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%I #16 Apr 30 2022 07:58:35
%S 1,2,5,7,9,7,10,20,18,14,22,19,18,24,26,24,30,30,28,37,25,30,35,35,34,
%T 38,47,52,49,54,40,49,49,69,57,67,78,67,67,68,67,64,65,86,76,81,92,79,
%U 83,83,95,82,85,80,84,95,92,91,121,105,100,108,111,109,118,105,110,88
%N Number of terms in the simple continued fraction for Sum_{k=1..n} 1/k^2.
%C Sum_{k>=1} 1/k^2 = zeta(2) = Pi^2/6.
%H Amiram Eldar, <a href="/A070985/b070985.txt">Table of n, a(n) for n = 1..10000</a>
%F Limit_{n ->infinity} a(n)/n = C =1.6....
%e The simple continued fraction for Sum_{k=1..10} 1/k^2 is [1, 1, 1, 4, 1, 1, 10, 4, 1, 2, 5, 2, 1, 24] which contains 14 terms, hence a(10) = 14.
%t lcf[f_] := Length[ContinuedFraction[f]]; lcf /@ Accumulate[Table[1/k^2, {k, 1, 100}]] (* _Amiram Eldar_, Apr 30 2022 *)
%o (PARI) for(n=1,100,print1(length(contfrac(sum(i=1,n,1/i^2))),","))
%Y Cf. A013679, A055573, A007406, A007407.
%K easy,nonn
%O 1,2
%A _Benoit Cloitre_, May 18 2002