A338629 Number of integers less than n with the same period of continued fraction for square root as n.

0, 0, 0, 1, 1, 1, 0, 2, 2, 2, 3, 4, 0, 1, 5, 3, 3, 6, 0, 7, 1, 2, 2, 8, 4, 4, 9, 3, 1, 10, 0, 4, 5, 6, 11, 5, 5, 12, 13, 14, 0, 15, 0, 1, 3, 0, 7, 16, 6, 6, 17, 4, 2, 5, 8, 18, 6, 0, 7, 9, 0, 10, 19, 7, 7, 20, 1, 21, 2, 8, 3, 22, 1, 3, 11, 1, 9, 12, 13, 23

Offset

1,8

Links

Robert Israel, Table of n, a(n) for n = 1..3000

Eric Weisstein's World of Mathematics, Periodic Continued Fraction

Formula

a(n) = |{j < n : A003285(j) = A003285(n)}|.

Example

a(11) = 3 because A003285(11) = 2 and also A003285(3) = A003285(6) = A003285(8) = 2.
More specifically,
sqrt(3)  = 1 + 1/(1 + 1/(2 + 1/(1 + 1/(2 + ...)))),
sqrt(6)  = 2 + 1/(2 + 1/(4 + 1/(2 + 1/(4 + ...)))),
sqrt(8)  = 2 + 1/(1 + 1/(4 + 1/(1 + 1/(4 + ...)))),
sqrt(11) = 3 + 1/(3 + 1/(6 + 1/(3 + 1/(6 + ...)))).

Maple

f:= proc(n) if issqr(n) then 0 else nops(numtheory:-cfrac(sqrt(n), periodic, quotients)[2]) fi end proc:
V:= map(f, [$1..200]):
seq(numboccur(V[n], V[1..n-1]), n=1..200); # Robert Israel, Nov 06 2020

Mathematica

Table[Length[Select[Range[n - 1], Module[{s = Sqrt[#]}, If[IntegerQ[s], 0, Length[ContinuedFraction[s][[2]]]]] == Module[{s = Sqrt[n]}, If[IntegerQ[s], 0, Length[ContinuedFraction[s][[2]]]]] &]], {n, 80}]

Crossrefs

Cf. A003285.

Sequence in context: A327746 A348626 A124492 * A057646 A238892 A238279

Adjacent sequences: A338626 A338627 A338628 * A338630 A338631 A338632

Keyword

nonn,look

Author

Ilya Gutkovskiy, Nov 04 2020

Status

approved