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A mediocre professor at a top-notch university making mediocre contributions to this top-notch community


Apr
16
revised Deleting any digit yields a prime… is there a name for this?
clarified that original number must also be prime
Apr
15
answered Deleting any digit yields a prime… is there a name for this?
Apr
15
comment Deleting any digit yields a prime… is there a name for this?
By the way, these number don't look random at all: they are heavily biased toward integers with repeated consecutive digits. Take 711110111, for example: such numbers appear often in our list because if deleting one 1 in a group yields a prime, then obviously deleting any other 1 in that group does as well.
Apr
15
comment Deleting any digit yields a prime… is there a name for this?
Here is heuristic evidence that there are infinitely many of these numbers: let $c(n)$ be the number of integers in $P$ with $n$ decimal digits. A quick computer program shows $c(2)=4$, $c(3)=11$, $c(4)=14$, $c(5)=16$, $c(6)=18$, $c(7)=13$, $c(8)=14$, $c(9)=18$; therefore $P$ is infinite. I'm joking, of course, but I would find it remarkable if such a list ended, in spite of your nice argument above.
Apr
15
comment Deleting any digit yields a prime… is there a name for this?
I see no reason to exclude 0 as a digit... seems rather artificial: there are infinitely-many primes with a 0 digit in their decimal expansion (101 is the first, then cf Dirichlet).
Apr
15
awarded  Nice Question
Apr
15
asked Deleting any digit yields a prime… is there a name for this?
Apr
12
comment Simple probability question, balls and bins
Douglas: Thanks for adding your answer to the mix. But I don't follow: you have your summation indexed by "$k$ bins known to be empty". How do you evaluate a summation based on what "is known"? I would suggest you improve your answer by clarifying this. Also, your answer doesn't have an answer explicitly stated anywhere (that I can find). I think it would be an improvement to highlight the answer for "exactly one bin empty" rather than leaving it to the reader to "multiply by $m$ and divide by $m^n$" and simplify. (Unless you were intending to give only a hint?!) Finally...
Apr
11
answered Decryption Problem
Apr
11
accepted How many sides does a circle have?
Apr
10
comment Difficult integral?
What have you tried? Where did you get stuck? Where is this problem from? Did you notice that $1-x^4 = (1-x^2)(1+x^2)$?
Apr
10
comment What does it mean to say a language is context-free?
See en.wikipedia.org/wiki/Context-free_language
Apr
9
comment How many sides does a circle have?
@Douglas Zare: C?
Apr
9
awarded  Nice Question
Apr
8
asked How many sides does a circle have?
Apr
8
comment $n=8\log_2(n)$, forgot basic math
The best way to do this is write a 10-line program to find the cross-over point: n=2; while (1) { if 2^n > n^8: {print n; exit;} else n++;}. Since the answer is 44, this would run in < 1 msec.
Apr
7
comment $n=8\log_2(n)$, forgot basic math
@Clash: Then you simply want to show that the growth rate of $2^n$ is higher than $n^8$. You can do this from the definition of $O()$, and you certainly don't need Lambert's W function.
Apr
7
answered $n=8\log_2(n)$, forgot basic math
Apr
7
comment $n=8\log_2(n)$, forgot basic math
I'm confused. The Wikipedia link you gave for the W func says $z=W(z)e^{W(z)}$ is the defining equation and asserts that $W(e)=1$. But in your answer you say that $W(x)=z$ iff $z=xe^x$ which implies $W(1)=e$, an inverse of the Wikipedia def.
Apr
6
comment Good resources (book or otherwise) to learn/study basic Combinatorics
Brualdi is decent but so error-ridden that it's annoying to read. That was 15 years ago, however... maybe he's finally cleaned it up in the latest editions.