A145300 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.

2, 7, 13, 13, 19, 23, 29, 31, 37, 43, 47, 53, 61, 61, 67, 73, 79, 83, 89, 89, 97, 103, 109, 113, 113, 131, 131, 137, 139, 139, 157, 163, 167, 173, 181, 181, 193, 199, 199, 211, 211, 211, 229, 233, 233, 239, 251, 263, 271, 271, 277, 283, 283, 293, 293, 307, 317, 317, 317, 317

Offset: 1

Author

Vladimir Shevelev, Oct 06 2008

Keywords

• nonn

Comments

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.

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

Crossrefs

Cf. A145236, A145281.

Programs

Mathematica

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 *)

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

Extensions

More terms from Michel Marcus, Dec 16 2018