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comment Combinatorial identity involving the square of $\binom{2n}{n}$
Did you actually sum $\binom{2n}{n}$ without squaring? I get your result with Sum[Binomial[2 k, k] /16^k, {k, 0, n}]
May
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answered Combinatorial identity involving the square of $\binom{2n}{n}$
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awarded  Supporter
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comment Are there infinitely many natural numbers not covered by one of these 7 polynomials?
To go a step away from the prime triple conjecture. $f_1$ and $f_2$ allow that $30n+11$ contains prime factors $p\equiv 1,11,19,29 \mod 30$. $f_3,f_5,f_6,f_7$ together are more restrictive, All possible prime factors are forbidden, For $f_4$: $30n-11$ may contain prime factors $p\equiv 1, 7, 19, 13\mod 30$. (Note: 1,11,19,29 and 1,7,19,13 are both multiplicative subgroups mod 30). Hence there is one prime condition, and 2 half-prime conditions, (sieve dimension 2), which is still undoable. (For sieve dimension 3/2 there is sometimes hope, Iwaniec half dimensional sieve).
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answered Polynomials with few prime factors
Dec
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comment runs of consecutive non squarefree integers
This answers the opposite: the density of runs of square-free integers has been worked out by L Mirsky, Note on an asymptotic formula connected with r-free integers. Quart. J. Math., Oxford Ser. 18, (1947). 178–182, (and some more related papers by Mirsky). I do not currently have access to the paper, but would hope that the same methods can be adapted to answer your question, maybe just by some combinatorial inclusion-exclusion.
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comment Sums of reciprocals of prime numbers: $p \equiv a \!\! \mod m$ vs. $p \equiv b \!\! \mod m$
[continued.] $\lim_{x \rightarrow \infty} \sum_p (-1)^{\frac{p+1}{2}} \exp(-p/x)=\infty$.
Dec
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comment Sums of reciprocals of prime numbers: $p \equiv a \!\! \mod m$ vs. $p \equiv b \!\! \mod m$
The part you refer is a historical comment on an informal letter of Chebychev. Indeed, Chebychev's comments are known to me not at the level of "proven results", see for example Narkiewicz (The development of prime number theory...), p. 122-124 books.google.at/… In this particular case it seems likely to me that a typos must have been copied from one source to another, and even the statemnet in Narkiewicz should read
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answered Sums of reciprocals of prime numbers: $p \equiv a \!\! \mod m$ vs. $p \equiv b \!\! \mod m$
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revised Reference for a conjecture on the first primes congruent to 1 modulo other primes
Added second part
Nov
25
revised Reference for a conjecture on the first primes congruent to 1 modulo other primes
Added second part
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revised Reference for a conjecture on the first primes congruent to 1 modulo other primes
edited body
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answered Reference for a conjecture on the first primes congruent to 1 modulo other primes
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