A145300 - OEIS

%I #20 May 24 2021 09:41:14

%S 2,7,13,13,19,23,29,31,37,43,47,53,61,61,67,73,79,83,89,89,97,103,109,

%T 113,113,131,131,137,139,139,157,163,167,173,181,181,193,199,199,211,

%U 211,211,229,233,233,239,251,263,271,271,277,283,283,293,293,307,317,317,317,317

%N a(n) is the maximal prime such that if p_n is the n-th prime then ceiling(sqrt(a(n)*p_n))^2 - a(n)*p_n is a perfect square.

%C Theorem. a(n) <= p_n + 2*sqrt(2*p_n) + 2. For example, for n=25, p_n=97. Using the theorem, we find: a(25) <= 126. Now, by the definition of the sequence, we verify that a(25)=113.

%C Or a(n) is the maximal prime q_n > p_n such that sqrt(q_n)-sqrt(p_n) < sqrt(2) [or (p_n+q_n)/2 < sqrt(p_n*q_n)+1]. I conjecture that lim_{n->infinity} (sqrt(q_n) - sqrt(p_n)) = sqrt(2). Note that in the considered case this conjecture is equivalent to the following: lim_{n->infinity} fract(sqrt(p_n*q_n)) = 0, where fract(x) denotes the fractional part of x. - _Vladimir Shevelev_, Oct 09 2008

%t a[n_] := Module[{pmax = 0, pn = Prime[n]}, p=2; While[p <= pn + 2*Floor[Sqrt[2*pn]] + 2, If[IntegerQ[Sqrt[Ceiling[Sqrt[p*pn]]^2-p*pn]], pmax = p]; p=NextPrime[p]]; pmax]; Array[a, 60] (* _Amiram Eldar_, Dec 16 2018 from the PARI code *)

%o (PARI) a(n) = {my (pmax = 0, pn = prime(n)); forprime(p=2, pn+2*sqrtint(2*pn)+2, if (issquare((ceil(sqrt(p*pn)))^2-p*pn), pmax = p);); pmax;} \\ _Michel Marcus_, Dec 16 2018

%Y Cf. A145236, A145281.

%K nonn

%O 1,1

%A _Vladimir Shevelev_, Oct 06 2008

%E More terms from _Michel Marcus_, Dec 16 2018