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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
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

Theorem. a(n)<=p_n+2sqrt(2p_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(sqrt(q_n)-sqrt(p_n))=sqrt(2), as n tends to infinity. Note, that in the considered case this conjecture is equivalent to the following: lim{sqrt(p_n*q_n}=0, where {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 A216525 A232637

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

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Last modified April 18 16:51 EDT 2019. Contains 322211 sequences. (Running on oeis4.)