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Sep
20 |
revised |
Upper bound on length of addition chain
added 386 characters in body |
Sep
20 |
revised |
Upper bound on length of addition chain
added 1565 characters in body |
Sep
18 |
comment |
Upper bound on length of addition chain
Yes, some minor correction just added. $C$ is close to 1, for quite large values of $n$. |
Sep
18 |
revised |
Upper bound on length of addition chain
Corrected. |
Sep
18 |
answered | Upper bound on length of addition chain |
Sep
1 |
revised |
Lower bounding the multiplicative order of 2 modulo p
added 1295 characters in body |
Aug
31 |
revised |
Lower bounding the multiplicative order of 2 modulo p
(edited jstor link). |
Aug
31 |
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 $a_k=\dfrac{3^{2^{k-1}}+1}{2}$?
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 $a_k=\dfrac{3^{2^{k-1}}+1}{2}$? |
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 $f_4$: $30n-11$ 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). |
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 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. |