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%I #13 Apr 29 2022 12:01:38
%S 1,2,5,9,14,10,15,16,14,20,22,19,33,25,27,31,41,33,52,47,36,54,38,56,
%T 50,56,65,68,81,76,77,84,75,80,88,76,90,89,98,89,113,100,110,101,104,
%U 103,121,122,131,121,131,127,124,134,127,132,143,131,143,144,161,141
%N Number of terms in the simple continued fraction for Sum_{k=1..n} 1/k^3.
%C Sum_{k>=1} 1/k^3 = zeta(3) = 1.2020... (A002117).
%H Amiram Eldar, <a href="/A070986/b070986.txt">Table of n, a(n) for n = 1..10000</a>
%F Limit_{n ->infinity} a(n)/n = C = 2.5....
%e The simple continued fraction for Sum_{k=1..10} 1/k^3 is [1, 5, 16, 135, 1, 5, 2, 3, 2, 1, 1, 1, 3, 3, 16, 1, 2, 1, 1, 2] which contains 20 terms, hence a(10)=20.
%t lcf[f_] := Length[ContinuedFraction[f]]; lcf /@ Accumulate[Table[1/k^3, {k, 1, 100}]] (* _Amiram Eldar_, Apr 29 2022 *)
%o (PARI) for(n=1,100,print1(length(contfrac(sum(i=1,n,1/i^3))),","))
%Y Cf. A002117, A055573.
%K easy,nonn
%O 1,2
%A _Benoit Cloitre_, May 18 2002