A071677 Numbers k such that the number of divisors of k equals the maximum number of elements among the continued fractions for k/1, k/2, k/3, k/4, ..., k/k.
1, 3, 8, 9, 14, 15, 22, 44, 45, 52, 63, 64, 105, 110, 130, 152, 154, 165, 174, 184, 189, 190, 195, 196, 208, 225, 230, 232, 256, 272, 310, 405, 442, 464, 496, 512, 567, 592, 656, 688, 820, 884, 891, 940, 976, 1036, 1068, 1125, 1148, 1210, 1215, 1284, 1305
Offset: 1
Keywords
Links
- Amiram Eldar, Table of n, a(n) for n = 1..1000
Programs
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Mathematica
q[k_] := DivisorSigma[0, k] == Max[Table[Length[ContinuedFraction[k/i]], {i, 1, k}]]; Select[Range[1500], q] (* Amiram Eldar, Apr 19 2025 *)
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PARI
for(n=1,1000, if( numdiv(n) == vecmax( vector(n,i,length(contfrac(n/i)))), print1(n,",")));
Comments