20151207, 19:05  #1 
Oct 2007
Manchester, UK
2×3×223 Posts 
Near repdigit primes on Numberphile
Thought some people on here may find this interesting, popularising of interesting primes.
https://www.youtube.com/watch?v=HPfAnX5blO0 
20151207, 20:20  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9,257 Posts 
It is nice for popularising, sure, but they still ought not to make up "new names" when established terms exist.
It is great to explain everything as well as they do (this episode is no outlier, they always very well done on the explanation and elementofsurprise level), but without reference to the body of existing work (perhaps at the very end) it comes across as crankish. Sorry, guys, if you will be reading this. Nice, yes, ... but crankish. They should invest some time in research before filming. Filming takes quite some effort and it is, in this case, done professionally, but what does a good production company have besides actors (be that in this case 'actor/author/producer/director') and an operator, a lighting and sound crew? There is always research. "Palindromic nearrepdigit primes". LMGTFY! UTM pages, as well as worldofnumbers.com, recmath and OEIS immediately show up in this search. No effort needed. 
20151207, 23:00  #3 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9,257 Posts 
* OEIS A265383 (pending approval)
* Kamada's NRR (these are "9v89w" in his notation, i.e. v w; the 9w89w are palindromes, 10^{2n+1}10^{n}1, studied much deeper by Darren Bedwell, with the largest known prime 10^13480910^674041). * UTM NRR primes * H. C. Williams, "Some primes with interesting digit patterns," Math. Comp., 32 (1978) 13061310. Corrigendum in 39 (1982), 759. MR 58:484 
20151208, 03:21  #5  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17·251 Posts 
I wonder if the database he referred to in the video was factordb.com?
Quote:
Quote:
In this video, Simon Pampena is the "actor" and Brady Haran is the "everything else" (as I understand it). Brady makes a lot of videos on math, computers, etc., but rarely stars in them. Last fiddled with by MiniGeek on 20151208 at 03:33 Reason: relevant xkcd 

20151208, 04:42  #7 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10010000101001_{2} Posts 
There is an invisible extra member in the sequence "1, 6, 9, 154, 253, 1114, 1390, 2618, 5611, 12871" (in draft).
Statistically, the next term is likely Asao's, I presieved and I am scanning up to 20,000++, too. Btw, prime UTM database remebers every little prime submitted many years ago. Use http://primes.utm.edu/bios/page.php?id=183 and press "All of this Person's primes"... P.S. You will find that Asao was also curious about "Cyclop"like series 2^{2n}2^{n}1 which unlike the series in the Numberphile episode (the palindromic one, 2^{2n+1}2^{n}1) is not algebraically factored. These are now, of course, also known as OEIS A098845. Last fiddled with by Batalov on 20151208 at 05:05 
20151208, 12:40  #8  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17×251 Posts 
Quote:
How did you presieve this sequence? Is there a tool out there that can take this form, or is it easy to tweak some siever to take a generalform number, maybe? I just ran PFGW with f so it tried factoring each number before doing its N+1 test. 

20151208, 13:02  #9 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
my first thought based on the numberphile video would be the polynomial form $ x^{2n}x^n1 $ all numbers in the sequence for x=10,can't be divisible by 2,3,5, mod 7 it becomes 3^{2n}3^{n}1

20151208, 14:15  #10  
Aug 2006
3×1,987 Posts 
Quote:
__________ P.S. (S.B.): just wanted to add so that this topic was not interrupted, here, in place: Only primes such that (5p) = 1 can be factors; that is, in other words, only p>=11 that end with 1 or 9. Similarly, for 10^{2n+1}10^{n}1, only p :: (41p) = 1 can be factors; that is, 23, 31, 37, 43, 59, 61, 73, 83, 103, 107, ... Last fiddled with by Batalov on 20151211 at 02:07 Reason: (P.S.) 

20151208, 14:31  #11 
Oct 2007
Manchester, UK
10100111010_{2} Posts 
This is true, but it's also obvious without using modular arithmetic. 998999 is clearly not divisible by 2 or 5, and since all but one of the digits is divisible by 3, the number as a whole is not divisible by 3.

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