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 A201267 a(1) = 1 and a(n) is the least integer such that the continued fraction for 1/a(1) + 1/a(2) + ... + 1/a(n) contains exactly n elements. 3
 1, 2, 3, 11, 2, 16, 4, 7, 4, 12, 5, 2, 41, 3, 11, 13, 3, 4, 22, 19, 2, 12, 27, 29, 9, 18, 8, 39, 94, 14, 13, 35, 101, 44, 122, 36, 2, 4, 60, 11, 7, 129, 4, 25, 18, 27, 19, 77, 62, 35, 14, 229, 74, 7, 29, 4, 32, 88, 132, 30, 2, 154, 511, 71, 59, 9, 127, 2, 47, 20, 34, 54, 22, 34, 57, 56, 68, 16, 45, 298, 57, 169, 13, 134, 45, 39, 120, 77, 109 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Conjecture: limsup_{n->infinity} a(n) = infinity. More precisely we claim that log(a(n))/log(n) is bounded and does not converge to zero (see related link). Does a(n) = 2 infinitely many times or does it exist M >= 2 such that a(n) <= M infinitely many times? LINKS Amiram Eldar, Table of n, a(n) for n = 1..10000 EXAMPLE 1/a(1)+1/a(2)+1/a(3)+1/a(4)=1+1/2+1/3+1/11 = 127/66 and the continued fraction expansion is [1, 1, 12, 5] containing 4 elements. Next k=2 is the smallest integer >=1 such that the continued fraction expansion of 127/66+1/k which is [2, 2, 2, 1, 4] contains 5 elements, thus a(5) = 2. MATHEMATICA a[1] = 1; a[n_] := a[n] = Module[{s = Total[1/Array[a, n - 1]], k = 2}, While[Length[ContinuedFraction[s + 1/k]] != n, k++]; k]; Array[a, 100] (* Amiram Eldar, Jun 05 2022 *) PROG (PARI) v=[1]; for(m=2, 100, k=1; while(abs(length(contfrac(1/k+sum(i=1, length(v), 1/v[i])))-m)>0, k++); v=concat(v, [k])); a(n)=v[n]; CROSSREFS Cf. A071012 (version with nondecreasing terms), A354742 (version with increasing terms). Sequence in context: A238455 A098929 A073098 * A046641 A336876 A083664 Adjacent sequences: A201264 A201265 A201266 * A201268 A201269 A201270 KEYWORD nonn AUTHOR Benoit Cloitre, Jan 09 2013 STATUS approved

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Last modified February 7 20:06 EST 2023. Contains 360129 sequences. (Running on oeis4.)