2:04 AM
I wonder what modding iterates looks like.

8 hours later…
10:07 AM
Clearly there is a prime that is $n$ digits long since by Bertrand's Postulate, there is $10^{n-1}<p<2 \cdot 10^{n-1}$
Clearly there is a semiprime that is $n$ digits long since by Bertrand's Postulate, we have $10^{n-2}<p<2 \cdot 10^{n-2} \implies 11 \cdot 10^{n-2} < 11p < 22 \cdot 10^{n-2}$
@Mathphile I doubt it is true, but it is interesting.
It can't be true actually. We could just find a cunningham chain $p,2p+1$ and show that $4p+3>35$ is divisible by $5$ and $7$. You need $p \equiv 8 \pmod{35}$. If $p =15k+8$, we basically need to find primes $15k+8,30k+17$
Oh sorry ^above is wrong at the last
It is supposed to be that $p=35k+8$ which shows that we need to find primes $35k+8$ and $70k+17$.
And as expected, that has a quite small solution. $43,87$ is a cunningham chain. The next element-to-be is $175=5 \cdot 35=5^2 \cdot 7$. Not semiprime!

10:52 AM
Disproved again!
although I too did not expect it to be true

@Mathphile What about this challenge : Find a polynomial f(n) , such that f(n) is semiprime for n = 0,1,...,k , such that k is as large as possible ?

$k$ can be arbitrarily large

How this ?

You can just set the degree of the polynomial to be $k+1$. The values $f(n)$ for $0 \leqslant n \leqslant k$ can take whatever values you assign them. Make them all semiprime.

Of course, no constant polynomial.
OK, I forgot that the coefficients should be integers.
And a small degree would be nice (in particular 2)
I just noticed that we can multiply the polynomial x^2+x+41 with some prime. So, to make it more interesting, the polynomial should be irreducible and primitive.

2 hours later…
12:50 PM
only 114 of the first 1000 survive to y=2
10y^2+10y+2 survives my code until y=6 even though it has coefficients that share a factor.
? parforvec(x=[[1,10],[1,10],[1,10]],polisirreducible(Pol(Vec(x)))&!isprime(c=vecsum(x))& bigomega(c)==2 & bigomega(substpol(Pol(Vec(x),y),y,2))==2&bigomega(substpol(Pol(Vec(x),y),y,3))==2&bigomega(substpol(Pol(Vec(x),y),y,4))==2&bigomega(substpol(Pol(Vec(x),y),y,5))==2&bigomega(substpol(Pol(Vec(x),y),y,6))==2& bigomega(substpol(Pol(Vec(x),y),y,7))==2&bigomega(substpol(Pol(Vec(x),y),y,8))==2,s,if(s,print(x)))

1:18 PM
only 3 with all coefficients under 100 pass up to y=29
replaced most Pol(Vec(x),y) with d
[2, 2, 82]
[2, 6, 86]
[2, 10, 94]
my guess is y=9,y=10 wipes most not a multiple of a prime producing polynomial away.

1:39 PM
1

Please, see the most updated version in this link. Thank you. There is a note in OEIS sequence A005563: "Erdős conjectured that $n^2 - 1 = k!$ has a solution if and only if $n$ is $5, 11, 71$ (when $k$ is $4, 5, 7$)." Here is one reasoning to solve it. Evaluating this Erdõs equation, we know t...

Hello

hello @MrPie

Havent had a very good week
Actually, past couple days in fact

were you in newfoundland with over 75 cm of snow ?

@RoddyMacPhee I was in Australia with the fires

1:51 PM
475x^2 - 255x + 451 is semiprime for the non-negative integers x upto 13

My mum lost her car from the fires, the air around my house has been just endless smoke for the past week or two and my grandma has had to live with us.
And now queensland is becoming flooded ffs
Every day is grey and every night is just deep red and orange.

@MrPie I'm really sorry to hear that

@Mathphile its fine dw about it

@MrPie nothing amazing
Just started packing for my flight back to US tommorrow

9413x^2 - 7619x + 6697 is semiprime for the non-negative integers x upto 16

1:58 PM
did you use my code ?

I have a 32 hour long journey so that's going to be tiring

@RoddyMacPhee I added some conditions to make the problem more interesting : f primitive and irreducible, only positive semiprimes count, but a square of a prime counts as a semiprime. Therefore, my code was slightly different.

the only condition of those mine doesn't show is being primitive.

Apparently, Brocards problem is still open. It has been tested upto n = 10^9 and someone claimed to have extended this to n = 10^12.

that and allowing only positive coefficients are my only drawbacks.

2:03 PM
OK, even more restrictive is that the coefficients must be positive. Good idea.
2365x^2 + 6273x + 2059 is an example holding upto n = 16
1174x^2 + 8088x + 3091 is an example holding upto n = 17
2414x^2 + 4760x + 3937 is an example holding upto n = 19

2:27 PM
you might surpass my newest code if you keep this up...

Ok im back, im feeling a bit better
Sorry if I downhearted anyone. I did some math, relaxed, and found something cool :)
(Hopefully something cool)
I believe I have found some kind of way to approximate $$\sqrt{3\prod_{i=3}^n i^2-1}$$ where the approximation is actually exact when $n=8$ (because the expression is, well, a square number!)
But first: @Peter how are you finding those polynomials? :P

I am just generating random polynomials until a new champion appears

Also I believe I have found a way to write $x$ in the polynomials $$x^2\pm N\cdot x \pm 1=0$$ and $$x^2+N\cdot x+1=0$$ as a continued fraction. But actually now that I think of it, it's probably straight-forward lol

I might give up on the n=28 attack lol

How have you been @RoddyMacPhee and @Peter?

2:39 PM
I just noticed that in the above lines it must of course be "upto x=..."

stupid like always

@Peter we knew what you meant :)

At the 27th January, I reach the magical age of 50 :)

Wait what
Happy early birthday! You are just about 3.125 times the age of me :)

Also, I factored 2^129+3^129+6^129 - spectacular !
FF 101 (show) 6^129+3^129+2^129<101> = 3971912810589834418346675207650457324695800288901<49> · 6060543782...91<52>

2:42 PM
I cannot believe that is semiprime
Well done :P
Thats a huge number jesus christ

Of course with siqs, ecm has very low chances to find such factors.
Current holes including the smallest composite without a known factor
5 CF 160 125 6^160+3^160+2^160<125> = 7 · 23 · 241 · 8229191006...53<120>
5 CF 163 127 6^163+3^163+2^163<127> = 44489160533<11> · 24900366203707<14> · 6225781393...21<103>
5 CF 166 130 6^166+3^166+2^166<130> = 7 · 3405326757902041647383<22> · 6249578574...29<107>
5 CF 168 131 6^168+3^168+2^168<131> = 37 · 4057 · 512543 · 6970645971...59<120>
5 CF 169 132 6^169+3^169+2^169<132> = 83 · 1069 · 34589 · 1048497769...97<123>
5 CF 170 133 6^170+3^170+2^170<133> = 7^2 · 479 · 660379 · 1245626151...61<123>
@MrPie Since I ruled out small factors with ECM , the likelihood that it would be a semiprime, was very high. But I did not expect such a high smallest prime factor.
I did not continue with the random C113 which is very tough indeed.

7

For any $n\in\mathbb N$, let $A=\{x\in\mathbb N \mid 1 \leq x \leq n\}$ and $B=\{x\in\mathbb N \mid n+1 \leq x \leq 2n\}$. Order the sets by greatest prime factor of each element, ascending. Let $A_1$ be the element in $A$ with the smallest gpf and $A_n$ be the largest, and likewise for $B$. ...

@Peter ah, even better then

hello @Haran
Here is something for you to disprove

I mean, in terms of the chances of it being semiprime. As regards C113, I don't know what else to say :P

2

Please, see the most updated version in this link. Thank you. There is a note in OEIS sequence A005563: "Erdős conjectured that $n^2 - 1 = k!$ has a solution if and only if $n$ is $5, 11, 71$ (when $k$ is $4, 5, 7$)." Here is one reasoning to solve it. Evaluating this Erdõs equation, we know t...

2:56 PM
I saw that beforehand
I couldnt be bothered reading it tbh xD

lol

In fact chances are a bit lower, since I have not run enough curves to rule out 37 digits with very high probability. We only can say that the number is composite and almost surely has no prime factor with 30 digits or less.
Chances for at least 40 digits are good, which would ensure that the number is semiprime.
@Mathphile My computer is too slow for a quadratic sieve for this number. But some freaks have factored even larger numbers , for example my weird C133 :
At the moment, I do not remember the weird expression ...
This was it !!
FF 133 (show) 331#+1013^13<133> = 13660997293197200778560512676198053659658942317253<50> · 4691179450...91<84>
I would never have expected to ever see this factorization !
Are currently games played in the Tata Steel chess tournament ?

4:00 PM
@Peter I have a new project

Which one ?

Find consecutive positive integers $a, b$ such that $b$ is even and $9 \mid b$ and $a=n!!/9^m$ where we choose $m$ such that $9 \nmid a$
@Peter
If we are able to prove that such $a, b$ cannot exist, then we have solved Brocard's problem!

Brocards problem deals with the factorial. How is this related ?

I have proved this to be equivalent
Can you write a program for this?

4:16 PM
Both cases a+1=b and b+1=a are allowed ?

so $18\mid b$ and $a\equiv$\pm 1\bmod 18$Can you show us the proof of the equivalence ? @Peter I think you started something with your n!! formula or a related one. True, we can express n!! with the factorials. The search limit of Brocard's problem is anyway insanely large. A counterexample is very hard to imagine. @Peter yes both cases are allowed @Peter This search is not to find a counter-example 4:28 PM What I mean : The evidence for no more counterexample is comparable with the evidence of the truth of Goldbach's conjecture. okay well for$n=(2k+1)\cdot 3$we have a minimum$m\geq k\$

@Peter Yes that's why I think that there are only 3 solutions to the equation

@Mathphile , @Peter means that he can't connect your statement with the conjecture.

My main doubt is whether this gives any better evidence than the calculation upto 10^9 that has already been done.

2 hours later…
6:14 PM
@Peter I will prove the connection of Brocard's problem with the statement as soon as I reach home.
For now could you try to run a program?

6:48 PM
n!! would need to be $$\pm 9^m\pmod {2d\cdot 9^{m+1}}$$ no ?

2 hours later…
8:20 PM
@MrPie :/
and there are still people who don't believe in climate change

1 hour later…
9:33 PM
newfoundland canada should, 39 foot 5 waves, 157 kph winds and 76+ cm of snow isn't normal.