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a(n) is the least positive integer k such that ceiling(sqrt(A046315(n)*k))^2 - A046315(n)*k is a square.
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%I #12 Nov 17 2019 01:56:29

%S 1,1,1,1,3,1,3,1,3,1,3,1,5,1,3,7,1,7,3,9,3,3,1,9,9,3,11,1,5,5,13,3,1,

%T 15,15,5,1,17,3,5,1,17,7,3,17,1,7,19,1,21,3,5,7,23,5,1,25,9,1,5,25,9,

%U 27,3,27,1,29,5,11,29,3,11,1,11,5,3,33,1,35,13

%N a(n) is the least positive integer k such that ceiling(sqrt(A046315(n)*k))^2 - A046315(n)*k is a square.

%C Fermat's factorization helper multiplier for the n-th odd semiprime.

%C a(n) is the least positive integer such that A046315(n) * a(n) can be factorized with a single iteration of Fermat's factorization method. Using the factorization of a(n), we can then deduce the prime factors of A046315(n). Example for n = 35490: A046315(n) = 199163 and a(n) = 40; ceiling(sqrt(199163*40)) = 2823; 199163*40 = 2823^2 - 2809 = 2823^2 - 53^2 = (2823+53)(2823-53) = 2876*2770, leading to 199163*(2*2*2*5) = (2*2*719)*(2*5*277) and eventually 199163 = 719*277.

%e a(18) = 7 because the 18th odd semiprime is A046315(18) = 93, ceiling(sqrt(93*7))^2 - 93*7 = 25 is a perfect square and 7 is the least positive integer for which this holds.

%Y Cf. A046315.

%K nonn

%O 1,5

%A _Arnauld Chevallier_, Jul 13 2018