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revised 
Lower bounding the multiplicative order of 2 modulo p
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1d

revised 
Lower bounding the multiplicative order of 2 modulo p
(edited jstor link). 
2d

answered  Lower bounding the multiplicative order of 2 modulo p 
Jul
7 
awarded  Yearling 
May
22 
revised 
For $k>3$ does there exist an odd prime $q_k$ such that $p_k=2^kq_k+1$ is prime and $p_k$ divides $a_k=\dfrac{3^{2^{k1}}+1}{2}$?
added 762 characters in body 
May
22 
answered  For $k>3$ does there exist an odd prime $q_k$ such that $p_k=2^kq_k+1$ is prime and $p_k$ divides $a_k=\dfrac{3^{2^{k1}}+1}{2}$? 
Aug
24 
awarded  Nice Answer 
Jul
7 
awarded  Yearling 
May
17 
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
17 
answered  Combinatorial identity involving the square of $\binom{2n}{n}$ 
Apr
30 
awarded  Supporter 
Apr
30 
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 $f_4$: $30n11$ 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 halfprime conditions, (sieve dimension 2), which is still undoable. (For sieve dimension 3/2 there is sometimes hope, Iwaniec half dimensional sieve). 
Mar
6 
awarded  Citizen Patrol 
Dec
12 
answered  Polynomials with few prime factors 
Dec
8 
comment 
runs of consecutive non squarefree integers
This answers the opposite: the density of runs of squarefree integers has been worked out by L Mirsky, Note on an asymptotic formula connected with rfree 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 inclusionexclusion. 
Dec
2 
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
2 
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. 122124 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 
Dec
1 
answered  Sums of reciprocals of prime numbers: $p \equiv a \!\! \mod m$ vs. $p \equiv b \!\! \mod m$ 
Dec
1 
awarded  Mortarboard 
Nov
25 
revised 
Reference for a conjecture on the first primes congruent to 1 modulo other primes
Added second part 