A336344 Number of k's, 1 <= k <= n where gcd(n,k) = 1, such that the continued fraction for n^2/k^2 has twice as many elements as the continued fraction for n/k.

0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 3, 1, 0, 2, 3, 1, 2, 2, 2, 2, 5, 1, 3, 0, 2, 1, 3, 1, 3, 3, 2, 5, 2, 2, 3, 4, 2, 2, 6, 3, 2, 2, 4, 1, 7, 2, 4, 0, 4, 5, 8, 3, 4, 5, 4, 4, 6, 1, 5, 4, 6, 4, 3, 0, 8, 5, 8, 0, 10, 4, 11, 3, 1, 6, 7, 2, 8, 6, 5, 3, 9, 1, 6, 10, 10, 2, 11, 3, 5, 7, 8, 6, 10, 4, 12, 3, 10, 2

Offset

1,13

Comments

Conjecture: a(n) >= 1 for n >= 71.

Links

Table of n, a(n) for n=1..100.

Formula

Conjecture: a(n) << n/log(n).

Mathematica

c[x_] := Length @ ContinuedFraction[x]; a[n_] := Count[Range[n], _?(CoprimeQ[n, #] && c[(n/#)^2] == 2*c[n/#] &)]; Array[a, 100] (* Amiram Eldar, Oct 04 2020 *)

Prog

(PARI) L(n, q)=length(contfrac(n/q));
a(n)=sum(k=1, n, if(L(n, k)*2-L(n^2, k^2), 0, if(gcd(n, k)-1, 0, 1)))

Crossrefs

Cf. A336090.

Sequence in context: A073200 A307765 A308077 * A352609 A198345 A104416

Adjacent sequences: A336341 A336342 A336343 * A336345 A336346 A336347

Keyword

nonn

Author

Benoit Cloitre, Oct 04 2020

Status

approved