Everything 2014++

# Blog from 2014-03-31 on

## Note 10, 2014-03-31

The month is ending, the week has just started (h.00:36 at this very moment). And tomorrow there will be April Fool. From time to time I would post a joke on an Internet group. Today the only contact I have is with Literotica. It's barely anything. I never post there anything on April 1. Right now I feel too tired to think about anything. My jokes were like solid articles, hey! When Lech Wałęsa was at the time the president of Poland, I claimed on Poland-L that he joined a monastery, even by jumping over a fence. A few years ago I so-to-speak wrote to sci.math about a young Russian genius, supposedly a descendant of Niels Abel sister. He, according to a coming news story, discovered that he found a contradiction in mathematics. Thus he was working secretly together with Grisha Perelman on saving mathematics. I also had a story on poewiki.org about Norwid writing haiku, and being appreciated in Japan. Etc.

Actually, I got an idea of an April Fool joke. I can use it with different audiences. Do I have energy for this? I'll see how I feel in 2 days.

Until now I was just entertaining myself with prime products (don't confuse them with the prime ribs). But a serious is following: show that for larger positive reals  $a\ b$ there is a rough logarithmic equality:

$$\Pi_a\times\Pi_b \sim \Pi_{a+b}$$

Would there be a wonderful identity which would make the above approximaste equality plausible?

The above approximate equality looks elegant but the following working version is perhaps better:

$$\Lambda_{s\cdot b} - \Lambda_{s\cdot a} \sim \Lambda_{t\cdot b} - \Lambda_{t\cdot a}$$

for positive  $a\ b\ s\ t$ (where  $s\ t$  are large), where

$$\Lambda(x) := \max\{{p\in\mathbb P}: \Pi_p\le x\}$$

## Note 11, 2014-04-01

Actually, I ve sent three different April Fool jokes to four different places. I may say more after the day is over :-)

I have more thoughts about prime products. I'll wait a little. At this moment let me at least write the exponential variation of the last formula above:

$$\frac{\Pi_{a+x}}{\Pi_x} \sim \frac{\Pi_{a+y}}{\Pi_y}\qquad (\mathbf {nonsens!})$$

where  $a\ x\ y$  are positive real numbers, and  $x\ y$  are large. Of course the idea is:

$$\lim_{x\rightarrow\infty}\ \frac{\Pi_{a+x}}x = \ldots\qquad (\mathbf {nonsens!})$$

That's a nonsense!

## Note 12, 2014-04-04

I wanted to leave at least one note each day. I skept two days though, too bad. And I had things to write down but hesitated. I should just keep writing. If there are more than one theme, just write about anyone. Just keep writing.

I certainly have things about prime product. And I was thinking about all my main topics: daba, , art of agreement, and even about Tangia (Tangia is written with an upper case T, and without any declination).

## Note 13, 2014-04-06

The links suddenly are not working?! Here, inside this file (blog). Ok, I've fixed it.

All these days I could do, as meant in September, a prime product argument involving just one but quite general prime. I have three variations:

• $n-p$   and    $n+p$
• $n-p$   and    $n-2\cdot p$
• $n-p$   and    $n-p^2$

We can assume that  $n\ge 7$  (use your fingers to compute anything about  $n\le 6$).  Let's consider the crucial moment  $n - p$,  where  $n=\Pi_s$  and  $p|n$,  i.e.  $p\le s$.  First we look at the easy case when  $n-p$  is not a power of  $p$  (it's also the most common case).  Then there exist a prime  $q\le s$,  and different from  $p$,  which divides  $n-p$,  i.e.  $q|n-p$.  Then  $q\le \frac{n-p}p = \frac np-1$. Obviously  $s<q$  since  $q$  does not divide  $n = (n-p) + p$.  Thus ultimately:

$$s < q \le \frac np-1$$

Now assume that  $n-p=p^d$  is a power of  $p$,  which is the second case. Along my first version, let's consider  $n+p$.  Since  $n-p$  is a power of  $p$  then  $n+p$  is not such a power. Then, as in the first case, we get

$$s < q \le \frac np+1$$

## Note 14, 2014-04-08

I don't feel much like writing here. At least, as of yesterday, I am done with tax2013. That heps.

I am thinking about the Art of Agreement. Especially that there is a chance, just tiny, that may be I'll talk to people about it (not too likely). I am also considering daba+dabanese. I need to get the ability of using a daba base here at ipage. I can either ask my family to help me or I can call ipage technical support. Thus I am not sure which one to choose :) .  In the meantime I'll write some more about the prime products. Actually, this time I got a small but sharp item:

Let  $p\ge 7$  be a prime. Let

• $n:=\Pi_p$
• $\nu := \frac np = \Pi_{p-1}$
• $m:= n-p^2$,  hence
• $\mu:=\frac mp = \nu-p$

It follows that  $\mu > 1$  (actually, this is true but still needs a proof; or I just may assume that  $\mu>1$).  Obviously, each prime divisor  $s$  of  $\mu$  is larger than  $p$; and there is at least one such  $s$.  Thus

$$p\ <\ s\ \le\ \frac np - p$$

## Note 15, 2014-04-08

Should I go soon to a bookstore nearby (Literati) or should I keep writing right here on 2014++? I should go out. It'd be good. And it would make sense to get a bite, say one JJ-sub "here", and 4 to take home, to have it for another 2-3 days. Perhaps going out would pump me with energy. I should leave already within about half an hour.

Theorem (Euclid++)  Let  $2\ \ldots\ q\ \,p$  be an initial interval of consecutive primes, where  $p\ge 7$.  Then

$$p\le \frac {\Pi_q}2 - 2$$

Proof  The first primes are  $2\ 3\ 5\ 7$.  Thus  $\frac {\Pi_q}2 - 2\ge 13 > 1$.  Thus there exists a prime divisor  $s|\frac {\Pi_q}2 - 2$.  Of course  $s$  is odd. If also  $s|\Pi_q$  then we would have

$$s\ |\ \Pi_q - 2\cdot(\frac {\Pi_q}2 - 2) = 4$$

which would mean that  the odd prime  $s$  divides  $4$,  a contradiction.  Thus  $s$  is not a divisor of  $\ \Pi_q$,   i.e.  prime  $s > q$.  Thus  $\frac {\Pi_q}2 - 2\ge s\ge p$.  End of proof

2014-04-08, 18:30:40

Should I stay or should I go? I feel like continuing my writing but the common sense tells me to go out. OK, I'll go.

2014-04-08, 20:25:00

I am back from the bookstore. It was a short session and a short walk. BTW, I have discovered today the datge+time feature in my on-line ipage editor. Going back to the poetry meeting, it was not with any associate professor from any obscure university. Instead, the room was filled up with about 20 (up to 25) udergraduate UM students, most of them women. Four authors were reading their poems just published in a student magazine called xylem, 2013-2014. All authors were women. The sensations from their own skin was a significant part of their poetry. They wanted a discussion after the reading but nobody acted. I wish I would, it'd make it more interesting. It was an exceptionally quiet reading. Only the last author encouraged any interaction. All that has happened at the very end was complement from her friend claiming that these poems were nice. It was even hard to hear it.

2014-04-08, 20:36:07

I'm truly enjoing this date+time feature :) . Now I'll get myself a coffee, and then I'll go back to the prime products (I hope).

2014-04-08, 23:52:16cI've wasted some time, too much. Now, below, I will provided the closing tiime, followed in the same line by the opening time of the next note. I need to remember this. Somehow, I don't feel what I had in me about prime products. A new, unpleasant sensation. I'll try to recover it (or else, I should rest a bit, even if there is nothing to rest after; but the rest may help).

No, it's nearly midnight. I'll stop for a few or more minutes. No need to record time since it's almost midnight. I'll just start a new note, under a new day. See you tomorrow.

## Note 16, 2014-04-11

I am slacking off.

2014-04-11 22:34:22

The Euclid++ theorem is in priciple a bit weaker than the previous one (except for the issue of  $\mu > 1$). However, by pushing the Euclid++ argument a bit further we can get a funny corollary:

THEOREM  Let  $2\ldots q\ \,p \ge 7$  be the initial consecutive primes. Let  $d$  be a natural number such that  $2^d<\frac{\Pi_q}2$.  Then

• $\frac{\Pi_q}2 - 2^d \ge p$
• $2^d + 3 \le \frac{\Pi_q}2$

2014-04-11 22:52:30 *** 23:01:50

This would require to prove something special but hard:

$$\frac{\Pi}2 \ne 2^d+1$$

Otherwise, I am sure that many people, and certainly myself years ago, had similar thoughts. In general, let  $2\ldots q\ \,p \ge 7$  be the initial consecutive primes. Let  $S\ P$  be realtively prime natural numbers such that  $x|S\cdot T$  for every prime number  $x\le q$,  and for no prime number  $x>q$.  Then  $|S-T|=1$  or  $|S-T|\ge p$.

2014-04-11 22:34:22

## Note 17, 2014-04-12

02:21:00

My poor brain is fried, I better start writing about poetry :-). At MathOverflow question and answers group I posted the diophantine equation:

$$|\frac{\Pi_q}2-2^d|=1$$

I got two comments, from the same person. (2014-04-12, 02:26:56).

## Note 18, 2014-04-15

h.01:11:13 -- I've posted another question on MO, and an answer to someone's question; actually formally two answers. First an elegant answer, which didn't fit the question for a trivial reason. The I posted also a modification, less elegant but an exact answer. Let me mention that right now I have 1623pt, not much. I often reside on 17 page, where each full page features 36 participants. There are a total of 700 full pages + 12 users on the last page 701 (the last ficticious(?), called Community).

h.01:23:55 -- I am about to open a part of "Everything 2014++", namely called Compuverse or Uniformatique. I will use both names. It seems crazy to open more and more parts, and not write many. But this way I still may write more these days or else nothing.

I couldn't successfully click on the editing icon. Perhaps I could after all but I need to point a bit to the left side of the pencil icon (or here or there, wherever it works). h.01:31:19.

h.01:31:51 -- Indeed, clicking around the pencil icon helps. Anyway, at this time I plan to write a bit about perl. That's an easy proposition for me, at least on a very elementasry level. I may come back to counting the pseudo-orders (and partial orders). Last September I got progress, and saw hbow to do it way better. It'd be more sweat thought, the program will be somewhat harder. I should also learn language J. So far I didn't learn much, just next to nothing. But it's an excellent language, it's a follop-up APL, which I loved. My problem is that the manuals and tutorials for J feel boring to me. Or I got old. h.01:37:50.

## Note 19, 2014-04-16

h.04:26:09 -- I'm something like tired. My day, or night, was not totally wasted. At least, before the midnight, I've written a poem, in Polish. Just 8 lines (plus title). But it's a heavy weight piece. I was thinking about a possible translation (into English). At first it seemed unlikely. After a while I could see a chance. I am not going to do it, not yet (or may be never).

I also did the realted 2014++ book keeping, most of it. OK, some of it. The poem is here, and the respective links around here. Not perfect but what can I say. Perhaps I should have a better system, should have better tools for such tasks.

I was a bit active on MO recently, and even have a bit to show for my so-to-speak effort. And even one of my somewhat nasty comment this time was received with 2 supporting votes. Wow, I am spoiled now. UNtil now, again and again, I was voted down on every question or answer which was not strictly mathematical. Of course it's all fun, or should be fun :-). h.04:37:12.

Before I left California, I got rid of all my Perl books--I can't believe I did. I guess I decided that I will buy new editions more up to date. That was a lot of books. Now, to write about perl, I need some books. I cannot do it by just reading perl pages on Internet. To find out this or that--yes; but for more systematic writing I need regular book references. h.04:41:48.