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 A350545 a(n) is the least k such that the continued fraction for sqrt(k) has period prime(n). 0
 3, 41, 13, 58, 61, 193, 157, 337, 586, 821, 601, 421, 1117, 1153, 1069, 1669, 2137, 2053, 1381, 3733, 3541, 1621, 4657, 2389, 4561, 6577, 3061, 4261, 5209, 6121, 6781, 8317, 7621, 6661, 6301, 7561, 7549, 15817, 9241, 9349, 12853, 8269, 11701, 16729, 14449, 23017, 31573 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Conjecture: All terms in this sequence (except a(4) = 58 and a(9) = 586) are primes. LINKS Table of n, a(n) for n=1..47. FORMULA a(n) = A013646(prime(n)). EXAMPLE a(5) = 61 because 61 is the least integer k whose period of the continued fraction for sqrt(k) is prime(5)=11, namely {1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14}. MATHEMATICA n=30; prm={}; fin={}; k=2; While[Length@prm

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Last modified March 3 03:04 EST 2024. Contains 370499 sequences. (Running on oeis4.)