Amiram Eldar, <a href="/A146348/b146348_1.txt">Table of n, a(n) for n = 1..2500</a> (terms 1..200 from Zak Seidov)
Amiram Eldar, <a href="/A146348/b146348_1.txt">Table of n, a(n) for n = 1..2500</a> (terms 1..200 from Zak Seidov)
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Primes p such that continued fraction of (1 + sqrt(p))/2 has period 3.
Amiram Eldar, <a href="/A146348/b146348_1.txt">Table of n, a(n) for n = 1..2500</a> (terms 1..200 from Zak Seidov)
Zak Seidov, Amiram Eldar, <a href="/A146348/b146348_1.txt">Table of n, a(n) for n = 1..2002500</a>
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The primes of the form p = n^2+1 for n>2 are in the sequence, and the continued fraction of (1+sqrt(p))/2 is [n/2; 1, 1, n-1, 1, 1, n-1, 1, 1, ...] with the period (1, 1, n-1).
We observe that the other primes {61, 317, 461, 557, 773, 1129, 1429,… ...} are prime divisors of composites numbers of the form k^2+1 where k = 11, 114, 48, 118, 317, 168, 620,… ... .
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