A338307 a(n) is the smallest positive integer m such that n appears as the denominator of a convergent to sqrt(m).

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

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-François 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