A368882 a(n) is the first number k such that the continued fraction for prime(k)/k has length n.

1, 2, 3, 8, 18, 23, 48, 49, 142, 128, 302, 714, 1584, 1809, 3733, 4641, 11281, 12533, 27170, 70759, 170717, 206956, 445933, 431440, 1096122, 909272, 2311327, 7107627, 7371755, 18882236, 18911330, 48297584, 120815756, 125176520, 299244034, 794635896, 1572930576, 1713395072, 839199226

Offset

1,2

Comments

a(n) is the first number k such that the continued fraction for k/prime(k) has length n+1.

Links

Table of n, a(n) for n=1..39.

Formula

A107901(a(n)) = n.

A107898(a(n)) = n + 1.

Example

a(4) = 8 because prime(8) = 23 and the continued fraction for 23/8 is [2; 1, 7] (i.e., 23/8 = 2 + 1/(1 + 1/7)) with 3 terms, and this is the first continued fraction of 3 terms that occurs.

Maple

V:= Vector(27): count:= 0: p:= 0:
for n from 1 while count < 50 do
  p:= nextprime(p);
v:= nops(Term(NumberTheory:-ContinuedFraction(p/n, all));
if v <= 27 and V[v] = 0 then V[v]:= n; count:= count+1 fi;
od:
convert(V, list);

Mathematica

a[n_]:=Module[{k=1}, While[Length[ContinuedFraction[Prime[k]/k]]!=n, k++]; k]; Array[a, 20] (* Stefano Spezia, Jan 09 2024 *)

Prog

(PARI) upto(n) = {my(res = [], t = 0); forprime(p = 2, oo, t++; if(t >= n, return(res)); c = #contfrac(p/t); if(c > #res, res = concat(res, vector(c - #res, i, oo)); res[c] = min(res[c], t)))} \\ David A. Corneth, Jan 09 2024

Crossrefs

Cf. A107898, A107901.

Sequence in context: A290878 A096254 A091765 * A036746 A232165 A129955

Adjacent sequences: A368879 A368880 A368881 * A368883 A368884 A368885

Keyword

nonn

Author

Robert Israel, Jan 08 2024

Extensions

More terms from David A. Corneth, Jan 09 2024

Status

approved