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A146330
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Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 5.
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3
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41, 149, 157, 181, 269, 397, 425, 493, 565, 697, 761, 941, 1013, 1037, 1325, 1565, 1781, 1825, 2081, 2153, 2165, 2173, 2465, 2477, 2693, 2725, 3181, 3221, 3533, 3869, 4253, 4409, 5165, 5213, 5273, 5297, 5741, 5837, 6485, 6757, 6949, 7045, 7325, 7465, 8021, 8069
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OFFSET
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1,1
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COMMENTS
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For primes in this sequence see A146350.
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LINKS
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EXAMPLE
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a(1) = 41 because continued fraction of (1+sqrt(41))/2 = 3, 1, 2, 2, 1, 5, 1, 2, 2, 1, 5, 1, 2, 2, 1, 5, 1, 2, ... has period (1,2,2,1,5) length 5.
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MAPLE
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isA146330 := proc(n) RETURN(A146326(n) = 5) ; end:
for n from 2 to 2000 do if isA146330(n) then printf("%d, ", n) ; fi; od: # R. J. Mathar, Sep 06 2009
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MATHEMATICA
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Select[Range[10^4], !IntegerQ @ Sqrt[#] && Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] == 5 &] (* Amiram Eldar, Mar 31 2020 *)
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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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