On the largest prime factors of consecutive integers in short intervals
Copyright policy )On the largest prime factors of consecutive integers in short intervals
For an integer n > 1 n>1 , let P ( n ) P(n) be the largest prime factor of n n . We prove that, for x → ∞ x\rightarrow \infty , there exists a positive proportion of consecutive integers n n and n + 1 n+1 such that P ( n ) > P ( n + 1 ) P(n)>P(n+1) in short intervals ( x , x + y ] (x, x+y] with x 7 / 12 > y ⩽ x . x^{7/12}>y\leqslant x. In particular, we have \[ | { n ⩽ x : P ( n ) > P ( n + 1 ) } | > 0.1063 x . \big |\{n\leqslant x: P(n)> P(n+1)\}\big |> 0.1063 x. \] This improves a previous result of La Bretèche, Pomerance and Tenenbaum.
- French National Centre for Scientific Research France
- University of Lorraine France
- Institut Élie Cartan de Lorraine France
- University of Lorraine France
Applications of sieve methods, consecutive integers, prime factors, Distribution of integers with specified multiplicative constraints, [MATH.MATH-NT] Mathematics [math]/Number Theory [math.NT]
Applications of sieve methods, consecutive integers, prime factors, Distribution of integers with specified multiplicative constraints, [MATH.MATH-NT] Mathematics [math]/Number Theory [math.NT]
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