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 A338629 Number of integers less than n with the same period of continued fraction for square root as n. 1
 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 (list; graph; refs; listen; history; text; internal format)
 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: A131704 A327746 A124492 * A057646 A238892 A238279 Adjacent sequences:  A338626 A338627 A338628 * A338630 A338631 A338632 KEYWORD nonn,look AUTHOR Ilya Gutkovskiy, Nov 04 2020 STATUS approved

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Last modified June 24 18:43 EDT 2021. Contains 345419 sequences. (Running on oeis4.)