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

1,1

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

Links

Table of n, a(n) for n=1..60.

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 *)

Prog

(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

Crossrefs

Cf. A145236, A145281.

Sequence in context: A231900 A323740 A130710 * A240029 A340104 A216525

Adjacent sequences: A145297 A145298 A145299 * A145301 A145302 A145303

Keyword

nonn

Author

Vladimir Shevelev, Oct 06 2008

Extensions

More terms from Michel Marcus, Dec 16 2018

Status

approved