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A336090
Number of k's, 1 <= k <= n, such that the continued fraction for n^2/k^2 has twice as many elements as the continued fraction for n/k.
1
0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 3, 1, 0, 3, 3, 2, 2, 3, 2, 3, 5, 2, 3, 3, 3, 3, 3, 1, 3, 6, 3, 8, 2, 5, 3, 6, 5, 5, 6, 6, 2, 6, 5, 6, 7, 6, 4, 3, 7, 9, 8, 7, 5, 8, 6, 7, 6, 5, 5, 7, 9, 10, 6, 5, 8, 14, 13, 3, 10, 10, 11, 6, 4, 13, 8, 7, 8, 13, 8, 9, 9, 9, 9, 12, 13, 8, 11, 10, 8, 14, 11, 13, 12, 13, 12, 8, 14, 8
OFFSET
1,13
COMMENTS
Conjecture: a(n) >= 1 for n >= 15.
FORMULA
Conjecture: a(n) << n/log(n) (see links for the graph of a(n)/n*log(n)).
Does the limit of a(n)*log(n)/n as n tends to infinity exist?
MATHEMATICA
c[x_] := Length @ ContinuedFraction[x] ; a[n_] := Count[Range[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, 1))
CROSSREFS
Cf. A336344.
Sequence in context: A036870 A036874 A256759 * A255123 A244483 A292727
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Oct 04 2020
STATUS
approved