Sa se arate ca daca \( \pi(x) \) reprezinta numarul numerelor prime mai mici ca un numar \( x \) atunci are loc inegalitatea
\( \pi(n!+2n)+\pi(n)\leq\pi(n!+n)+\pi(2n), \forall n\geq 1 \).
Inegalitate cu functia prime-counter pi
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Inegalitate cu functia prime-counter pi
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
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Dragos Fratila
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It is equivalent to prove that a prime between \( n!+n \) and \( n!+2n \) remains prime when we take it modulo \( n! \).
Suppose it doesn't. Then it must have a divisor less than or equal to \( n \). That is \( p=n!+q\cdot k \), where \( q \) is a prime less than \( n \). Then \( p =0 \) modulo \( q \) since \( q \) divides \( n! \), hence \( p \) is not prime. Contradiction.
ma scuzati va rog ca am scris in engleza ... citeam ceva in engleza cand mi-a venit ideea... abia dupa ce am terminat m-am prins, si nu mai rescriu acu... scuze inca o data
Suppose it doesn't. Then it must have a divisor less than or equal to \( n \). That is \( p=n!+q\cdot k \), where \( q \) is a prime less than \( n \). Then \( p =0 \) modulo \( q \) since \( q \) divides \( n! \), hence \( p \) is not prime. Contradiction.
ma scuzati va rog ca am scris in engleza ... citeam ceva in engleza cand mi-a venit ideea... abia dupa ce am terminat m-am prins, si nu mai rescriu acu... scuze inca o data