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5:00 PM
@Mathphile Hi, I continue the search for the prime pairs $2^n\pm n$
 
@Peter Hello, I came back to your answer to "arbitrarily long 2-palindromes" and wanted to prove it. I looked at the digits, and found some binomial patterns. Hence, my recent comment there, if you have seen it.
 
@TheSimpliFire I have a new project : Find a prime factor of the composite number $$39^{39^3}+39^{39^2}+1$$ with $94381$ digits
@Vepir I have seen it and also arrived a few minutes ago. Good timing indeed !
 
For $k=2n$, The middle digits are also equal $a(n)$ for $n$ even, and equal $b-a(n)$ for odd n.
in base $b+1$
 
@Vepir Is there a formula for $a(n)$ ?
 
Yes, it is the

$$
\sum_{k=1}^n \binom{n}{k}^2
$$
 
5:09 PM
And this is proven, right ?
 
This is just an observed pattern that I commented (and observed) just now.
 
Ah, OK, no proof yet ...
 
Yes.. I need to find the pattern for all of the digits, and show it sums to the given expression.
That is just the middle digit. The others seem to have similar patterns.
 
Maybe, with induction we can prove the patterns.
See here : wolframalpha.com/input/… for a closed form of the above sum
 
We need to conjecture a formula for the $(x,y)$th digit in a pascal-like triangle.
that sum is the rightmost diagonal
Then we can prove rows sum to your expression when multiplied by the base
It should start like:
$$
1\\
3\space\space\space\space5\\
5\space\space\space\space14\space\space\space19\\
7\space\space\space\space27\space\space\space55\space\space\space69\\
9\space\space\space\space44\space\space\space119\space\space\space209\space\space251\\
11\space\space\space65\space\space\space219\space\space494\space\space791\space\space923
$$
 
5:20 PM
1,3,5,7,9,11,... is an obvious pattern, but how do we get the other entries ?
 
Thats the question
Second row gives $(b-3,5,b-3)$ representation in $b+1$.

Third row gives $(b-5,14,b-19,14,b-5)$ representation in $b+1$.

Etc.
Other than putting diagonals in the OEIS, I'm not sure yet how to see a pattern for the entire triangle.
 
3*1+2 = 5
5*2+4=14
7*3+6=27
9*4+8=44
or even easier
3*2-1 = 5
 
Maybe its more natural to show it like:

$$
1\\
3\space\space\space\space5\space\space\space\space3\\
5\space\space\space\space14\space\space\space19\space\space\space14\space\space\space5\\
7\space\space\space\space27\space\space\space55\space\space\space69\space\space\space55\space\space\space27\space\space\space7\\
9\space\space\space\space44\space\space\space119\space\space\space209\space\space251\space\space209\space\space119\space\space44\space\space\space9\\
11\space\space\space65\space\space\space219\space\space495\space\space791\space\space923\space\space791\space\space495\space\sp
 
5*3-1=14
7*4-1=27
and so on
Seems that the second (and the second last) entries are completely determined by my pattern.
 
Btw, this triangle represents just one sequence of 2-palindromes. The following pattern represents "templates" for all possible 2-palindromes: i.stack.imgur.com/qieyK.png
(Where c_0 and c are placeholders for expressions depending on $b$ and are not equal)
 
5:47 PM
Here is the triangle extended:

{1}
{3, 5, 3}
{5, 14, 19, 14, 5}
{7, 27, 55, 69, 55, 27, 7}
{9, 44, 119, 209, 251, 209, 119, 44, 9}
{11, 65, 219, 494, 791, 923, 791, 494, 219, 65, 11}
{13, 90, 363, 1000, 2001, 3002, 3431, 3002, 2001, 1000, 363, 90, 13}
{15, 119, 559, 1819, 4367, 8007, 11439, 12869, 11439, 8007, 4367, 1819, 559, 119, 15}
{17, 152, 815, 3059, 8567, 18563, 31823, 43757, 48619, 43757, 31823, 18563, 8567, 3059, 815, 152, 17}
{19, 189, 1139, 4844, 15503, 38759, 77519, 125969, 167959, 184755, 167959, 125969, 77519, 38759, 15503, 4844, 1139, 189, 19}
If im not mistaken, @Peter, The problem is actually to prove these coefficients correspond to $a_n(i)$ in:

$$
\frac{b^{2n}-1}{b+1}=\sum_{i=1}^{2n-1}A_n(i)(b+1)^{2n-1-i}=(A_n(i))_{b+1}
$$

Where $A_n(i)=a_n(i)$ if $i$ is even, and $A_n(i)=b-a_n(i)$ for $i$ odd.

For all $b\ge \sum_{k=1}^n \binom{n}{k}^2$.
 
6:45 PM
Third diagonal is $(2*n*(n+1)*(2*n+1)/3)-1$
@Peter I see it now! Diagonals are given by:

$$\binom{2k+2}{d}-1$$
 
7:20 PM
This gives the entire triangle with: -1 + Binomial[2 n, -i + 2 n]
$$
 
@Mathphile I extended the search of primw pairs $2^n\pm n$ upto $n=220\ 000$ and found no other pair.
 
$$
\binom{2n}{2n-i}-1
$$
 
@Vepir Interesting !
 
Now we should be able to substitute that into sum and observe two cases, $n$ odd and even, and prove via induction!
 
@Peter hey its been a rather long time since we have talked
 
7:23 PM
@Vepir In fact, the rest should be no problem
@Mathphile In fact
 
@Peter nice that's a large range
 
First I determined the candidates with trial division, then I tested just $2^n-n$ (which was never prime yet for the candidate $n$ !)
 
@Peter have we checked the primality of the expression $n$#$+k$ where $k$ is prime?
 
Finding another pair would be similar to find a twin prime. In this range, such a twin prime would be at rank $4$ in the list of the largest known twin primes !
 
@Peter yes
 
7:31 PM
I do not remember exactly what we verified.
 
@Peter I think we only verified that for each prime $k$ there is at least one prime of the form $n! \pm k$
15
Q: Conjecture: "For every prime $k$ there will be at least one prime of the form $n! \pm k$" true?

MathphileUsing PARI/GP, I searched for primes of the form $n!\pm k$ where $k \ne 2$ is prime and $n\in \Bbb{N}$. With the help of user Peter, we covered a range of $k \le 10^7$ and couldn't find a prime $k$ for which $n!\pm k$ has no primes. Observations: $(1)$ When $n \ge k$, $n! \pm k$ cannot be p...

@Peter can you search for what prime $k$, there is at least one prime of the form $n$#$+k$?
 
I think, I also verified n# + k , but I do not know how far. Anyway, I have some new projects, including to find a factor of $$39^{39^3}+39^{39^2}+1$$ and verifying whether $$96^{96^3}+96^{96^2}+1$$ is prime.
 
Hello
 
hello @TheSimpliFire
 
@TheSimpliFire Hi
 
7:44 PM
you done with your dissertation @TheSimpliFire?
 
@Mathphile Nope, I got an extension to the 25th. But I have a bit of time now
 
is much work left?
 
Considerably, I guess. I've written 45/60 pages so still not a full first draft!
 
What is the theme of the dissertation ?
 
@TheSimpliFire that does sound like a lot
 
7:47 PM
Indeed...
@Peter Medical statistics
 
i hope you get it complete soon
 
Interesting, diagnostic with Bayes' rule ?
 
@Peter I'll post the proof as a self answer linking to your proposed form: Thanks for finding $(b^k-1)/(b+1)$.
 
In particular, determining how many participants you need in RCTs whilst taking account of withdrawals
@Peter Two of my chapters use Bayesian statistics
 
@TheSimpliFire after this are you done with your masters?
 
7:50 PM
Statistic was always my favorite topic concerning "practical mathematics"
 
@Mathphile Yes :)
moving on to a PhD next Sep
@Peter Nice, didn't know you were interested in stats
 
@TheSimpliFire If you are still interested in consecutive palindromes, we now know for sure that infinitely many 2-palidndromes exits, for every case of digits ($d=2l+1$).

But we still do not know all of them ($d\ge 9$).
 
I remember them @Vepir
I have a little section on that in my notebook
 
The identity Peter proposed $(b^{2n}-1)/(b+1)$ is a 2-palindrome in $b,b+1$ for all $b\ge \sum_{k=1}^n \binom{n}{k}^2$.
 
Has that been proven?
 
7:58 PM
Yes, I found explicitly digits in $b+1$. The $b$ is easy to see.
 
Great
 
0
A: Arbitrarily long palindromes in two consecutive number bases

VepirThanks to @Peter's answer for conjecturing a pattern that should give such a sequence. Here, I managed to prove his proposed identity. The linked answer proposed that the following gives $(b,b+1)$ 2-palindromes for even $k$ and large $b$: $$ \frac{b^k-1}{b+1} $$ For large $k$, we have arbit...

 
One more (+1) and you will get the Self-Learner badge
 
I wonder if I can find more such patterns.
 
@TheSimpliFire I was never aware of this badge :)
@Vepir Maybe this helps for your 3-palindrome problem.
 
8:04 PM
Perhaps this will help extend the general 2-palindrome table when I find time for it.
 
@Mathphile I am almost at $n=230\ 000$
 
@Vepir Note: for single-dollar (inline) expression, \sum\limits_{}^{} works quite nicely
 
I knew something for inline formatting exists, but I forgot it is "\limits". Thanks
 
8:21 PM
Btw, Is there a nice way to format "triangles" of numbers? I used here '\space' a lot to make it resemble a triangle.
 
@Peter There's a lot of badges where I know them only after I get them
@Vepir How about \qquad
Actually, I didn't know about \space
 
Hey would it be okay to post this here? The room's description read:
"Discussions/answers to conjectures (may or may not be original), interesting maths problems in/out of MSE, and any random stuff"
And I think it's an interesting math problem :)))

https://math.stackexchange.com/questions/2888976/the-definite-integral-problem-with-a-twist
 
Hello @MoreAnonymous
 
@TheSimpliFire yo!
 
That's a long post, give me a sec
 
8:26 PM
You don't need to go through the heuristic proof
That involves black magic
 
I was going to suggest Mobius inversion but I see you've used that
 
Ur going through the Heuristic proof which involves black magic :P
:O
 
I think there is some merit in parts of it
though it is what it is
 
@TheSimpliFire Haha.. black magic :P
Do you think the "The Definite Integral Problem (with a twist)"
is an interesting problem ... I'm suprised if I'm the only one who has worked on it ...
@TheSimpliFire I'm wondering if I should put a regularization tag (won't be surprised if the are other getting that limit $s \to 1$ on the RHS
 
$1/\zeta(s)=\sum\limits_{r=1}^\infty\frac{\mu(r)}{r^s}$
 
8:34 PM
so?
 
Since it's limit I can also use an asymptotic version of the zeta version (I think $1/(s-1)$)
I noticed "Modern Abstract Analysis" ... U think they might be interested?
 
@MoreAnonymous You mean $\lim\limits_{s\to1}\left(\zeta(s)-\frac1{s-1}\right)=\gamma$
 
@TheSimpliFire I was thinking $\lim_{s \to 1} \zeta(s) (s-1) = 1$ But urs is more powerful
 
How many examples have you tested it for?
 
8:45 PM
@TheSimpliFire here's a crazy non-periodic $d_r = a_r$ example :)
1
Q: What is the limit of this Dirichlet series?

More AnonymousBackground & Motivation I'm trying to verify/disprove the conjectured formula of the weighted integral of $f(x)$: A rough proof for infinitesimals? $$ \lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n d_r \left( f(\frac{k}{n}r)\frac{k}{n} \right) = \lim_{s \to 1} \! \underbrace{\frac{1}...

It's answered via the comments :)
 
Regarding the literature thing, I think it's overkill to try and write $\int f$ as a quotient of those two limits. Riemann summation is much simpler.
 
Hmmm ... I wouldn't know the formula if we were considering a Rieman sum (finite strips)
I feel it's a super cool formula u have:

A measure theory (L.H.S) = A limit of a dirichlet series (Number theory) times a constant (the normal integral)
@TheSimpliFire Also do u think my version of a formula can exist with finite strips?
 
For a finite version some of the limits wouldn't really make sense
 
@TheSimpliFire Hmm ... I felt I cheated plenty during the "heuristic proof" ... Maybe if I cheat again I can get a finite strip :P
@TheSimpliFire gotta catch up on shut eye .. lemme know if u make progress? Also do u think the "Modern Abstract Analysis" chat room would be interested ?? (I read measure theory over there)
 
9:47 PM
@TheSimpliFire nice what would it be in?
 
@Mathphile Population Health Sciences (interestingly it's a relatively new discipline of research)
 
is that even related to math?
 
Academically I've started to focus on the medical applications of mathematics rather than studying maths itself
@Mathphile Yes, statistics
all applied
 
nice
 
I'd congratulate anyone who can find a link between the Riemann zeta function and randomised controlled trials :)
 
9:51 PM
do you have an estimate on how long it would take?
 
Well I get paid for 3 years - I'm not expected (and I don't expect) to finish it past that period of time
 
i thought phd's take over 5 years to complete
 
On average, but this one's got more workload/day
They don't want to pay me for too long ;)
 
wow so you are getting paid to study
thats pretty nice
 
@Mathphile Not really study, it's mostly research. Make proposals, write papers, that kind of thing
 
9:55 PM
yeah
 
@Mathphile I'm sure you'll get there soon enough :)
 
what's even better is that you'll be finishing off your phd before people even start undergrad
 
In the meantime I'll just rely on Math.SE to get more knowledge on "pure" maths
@Mathphile Technically they'd have finished their 1st year undergrad by the time I finish
Also, it's possible to go from undergrad directly to PhD without a masters
 
@TheSimpliFire lol i don't think i'll ever do a phd tho
 
@Mathphile Do you want to go into research?
 
9:59 PM
maybe i'll be able to do some light research in computer science in my undergrad
 
I guess for your bachelor paper dissertation
 
seems like you're even brighter than Terence Tao who got his phd at 21
@TheSimpliFire i don't think we have a dissertion here for bacheolrs
 
younger than me :P
@Mathphile Fun fact: Karl Witte earned his PhD when he was 13
 
im pretty sure you might be in the top 10 most youngest people who are starting their phd
 
perhaps...
perhaps not...
but I don't really care :P
 
10:07 PM
but still it is amazing how far you have reached for your age
 
Thanks @Mathphile
Oh, btw
You have no idea how relatable these are to me (I might have mentioned procrastination one too many a time here :P)
 
lol
procrastination is a huge problem for me too
 
btw do you sit in any math classes at high school anymore?
@TheSimpliFire yup lol
 
@Mathphile I help out when I turn up
 
10:14 PM
teaching the teacher? :P
 
It's a small class and I'm friends with many of them, so I thought why not
@Mathphile I'm more of a TA
except not at uni
 
nice
i gtg now to study for my history test tomorrrow
bye @TheSimpliFire
 
Well I gtg to bed as it's quite late now
see you @Mathphile
 

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