A190326 - OEIS

A190326

n + [n*s/r] + [n*t/r]; r=1/2, s=sinh(Pi/2), t=cosh(Pi/2).

10, 21, 31, 42, 53, 63, 74, 84, 95, 106, 116, 127, 137, 148, 159, 169, 180, 190, 201, 212, 222, 233, 243, 254, 265, 275, 286, 296, 307, 318, 328, 339, 349, 360, 371, 381, 392, 402, 413, 424, 434, 445, 455, 466, 477, 487, 498, 508, 519, 530, 540, 551, 561, 572

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

OFFSET

1,1

COMMENTS

This is one of three sequences that partition the positive integers. In general, suppose that r, s, t are positive real numbers for which the sets {i/r: i>=1}, {j/s: j>=1}, {k/t: k>=1} are pairwise disjoint. Let a(n) be the rank of n/r when all the numbers in the three sets are jointly ranked. Define b(n) and c(n) as the ranks of n/s and n/t. It is easy to prove that

f(n) = n + [n*s/r] + [n*t/r],

g(n) = n + [n*r/s] + [n*t/s],

h(n) = n + [n*r/t] + [n*s/t], where []=floor.

Taking r=1/2, s=sinh(Pi/2), t=cosh(Pi/2) gives

f=A190326, g=A190327, h=A190328.

LINKS

G. C. Greubel, Table of n, a(n) for n = 1..10000

FORMULA

A190326: f(n) = n + [2*n*sinh(Pi/2)] + [2*n*cosh(Pi/2)].

A190327: g(n) = n + [n*csch(Pi/2)/2] + [n*coth(Pi/2)].

A190328: h(n) = n + [n*sech(Pi/2)/2] + [n*tanh(Pi/2)].

MATHEMATICA