A320773 -id:A320773 - cp's OEIS frontend

A331960 Integers whose square root has a continued fraction [b(0);[b(1),...,b(p)]] with a period p > 2 such that b(1)=b(2)=...=b(p-1).

7, 13, 32, 41, 55, 58, 74, 75, 130, 135, 136, 180, 185, 215, 269, 312, 335, 346, 370, 377, 425, 427, 458, 557, 560, 646, 697, 711, 818, 819, 880, 925, 986, 987, 1064, 1067, 1129, 1130, 1272, 1313, 1325, 1326, 1400, 1462, 1490, 1495, 1613, 1714, 1736, 1885

Offset: 1

Gerhard Kirchner, Feb 02 2020

nonn

Any periodic continued fraction represents a rational number, in particular [b(0);[c,c,...,c,b(p)]]. An integer requires b(p)=2*b(0). The exclusion of p < 3 makes sense because there should be at least two constant c-terms. Note that, with m=a0, the terms associated with the continued fractions [m;[2m]] (p=1) and [m;[c,2m]] (p=2) are those in A320773.
General aspect: If [m;[c,c,...,c,2m]] is an integer, it belongs to a quadratic subsequence, see link "Special periodic continued fractions".
The four sequences below, see formula, cover 336 of the first 500 terms.

			7  = [2; [1, 1, 1, 4]]
13 = [3; [1, 1, 1, 1, 6]]
32 = [5; [1, 1, 1, 10]]
41 = [6; [2, 2, 12]]
55 = [7; [2, 2, 2, 14]]

Gerhard Kirchner, Special periodic continued fractions

Cf. A320773.

a:={};For[k=0, k<2000, k++, b:=Last[ContinuedFraction[Sqrt[k]]]; p:=Length[b]; If[p>2, For[i=2, iStefano Spezia, Feb 04 2020 *)

block([an: 2,  n: 0,  nmax: 100],
/*transfers the first nmax terms to a file in the current directory*/
   fl: openw(concat("terms-A331960-",nmax, ".txt")),
    while n m and mod(2*m,an-m^2)>0 then
       (a: m, i: 0, x: w, ok: true,
        while a<2*m and ok do
         (i: i+1, x: 1/(x-floor(x)),
          a: floor(x),
          if i=1 then c: a
          elseif a # c and a<2*m then ok: false),
     if ok then(n: n+1, printf( fl, "~d, ", an)))),
   close(fl));

Formulas for some quadratic subsequences:
p,c        formula       first term  a(1) thru a(500)
                            (k=1)       frequency
4,1   (3k-1)^2 + 4k-1     a(1) =  7        125
5,1   (5k-2)^2 + 6k-2     a(2) = 13         75
3,2   (5k+1)^2 + 4k+1     a(4) = 41         74
4,2   (6k+1)^2 + 5k+1     a(5) = 55         62

cp's OEIS Frontend

A331960 Integers whose square root has a continued fraction [b(0);[b(1),...,b(p)]] with a period p > 2 such that b(1)=b(2)=...=b(p-1).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Maxima

Formula