, 41, 43, 47, 53, 59];  \\ 262
pors8[54] = [23, 29, 31, 37, 41, 43, 47, 101];  \\ 263
pors8[55] = [23, 29, 31, 37, 41, 43, 67, 71];  \\ 264
pors8[56] = [23, 29, 31, 41, 43, 47, 53, 71];  \\ 268
pors8[57] = [23, 29, 31, 37, 43, 47, 53, 79];  \\ 271
pors8[58] = [23, 29, 31, 37, 43, 47, 59, 71];  \\ 271
pors8[59] = [23, 29, 31, 37, 41, 47, 53, 83];  \\ 272
pors8[60] = [23, 29, 31, 37, 41, 43, 61, 79];  \\ 274
pors8[61] = [23, 29, 31, 37, 41, 43, 47, 103];  \\ 277
pors8[62] = [23, 29, 31, 37, 41, 47, 61, 73];  \\ 281
pors8[63] = [23, 29, 31, 37, 41, 53, 59, 67];  \\ 282
pors8[64] = [23, 29, 31, 37, 41, 43, 67, 73];  \\ 285
pors8[65] = [23, 29, 31, 37, 41, 43, 59, 83];  \\ 286
pors8[66] = [23, 29, 37, 41, 43, 47, 53, 61];  \\ 287
pors8[67] = [23, 29, 31, 41, 43, 47, 53, 73];  \\ 289
pors8[68] = [23, 29, 31, 37, 43, 47, 59, 73];  \\ 292
pors8[69] = [23, 29, 31, 37, 43, 47, 61, 71];  \\ 297
pors8[70] = [23, 29, 31, 41, 43, 47, 59, 67];  \\ 306
pors8[71] = [23, 29, 31, 37, 41, 43, 47, 107];  \\ 307
pors8[72] = [23, 29, 31, 37, 41, 53, 61, 67];  \\ 308
pors8[73] = [23, 29, 31, 37, 43, 47, 53, 83];  \\ 309
pors8[74] = [23, 29, 31, 37, 41, 43, 61, 83];  \\ 313
pors8[75] = [23, 31, 37, 41, 43, 47, 53, 59];  \\ 313
pors8[76] = [23, 29, 31, 37, 47, 53, 59, 61];  \\ 316
pors8[77] = [23, 29, 31, 37, 41, 47, 59, 79];  \\ 318
pors8[78] = [23, 29, 31, 37, 43, 47, 61, 73];  \\ 319
pors8[79] = [23, 29, 31, 37, 43, 53, 59, 67];  \\ 320
pors8[80] = [23, 29, 31, 37, 41, 43, 47, 109];  \\ 322

Справа указан прирост шага в %. Видно же, что уже здесь начали появляться числа превышающие 100:
В 54-й расстановке — 101.
В 61-й расстановке — 103.
В 71-й расстановке — 107.
В 80-й расстановке — 109.

[Huz]

Dmitriy40 в сообщении #1719841 писал(а):

Например для M(60) и M(84) каковы U(60) и U(84)?

Assuming $U(n)$ means the value we have proven $M(n)$ cannot exceed, my records say $U(60) = 23$ and $U(84) = 15$. For $12n < 200$, I have: $U(12n) = [ 15, 31, 15, 31, 23, 31, 15, 31, 21, 107, 21, 31, 21, 31, 26, 31 ]$.

-- 11.03.2026, 02:03 --

Yadryara в сообщении #1719868 писал(а):

Пока насчёт замеченных ошибок по датам напишу.

На 1-й странице.
D(48, 23) найдена в январе 26-го.

В Марафоне
D(96, 19) найдена в марте 26-го.

I'm confused, I see $D(48,23)$ reported by Eugene on 31st January; and we haven't yet reached 26th March. Is this a translation error? Yandex is showing them as the 26th day of January/March, but perhaps they were supposed to be "January/March 2026"?

[EUgeneUS]

Huz в сообщении #1719871 писал(а):

Yandex is showing them as the 26th day of January/March, but perhaps they were supposed to be "January/March 2026"?

Yandex is wrong here.
Right:
"в январе 26-го" -> "in January 2026"
but:
"26-го января" -> "on 26th day of January"

-- 11.03.2026, 07:00 --

Yadryara в сообщении #1719870 писал(а):

Приплыли.

Yadryara в сообщении #1719870 писал(а):

это совсем глупо.

Yadryara в сообщении #1719870 писал(а):

странно. Разве не очевидно

Идите к чёрту.

-- 11.03.2026, 07:29 --

Huz
About FORMULA in the A119479
Currently, the OEIS lists two long-known cases:

Цитата:

a(2n+1) = 1, since numbers with an odd number of divisors must be squares. If n is not divisible by 3, a(2n) <= 7.

But in 2022, other important cases were proven by Denis Shatrov. What is reflected on the first page of this topic

Specifically:
1. $a(12n+6) \leqslant 5$
2. For $a(12n \pm 2)$: $a(2P) \leqslant 3$, where $P=\prod_{i=1}^s p_i$, and $p_i$ - prime numbers such that $gcd_{i=1}^s(p_i-1) > 2$.

There are also known restrictions that complement point 2 above (for $a(12n \pm 2)$):
2.1. $a(2p) \le 3$, where $p$ - prime number, greater than 3;
2.2. $a(2pq) \le 3$, where $p,q$ - prime numbers greater than 3 (not necessarily distinct);

This gives a nearly complete picture of the upper bounds, with two exceptions:
1. For $a(12n \pm 2)$, where $P=\prod_{i=1}^s p_i$, and $p_i$ - prime numbers such that $gcd_{i=1}^s(p_i-1) = 2$.
2. For $a(12n)$, the maximum length of such chains is not bounded from above according to the Erdős conjecture. AFAIK.

Perhaps it makes sense to reflect these results in the OEIS?

[EUgeneUS]

EUgeneUS в сообщении #1719873 писал(а):

This gives a nearly complete picture of the upper bounds, with two exceptions:
1. For $a(12n \pm 2)$, where $P=\prod_{i=1}^s p_i$, and $p_i$ - prime numbers such that $gcd_{i=1}^s(p_i-1) = 2$.

Correction:
1. For $a(12n \pm 2)$: $a(2P)$ where $P=\prod_{i=1}^s p_i$, and $p_i$ - prime numbers such that $gcd_{i=1}^s(p_i-1) = 2$, and $p_i > 3$

[VAL]

Huz в сообщении #1719871 писал(а):

Assuming $U(n)$ means the value we have proven $M(n)$ cannot exceed, my records say $U(60) = 23$ and $U(84) = 15$. For $12n < 200$, I have: $U(12n) = [ 15, 31, 15, 31, 23, 31, 15, 31, 21, 107, 21, 31, 21, 31, 26, 31 ]$

Добавлю, что числа 15 и 31 - наверняка точные значения соответствующих $M(k)$. Это следует из гипотезы Диксона, в справедливости которой серьезные (а также несерьезные, вроде, меня) математики не сомневаются.
В то же время, полагаю, $M(60)=17$. Более длинные цепочки могут получиться только при условии разрешимости экзотических систем диофантовых уравнений, разрешимость которых вызывает большие сомнения.

[EUgeneUS]

VAL
По оформлению первого поста предложение: вытащить в таблицу все $k=12n$, например, до $k < 200$ без пропусков.

[VAL]

EUgeneUS в сообщении #1719895 писал(а):

VAL
По оформлению первого поста предложение: вытащить в таблицу все $k=12n$, например, до $k < 200$ без пропусков.

Чуть выше уже писал, что таблицу пришлось урезать вынужденно. Иначе весь пост корежится.
В принципе, можно выложить ссылку на скачивание такой таблицы. Ссылка на другую таблицу там уже есть.

[Huz]

VAL в сообщении #1719954 писал(а):

Чуть выше уже писал, что таблицу пришлось урезать вынужденно. Иначе весь пост корежится.
В принципе, можно выложить ссылку на скачивание такой таблицы. Ссылка на другую таблицу там уже есть.

It might be sensible to consider hosting some tables as wiki pages under the divrep github repository. I have a local mediawiki instance in which I maintain various tables of progress, which I was also thinking of putting there.

If I can work out how to set permissions appropriately, I was also planning to open it up for one or more people to maintain Russian-language versions of things like the divrep documentation.

-- 11.03.2026, 18:39 --

EUgeneUS в сообщении #1719873 писал(а):

Perhaps it makes sense to reflect these results in the OEIS?

I agree, are the proofs published anywhere?

For $n = \pm 2 \pmod{12}$, I think you need to refer to $s$ as well: if I have it right, given $n = 2\prod_1^s{p_i}$ we know $M(n) \le 3$ when $s < 3$ or $\gcd(\{ p_i - 1 \}) > 2$, and so the first unknown case is $n=350$. $p_i > 3$ already follows from $n = \pm 2 \pmod{12}$.

[VAL]

25-я дрофа:

680170245157095730628531974360935807213746617142556240181240

Huz в сообщении #1719962 писал(а):

It might be sensible to consider hosting some tables as wiki pages under the divrep github repository.

I have posted materials on dxdy many times using external repositories. It almost always ended with the materials becoming unavailable over time.

[EUgeneUS]

Huz в сообщении #1719962 писал(а):

are the proofs published anywhere?

This is a good question and a sore subject.

EUgeneUS в сообщении #1719873 писал(а):

2.1. $a(2p) \le 3$, where $p$ - prime number, greater than 3;

This is a long-known result from Duentsch, R. B. Eggleton. Equidivisible consecutive integers, 1990., http://www.cosc.brocku.ca/~duentsch/archive/equidiv.pdf.

The remaining results, as you know, were obtained by the participants in this topic in the summer of 2022.
It was planned to present them in the form of an article that could be posted on arXiv.org. But this work was not completed.

And I already doubt that it will be fulfilled.
However, drafts can be posted publicly, for example, here on the forum. Including the possibility of verification of evidence by interested parties.

I restored access to papeeria to the project with drafts of the article, and checked the current state of affairs.

1. $a(2pq) \le 3$, where $p,q$ - prime numbers greater than 3 (not necessarily distinct);

This is my result. The proof is presented in a separate document. There are versions in English and Russian. And even somewhere in the depths of the topic there are links to Google Disk, where the latest Russian and English versions are posted. There is no problem posting it in any other place, except for one thing - the text is not formatted according to the requirements for a scientific article.

2a. $a(12n+6) \leqslant 5$

I participated in the proof of this statement, in part reducing it to the proof of the unsolvability of 16 Diophantine equations. But the main work - the actual proof of unsolvability this Diophantine equations - was done by Denis Shatrov.

2b. For $a(12n \pm 2)$: $a(2P) \leqslant 3$, where $P=\prod_{i=1}^s p_i$, and $p_i$ - prime numbers such that $gcd_{i=1}^s(p_i-1) > 2$.

This result belongs entirely to Denis Shatrov

Before stopping work on the article, Denis managed to draw up evidence of these results (2a, 2b).

It would be incorrect, of course, to post these drafts in the public domain without Denis’ permission.

But I can try to contact Denis through contacts available since 2022 in order to obtain permission to do this.

If that makes any sense, of course.

-- 12.03.2026, 19:25 --

VAL в сообщении #1719969 писал(а):

It almost always ended with the materials becoming unavailable over time.

GitHub hosts a lot of processes in a huge number of companies around the world. As a platform, GitHub is one of the most stable and reliable in the world.
Without considering political risks, the worst that can happen is that the account owner stops updating information.

[EUgeneUS]

Huz
BTW, Denis also has a derivation of the hypothesis $a(2P) \leqslant 3$, where $P=\prod_{i=1}^s p_i$, and $p_i$ - prime numbers such that $gcd_{i=1}^s(p_i-1) = 2$, as a consequence of the abc conjecture.
Which is also very interesting.

[Yadryara]

Ласточки от Демиса для D(96, 20) :

17882636301316551372326583218834641574747613582642367783561208        1 111111111111111 11     18
81008598805027982227985584468019740960549303479915324245026808        11 11111 11111111111     18
91132039717277785646193652415770263152675632825692678014089208        11 1111 1111111111111    19
51906864054094285732336117271818729070968130611277162558396408    1   1111  11111111111111     19
88569547215544213733348032610081035114819014564570430983510008        1111 11 111111111111     18
3015976881568149595029624025634377851444336746571289754761208         1  11111111111111111     18
18884308211399133171560809061962726782483245195210805759010808        111111111111111 11 1     18
65286103304398022810199499908144628068112419302967246299170808         11111 1111111111111     18
50971949892577166574885130955498868223258466907863669541308408        11111111 111111 1111     18

Это старый способ поиска. Посему и числа не более чем 62-значные.

Программа поиска для не более чем 58-значных всё ещё тестируется.

[VAL]

26-я и 27-я дрофы:

340476555032973904658742243978816190935038086014234806581240
651190466521177566390141992017313598013295700023139830581240

Последняя цепочка пока единственная, в которой все 15 чисел имеют не менее 192 делителей.

И 28-я:

47249548618339929465581601087651938220083955911765136181240

[VAL]

29-я, 30-я, 31-я и 32-я:

387431201429151567610422586389361942272692068014215107381240
483462492335183964502972066419315702040044110982605059381240
367009459600109476621817073867271197171093181371296860981240
457730216675148221515234869645360060178758530324870236981240

Конечна ли эта стая?!

Уточню. В том что, стая этих дроф бесконечна, я уверен.
Меня смущает, не встретится ли она целиком, до появления первой пятнашки.