A190002 - OEIS

A190002

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

4, 8, 14, 18, 22, 28, 32, 38, 42, 46, 52, 56, 60, 66, 70, 76, 80, 84, 90, 94, 100, 104, 108, 114, 118, 122, 128, 132, 138, 142, 146, 152, 156, 160, 166, 170, 176, 180, 184, 190, 194, 200, 204, 208, 214, 218, 222, 228, 232, 238, 242, 246, 252, 256, 260, 266, 270, 276, 280, 284, 290, 294, 300, 304, 308, 314, 318, 322, 328, 332, 338

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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

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

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

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

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

a=A190002, b=A190003, c=A190004.

LINKS

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

FORMULA

A190002: a(n) = n + [n*(sinh(1))^2] + [n*(cosh(1))^2].

A190003: b(n) = n + [n*(csch(1))^2] + [n*(coth(1))^2].

A005408: c(n) = 2*n - 1.

MATHEMATICA