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Oct 21, 2014 at 9:07 vote accept Włodzimierz Holsztyński
Oct 16, 2014 at 5:45 comment added Włodzimierz Holsztyński The number theoretical 1-summation formulas perhaps are compact but in my case they lead to strain on my eyes. Also, non-specialists especially may occasionally get confused when it is hard or impossible to decode the 1-notation formula.
Oct 16, 2014 at 5:40 comment added Włodzimierz Holsztyński Thank you a lot. @Lucia, I enjoyed your answer. To follow the calculations I'd like to ask a volunteer (you?--I don't dare :-), if possible, to translate the common number-theoretical notation into set-theoretical way. I am talking about the style translation along the line: $\ \sum_{x\in A} 1 = |A|.\ $ Or, say, $\ \sum_{1\le k\le n,\ k\cdot n\le M} 1\ =\ \left|\{(k\ n)\in \mathbb N^2:k\cdot n\le M\ \ and\ \ k\le n\}\right|.\ $ And if I see $\ \sum_{k\ n\ge 1,\ k\cdot n\le M} 1\ $ then I am lost, I don't know how to read it uniquely.
Oct 16, 2014 at 5:28 comment added The Masked Avenger Suppose I gave you the k primes (and, strangely, asked only for odd k). Other than that 2n was smaller than the product of the k primes, could you tell me how small n could be so that each of n+1,..,n+k was divisible by exactly one of those primes? Failing that, can you give bounds on n if the product (n+1)...(n+k) isnonzero and is divisible by the product of the k given primes?
Oct 16, 2014 at 3:36 history answered Lucia CC BY-SA 3.0