A172265 Partial sums of A024810(n) = floor(2^(n+1)/Pi).

1, 3, 8, 18, 38, 78, 159, 321, 646, 1297, 2600, 5207, 10422, 20852, 41712, 83433, 166876, 333762, 667534, 1335078, 2670166, 5340342, 10680695, 21361402, 42722816, 85445645, 170891304, 341782622, 683565259, 1367130534, 2734261085, 5468522187, 10937044391

Offset: 1

Author

Michel Lagneau, Jan 30 2010

Keywords

• nonn

Comments

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.

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

Examples

			From _M. F. Hasler_, Oct 25 2019: (Start)
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.
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.
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)
		

Crossrefs

Cf. A024810.

Programs

PARI

my(s=0); for(n=0, 29, s+=floor(4*2^n/Pi); print1(s, ", ")) \\ Hugo Pfoertner, Oct 24 2019

Extensions

Edited by M. F. Hasler, following a remark by Kevin Ryde, Oct 24 2019

Data corrected and extended by M. F. Hasler, Oct 25 2019