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%I #34 Oct 26 2019 07:36:21
%S 1,3,8,18,38,78,159,321,646,1297,2600,5207,10422,20852,41712,83433,
%T 166876,333762,667534,1335078,2670166,5340342,10680695,21361402,
%U 42722816,85445645,170891304,341782622,683565259,1367130534,2734261085,5468522187,10937044391
%N Partial sums of A024810(n) = floor(2^(n+1)/Pi).
%C A024810(n) is the number of x in the interval (1/2^(n+1), 1) such that f(x) = sin(1/x) = 0, for n >= 1. It is well known that the function f(x) oscillates indefinitely around 0 as x approaches zero.
%C Equivalently, the number of roots of the sine function on [1, 2^(n+1)], given by floor(2*2^n/Pi). - _M. F. Hasler_, Oct 25 2019
%e From _M. F. Hasler_, Oct 25 2019: (Start)
%e The first nonzero root of the sine function is at Pi ~ 3.14, so there is one on [1, 4] = [1, 2^(1+1)], whence a(1) = A024810(1) = 1.
%e On [1, 8] = [1, 2^(2+1)], there is one more root, x = 2*Pi ~ 6.28. Therefore A024810(2) = 2 (number of roots) and a(2) = a(1) + 2 = 3.
%e On [1, 16] = [1, 2^(3+1)], there are a total of five roots, x = k*Pi with k = 1,...,5. Therefore A024810(3) = 5 and a(3) = a(2) + 5 = 8. The formula floor(2^(n+1)/Pi) for A024810(n), by definition equal to the increment a(n) - a(n-1), becomes obvious. (End)
%o (PARI) my(s=0); for(n=0, 29, s+=floor(4*2^n/Pi); print1(s, ", ")) \\ _Hugo Pfoertner_, Oct 24 2019
%Y Cf. A024810.
%K nonn
%O 1,2
%A _Michel Lagneau_, Jan 30 2010
%E Edited by _M. F. Hasler_, following a remark by _Kevin Ryde_, Oct 24 2019
%E Data corrected and extended by _M. F. Hasler_, Oct 25 2019
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