156 reputation
18
bio website
location Chennai, India
age 19
visits member for 2 years, 4 months
seen Jun 5 '13 at 11:07

I'm a college student of Computer Science, who likes Number Theory (like you haven't heard that one before!).

You can probably find me online on GMail (furlox.mod@gmail.com) but I wouldn't know why you'd want to do that. I have a thing for four-leggeds, too.


Sep
24
awarded  Autobiographer
Aug
31
awarded  Popular Question
Aug
2
awarded  Good Question
Dec
30
awarded  Tumbleweed
Aug
26
revised Primality using $\Gamma(x)$
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Aug
26
revised Primality using $\Gamma(x)$
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Aug
26
comment Primality using $\Gamma(x)$
@Qiaochu Yuan My motive was not Wilson related so much as exploring the Gamma function, being new to it and all. Perhaps 'primality' is an unsuitable term? To Alex, I'll try to be careful =P $a+bi$ where $b$ is non-zero.
Aug
26
awarded  Commentator
Aug
26
asked Primality using $\Gamma(x)$
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revised A prime number pattern
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Jul
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comment A prime number pattern
Of course, it would also have to happen that the only $1s$ I have encountered in odd number chains correspond to $Z_t$ values, which end the sequence. If this is the only place $1s$ might occur (for odd numbers), the conjecture would fall to induction.
Jul
30
comment A prime number pattern
My real bad. That aside, even numbers only get $Z_t\in\{-1,0,1\}$ while odd numbers have $\{0,1,2\}$. Let us assume above observation is true till $2n+1$. Then, we can prove if $2n+1$ never reaches $1$, it will not reach $-1$. Follows from the fact that $2n$ would have the same chain as $2n+1$, but for a displacement of $1$ in values. Hence, at the terminus, $Z_t({2n})$ would displace to $2$ (from $-1$) instead of $-2$. This won't work for for any odd number reaching $1$ in the middle of the sequence, of course.
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revised A prime number pattern
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29
awarded  Nice Question
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revised A prime number pattern
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revised A prime number pattern
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revised A prime number pattern
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Jul
29
revised A prime number pattern
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Jul
29
comment A prime number pattern
An elementary result that would follow if odd numbers don't have $−1$ as a terminal is that there always exists a prime between $p_2n+1$ and $−1+\sum_{i=1}^{{2n+}1}p_i$ where $p_n$ is the $n-th$ prime. For example, $-1 + 2 + 3 + 5 = 9$, if there is no prime $p$ such that $5<p<9$ then we have a contradiction.
Jul
29
awarded  Editor