A062389 - OEIS

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A062389

a(n) = floor( (2n-1)*Pi/2 ).

1, 4, 7, 10, 14, 17, 20, 23, 26, 29, 32, 36, 39, 42, 45, 48, 51, 54, 58, 61, 64, 67, 70, 73, 76, 80, 83, 86, 89, 92, 95, 98, 102, 105, 108, 111, 114, 117, 120, 124, 127, 130, 133, 136, 139, 142, 146, 149, 152, 155, 158, 161, 164, 168, 171, 174, 177, 180, 183, 186

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

In general, the complement of a nonhomogenous Beatty sequence [n*r + h] is given by [n*s + h - h*s], where s = r/(r - 1). As an example, the complement of this sequence is A246046. This sequence gives the positive integers k satisfying tan(k) > tan(k + 1), and A246046 gives those satisfying tan(k) < tan(k + 1). - Clark Kimberling, Aug 24 2014

Excluding a(1), a(n) = positive floored solutions to tan(x) = x. - Derek Orr, May 30 2015

REFERENCES

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 223.

LINKS

Harry J. Smith, Table of n, a(n) for n=1..1000

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

MAPLE

seq(floor((2*n-1)*Pi/2), n=1..1000); # Robert Israel, Jun 01 2015

MATHEMATICA

r = Pi; s = Pi/(Pi - 1); h = -Pi/2; z = 120;

u = Table[Floor[n*r + h], {n, 1, z}] (* A062389 *)

v = Table[Floor[n*s + h - h*s], {n, 1, z}] (* A246046 *)

(* Clark Kimberling, Aug 24 2014 *)

PROG

(PARI) j=[]; for(n=1, 150, j=concat(j, floor(1/2*(2*n-1)*Pi))); j

(PARI) { default(realprecision, 50); for (n=1, 1000, write("b062389.txt", n, " ", (2*n - 1)*Pi\2); ) } \\ Harry J. Smith, Aug 06 2009

CROSSREFS

Cf. A246046.