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A146348
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Primes p such that continued fraction of (1 + sqrt(p))/2 has period 3.
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39
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17, 37, 61, 101, 197, 257, 317, 401, 461, 557, 577, 677, 773, 1129, 1297, 1429, 1601, 1877, 1901, 2917, 3137, 4357, 4597, 5417, 5477, 6053, 7057, 8101, 8761, 8837, 10733, 11621, 12101, 13457, 13877, 14401, 15277, 15377, 15877, 16333, 16901, 17737, 17957, 18329, 21317, 22501, 23593, 24337, 25601, 28901, 30137, 30977, 32401, 33857, 41453, 41617, 42437, 44101
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OFFSET
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1,1
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COMMENTS
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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, ... .
(End)
Another possibly infinite subset of the sequence is primes of the form 100*k^2-44*k+5, where the continued fraction is [5*k-1; 2, 2, 10*k-3, ...] with period [2, 2, 10*k-3]. This includes {61, 317, 773, 1429, 4597, 6053, ...}. - Robert Israel, Sep 03 2014
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LINKS
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MAPLE
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A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic', 'quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146348 := proc(n) RETURN(isprime(n) and A146326(n) = 3) ; end: for n from 2 to 4000 do if isA146348(n) then printf("%d, \n", n) ; fi; od: # R. J. Mathar, Sep 06 2009
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MATHEMATICA
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okQ[n_] := Length[ContinuedFraction[(1 + Sqrt[n])/2][[2]]] == 3; Select[Prime[Range[100]], okQ]
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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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