A031415 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.

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

Offset: 1

Author

David W. Wilson

Keywords

• nonn

Comments

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

Examples

			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
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000 (terms n = 1..999 from T. D. Noe)

Crossrefs

Cf. A031404-A031423.

Subsequence of A003814.

Programs

Mathematica

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 *)

Python

from sympy.ntheory.continued_fraction import continued_fraction_periodic
A031415_list = []
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:
        A031415_list.append(n) # Chai Wah Wu, Sep 16 2021

Extensions

a(1) corrected by T. D. Noe, Apr 03 2014