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A343515
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a(n) is the number of real solutions to the equation sin(x) = x/n.
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0
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1, 3, 3, 3, 3, 3, 3, 7, 7, 7, 7, 7, 7, 7, 11, 11, 11, 11, 11, 11, 15, 15, 15, 15, 15, 15, 19, 19, 19, 19, 19, 19, 23, 23, 23, 23, 23, 23, 23, 27, 27, 27, 27, 27, 27, 31, 31, 31, 31, 31, 31, 35, 35, 35, 35, 35, 35, 35, 39, 39, 39, 39, 39, 39, 43, 43, 43, 43, 43
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OFFSET
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1,2
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COMMENTS
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All terms are odd.
All terms are congruent to 3 modulo 4 after the first term. Proof: define sin(x)/x to be 1 at x = 0. If a(n) == 1 (mod 4), then the horizontal line y = 1/n is tangent to the curve y = sin(x)/x at (x_n, sin(x_n)/x_n) for some x_n >= 0. We have tan(x_n) = x_n and sin(x_n)/x_n = 1/n, so cos(x_n) = 1/n. By the Lindemann-Weierstrass theorem, we have either x_n = 0 or x_n must be transcendental (if x is a nonzero algebraic number, then exp(x) is transcendental). On the other hand, x_n = tan(x_n) = sqrt(n^2-1) is algebraic, so the only possibility is n = 1. - Jianing Song, Jul 13 2021
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LINKS
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FORMULA
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a(n) ~ 4*(floor((n-Pi/2)/(2*Pi))+1)-1.
For n > 1, a(n) = 4*(floor((n-Pi/2)/(2*Pi))+1)-1 + r(n), where r(n) is an error term defined as follows: let E be the system of equations given by cos(sqrt(n^2-1)) = 1/n and sin(sqrt(n^2-1)) = sqrt(n^2-1)/n; r(n) = 4 if the closest solution of E from the left to Pi/2 + 2*Pi*(floor((n-Pi/2)/2*Pi)+1) is smaller than n; r(n) = 0 otherwise.
Define x_k to be root of tan(x) = x in [k*Pi, (k+1)*Pi), k >= 0. For n > 1, if sec(x_(2*k)) < n < sec(x_(2*k+2)) (or equivalently, x_(2*k) < sqrt(n^2-1) < x_(2*k+2)), then a(n) = 4*k + 3.
For n >= 2, a(n+1) - a(n) is either 0 or 4. a(n+1) - a(n) = 4 if n is of the form floor(sec(x_(2*k))) = floor(sqrt((x_(2*k))^2+1)) for some k > 0. (End)
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EXAMPLE
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a(3) = 3 because the equation sin(x) = x/3 has 3 real solutions: {-2.27886..., 0, 2.27886...}.
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MATHEMATICA
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Join[{1}, Table[CountRoots[n*Sin[x] - x, {x, -n, n}], {n, 2, 100}]] (* Vaclav Kotesovec, Jun 25 2021 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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