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Timeline for Pascal triangle and prime numbers

Current License: CC BY-SA 4.0

12 events
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Mar 24, 2023 at 8:20 history edited Martin Sleziak CC BY-SA 4.0
http -> https (the question was bumped anyway)
Oct 12, 2016 at 21:00 history edited TRiG CC BY-SA 3.0
Link text. "this wonderfully-titled paper" is great when the link goes down, isn't it? Also Mathjax. And curled apostrophes, for the Hell of it.
Apr 25, 2014 at 21:37 comment added TRiG And this is why it's important not to have link text like "click here" or "this paper". Not only is useless link text inaccessible, it also offers no clues whatsoever when the link breaks, as it has done. To what wonderfully titled were you referring? And if the title is so wonderful, why not include it in your answer?
Jun 8, 2013 at 1:00 comment added Gerry Myerson The link I gave in 2010 is kaput, but oeis.org/A088164 works now, and there are still only two primes listed.
Mar 17, 2010 at 5:27 comment added Gerry Myerson @Harrison Brown: the "certain property" is that the binomial coefficient $2p$-choose-$p$ be 2 modulo $p^4$. Such primes form sequence A088164, Wolstenholme primes, at the Online Encyclopedia of Integer Sequences, research.att.com/~njas/sequences/A088164. Only the two primes 16843 and 2124679 are given, so, if the third is known, it hasn't made it to the OEIS.
Dec 17, 2009 at 19:28 comment added Harrison Brown @Greg Kuperberg: Looking at the link I see that problem 10 is attributed to you, and it mentions that (whenever this was written -- when was this written?) we knew of two primes with a certain property, and you thought the third was out of reach. Do we know the third yet? :P
Dec 17, 2009 at 18:58 comment added Greg Kuperberg Dmitry Fuchs mentions a congruence that implies that $\binom{ap^n}{bp^n}$ converges $p$-adically as $n \to \infty$. It's a good question to understand what this $p$-adic limit could mean. mathcircle.berkeley.edu/BMC6/ps0405/binom.ps
Dec 17, 2009 at 17:17 comment added Harrison Brown I just realized that my added paragraph also explains why composites are not as nice as primes. For p prime, if p^m is the largest power that divides a and p^n is the largest power that divides b, then p^(m-n) is the largest power that divides a/b. But this isn't true in general for composites! Think about 1000/250, for example. The key is that divisibility by 10 is dependent on divisibility both by 2 and by 5, and we have the freedom to mess around with one of those without changing the other.
Dec 17, 2009 at 17:16 comment added Michael Lugo If you know what happens for powers of primes, then you can figure out what happens for any number; the binomial coefficients divisible by, say, 36 are just those divisible by both 4 and 9.
Dec 17, 2009 at 17:13 vote accept Alix Axel
Dec 17, 2009 at 17:10 history edited Harrison Brown CC BY-SA 2.5
added 750 characters in body
Dec 17, 2009 at 17:04 history answered Harrison Brown CC BY-SA 2.5