

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.


1



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
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OFFSET

1,2


COMMENTS

Conjecture: limsup n>infty a(n)=infinity. More precisely we claim that log(a(n))/log(n) is bounded and doesn't 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

Table of n, a(n) for n=1..89.
Benoit Cloitre, Plot of log(a(n))/log(n) (black dot) vs arithmetic mean of log(a(n))/log(n) (red line)


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.


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 increasing terms).
Sequence in context: A238455 A098929 A073098 * A046641 A083664 A083125
Adjacent sequences: A201264 A201265 A201266 * A201268 A201269 A201270


KEYWORD

nonn


AUTHOR

Benoit Cloitre, Jan 09 2013


STATUS

approved



