275 reputation
311
bio website
location Iran
age
visits member for 1 year, 4 months
seen Apr 9 at 8:50

Originally started programming by ACM contests and then learning how to code some HPC in Prime number category.

Then switching to .NET world and C# as a language to express myself.

Plus that, translating CLR via C#, 3rd Ed by Jeffery Ritcher to Persian language were my great moments of coding.

Add to them the wows of MVC 4 and Android experience and you'll get a mixture of all. ;-)


Dec
3
awarded  Yearling
Jul
19
awarded  Nice Answer
May
19
awarded  Constituent
May
11
awarded  Caucus
Apr
22
comment Last 2 digits of modular exponentiation
@Mike,Yes, I want the last two digits of the first modular operation so it requires another (mod 100) on the result of the first mod.
Apr
22
revised Last 2 digits of modular exponentiation
added 12 characters in body; edited title
Apr
22
asked Last 2 digits of modular exponentiation
Apr
3
awarded  Citizen Patrol
Apr
3
accepted Proof of Generalized Primorial Primes
Apr
3
comment Numerical upper bound of $o(1)$
f(x) is o(g(x)) means that growth of f(x) is nothing compared to g(x). (Correct: it is upper bound) So maximum of $o(1)$ is $1$ so $x<2n\log n$. OK?
Apr
3
comment Numerical upper bound of $o(1)$
No confusion. It is smallOh. I want to replace that smallO in upper bound (although smallO is usually for lower bound), with a number
Apr
3
asked Numerical upper bound of $o(1)$
Mar
19
awarded  Enthusiast
Mar
6
comment Proof of Generalized Primorial Primes
Thank you, brilliant explanation ...
Mar
6
asked Proof of Generalized Primorial Primes
Mar
3
revised Proof of Prime Maker Conjecture
added more samples
Mar
3
comment Proof of Prime Maker Conjecture
@Hans Engler, I've updated the algorithm with correct explaination
Mar
2
revised Proof of Prime Maker Conjecture
added 130 characters in body
Mar
2
revised Proof of Prime Maker Conjecture
added 17 characters in body
Mar
2
comment Proof of Prime Maker Conjecture
$d_0$ is a prime factor of $n$ or $m+1$, so $d_0 \mid m+1$, and so then we have $d_0 \not\mid d_0\times m + 1$. Got it?