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A031415
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Numbers k such that the continued fraction for sqrt(k) has odd period and if the last term of the periodic part is deleted then there are a pair of central terms both equal to 2.
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1
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41, 61, 113, 130, 181, 202, 265, 269, 313, 317, 394, 421, 458, 586, 613, 617, 685, 697, 761, 773, 853, 925, 929, 937, 986, 1013, 1066, 1109, 1117, 1201, 1213, 1301, 1325, 1354, 1409, 1417, 1429, 1466, 1586, 1625, 1637, 1649, 1714, 1741, 1745, 1753, 1861
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OFFSET
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1,1
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COMMENTS
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In general, the simple continued fraction expansion of sqrt(m) is a periodic palindromic sequence. That is, contfrac( sqrt(m) ) = [c(0); c(1), c(2), ..., c(p), c(p+1), ...] where p is the period. c(p) = 2*c(0), c(k) = c(p+k) for k>0, c(k) = c(p-k) for p>k>0. If the period p is odd, then p = 2*k+1 and c(k) = c(k+1) can be considered a pair of equal central terms. If the period is even, then p = 2*k and the unique central term is c(k). - Michael Somos, Apr 04 2014
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LINKS
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EXAMPLE
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The simple continued fraction expansion of sqrt(41) = [6; 2, 2, 12, 2, 2, 12, 2, 2, 12, ...] with odd period 3 and two terms equal to 2. Another example is sqrt(202) = [14; 4, 1, 2, 2, 1, 4, 28, 4, 1, 2, 2, 1, 4, 28, 4, 1, 2, 2, 1, 4, 28, ...] with odd period 7 and two terms equal to 2. - Michael Somos, Apr 03 2014
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MATHEMATICA
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n = 1; t = {}; While[Length[t] < 50, n++; If[! IntegerQ[Sqrt[n]], c = ContinuedFraction[Sqrt[n]]; len = Length[c[[2]]]; If[OddQ[len] && c[[2, (len + 1)/2]] == 2, AppendTo[t, n]]]]; t (* T. D. Noe, Apr 03 2014 *)
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PROG
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(Python)
from sympy.ntheory.continued_fraction import continued_fraction_periodic
for n in range(1, 10**3):
cf = continued_fraction_periodic(0, 1, n)
if len(cf) > 1 and len(cf[1]) > 1 and len(cf[1]) % 2 and cf[1][len(cf[1])//2] == 2:
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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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