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A336090
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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.
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1
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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
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OFFSET
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1,13
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COMMENTS
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Conjecture: a(n) >= 1 for n >= 15.
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LINKS
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FORMULA
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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?
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MATHEMATICA
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c[x_] := Length @ ContinuedFraction[x] ; a[n_] := Count[Range[n], _?(c[(n/#)^2] == 2 * c[n/#] &)]; Array[a, 100] (* Amiram Eldar, Oct 04 2020 *)
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PROG
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(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))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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