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A338307 a(n) is the smallest positive integer m such that n appears as the denominator of a convergent to sqrt(m). 4
1, 2, 3, 3, 2, 8, 15, 15, 6, 24, 3, 2, 12, 7, 3, 61, 5, 80, 11, 6, 30, 59, 46, 28, 42, 127, 14, 12, 2, 44, 7, 32, 13, 233, 8, 20, 10, 13, 19, 77, 3, 22, 33, 23, 132, 113, 33, 7, 23, 75, 27, 42, 75, 71, 15, 3, 73, 61, 60, 11, 43, 41, 15, 159, 17, 54, 32, 72, 59, 2, 13, 5, 38, 116, 69, 90, 133, 300, 95, 207, 105, 302, 74, 110, 173, 97, 93, 74, 6, 47, 138, 132, 107, 96, 86, 114, 47, 457, 24, 156 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

For n > 2, a(n) <= ceiling(n/2)^2 - 1, since for m > 1, sqrt(m^2 - 1) has convergents (2m^2 - m - 1)/(2m - 1) and (2m^2 - 1)/(2m). Apparently, this bound is achieved only for n in {3, 4, 6, 7, 8, 10, 18}.

Is there an efficient algorithm for computing a(n)?

LINKS

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

MATHEMATICA

a[n_] := Module[{a, b, r, q, v}, If[n == 1, Return[1]]; For[m = 1, True, m++, If[IntegerQ[Sqrt[m]], Continue]; r = Sqrt[m] // Floor; a = 0; b = 1; q = {1, 0}; While[q[[2]] < n && b != 0, v = Quotient[r+a, b]; a = b v-a; b = (m - a^2)/b; q = {{0, 1}, {1, v}}.q]; If[q[[2]] == n, Return[m]]]];

Array[a, 100] (* Jean-Fran├žois Alcover, Oct 23 2020, after PARI code *)

PROG

(PARI) { A338307(n) = my(a, b, r, q, v); if(n==1, return(1)); for(m=1, oo, if(issquare(m), next); r=sqrtint(m); a=0; b=1; q=[1, 0]~; while(q[2]<n, v=(r+a)\b; a=b*v-a; b=(m-a^2)/b; q=[0, 1; 1, v]*q; ); if(q[2]==n, return(m)); ); }

CROSSREFS

Cf. A338308.

Sequence in context: A304311 A175393 A184829 * A153290 A153516 A153311

Adjacent sequences:  A338304 A338305 A338306 * A338308 A338309 A338310

KEYWORD

nonn

AUTHOR

Max Alekseyev, Oct 21 2020

STATUS

approved

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Last modified April 22 12:22 EDT 2021. Contains 343177 sequences. (Running on oeis4.)