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%I #14 Apr 29 2022 12:01:34
%S 1,3,4,9,5,9,14,10,10,19,16,21,22,22,24,20,19,24,28,28,29,30,39,31,44,
%T 40,44,33,41,47,44,48,54,48,60,49,63,51,65,72,64,70,78,64,79,77,74,87,
%U 75,86,82,94,88,106,106,94,104,108,87,107,86,106,98,110,115,110,105,115
%N Number of terms in the simple continued fraction expansion of Sum_{k=0..n}(-1)^k/(2k+1), the Leibniz-Gregory series for Pi/4.
%C Pi/4 = Sum_{k=>0} (-1)^k/(2k+1).
%H Amiram Eldar, <a href="/A070154/b070154.txt">Table of n, a(n) for n = 0..10000</a>
%F Limit_{n -> infinity} a(n)/n = C = 1.6...
%e The simple continued fraction for Sum(k=0,10,(-1)^k/(2k+1)) is [0, 1, 4, 4, 1, 3, 54, 1, 2, 1, 1, 4, 11, 1, 2, 2] which contains 16 elements, hence a(10)=16.
%t lcf[f_] := Length[ContinuedFraction[f]]; lcf /@ Accumulate[Table[(-1)^k/(2*k + 1), {k, 0, 100}]] (* _Amiram Eldar_, Apr 29 2022 *)
%o (PARI) for(n=1,100,print1( length(contfrac(sum(i=0,n,(-1)^i/(2*i+1)))),","))
%Y Cf. A003881, A007509, A055573, A069880, A069887.
%K easy,nonn
%O 0,2
%A _Benoit Cloitre_, May 06 2002
%E Offset changed to 0 and a(0) inserted by _Amiram Eldar_, Apr 29 2022
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