A190329 - OEIS

A190329

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

2, 5, 9, 11, 15, 18, 20, 24, 27, 31, 33, 36, 40, 42, 46, 49, 53, 55, 58, 62, 64, 68, 71, 73, 77, 80, 84, 86, 90, 93, 95, 99, 102, 106, 108, 111, 115, 117, 121, 124, 126, 130, 133, 137, 139, 143, 146, 148, 152, 155, 159, 161, 164, 168, 170, 174, 177, 181, 183, 186, 190, 192, 196, 199, 201, 205, 208, 212, 214, 217, 221, 223, 227, 230

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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=sqrt(2), t=1/s gives

a=A190326, b=A190327, c=A190328.

LINKS

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

FORMULA

A190329: a(n) = n + [n*sqrt(2)] + [n/sqrt(2)].

A190330: b(n) = n + [n/sqrt(2)] + [n/2].

A187338: c(n) = 3*n + [n*sqrt(2)].

MATHEMATICA