Reply from On-Line Encyclopedia of Integer Sequences (fwd)

[Don Reble]

> values of  n  for which length of
> period of continued fraction for sqrt(3n^2) was 2.
>
> 1 2 3 4 7 11 14 15 26 28 41 52 56 79 98 ...

Actually, 79 and 98 aren't in the sequence; 78 and 97 are.

> Is it connected with the convergents to the continued
> fraction for  sqrt(3) ? :
>
>     0  1  1  2  5  7  19  26  71  97
>     -  -  -  -  -  -  --  --  --  --  ...
>     1  0  1  1  3  4  11  15  41  56

Yes. Let b/a be one such convergent. Then either
    b^2 = 3a^2 - 2, or
    b^2 = 3a^2 + 1;
those two conditions alternating in the convergents.

If b^2 = 3a^2 - 2, then sqrt(3a^2) = << b,   {b, 2b}...   >>.
If b^2 = 3a^2 + 1, then sqrt(3a^2) = << b-1, {1, 2b-2}... >>.

If b^2 = 3a^2 + 1, then 3b^2 = (3a)^2 + 3, and
                        sqrt(3b^2) = << 3a, {2a, 6a}... >>

So the bottom row, and half of the top row, are in the sequence.

Also:
If b^2 = 3(ja)^2 + j,  then sqrt(3b^2) = << 3ja, { 2a, 6ja }... >>
If b^2 = 3(ka)^2 + 2k, then sqrt(3b^2) = << 3ka, {  a, 6ka }... >>

That explains the other numbers:
    14 (j=4 a=2), 28 (j=16 a=1), 52 (k=2 a=15), 78 (j=9 a=5).

--
Don Reble       djr at nk.ca