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A145300
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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.
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2
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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
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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