

A190265


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


4



6, 12, 19, 25, 31, 38, 44, 50, 57, 63, 69, 77, 83, 89, 96, 102, 108, 115, 121, 128, 134, 140, 147, 154, 160, 167, 173, 179, 186, 192, 198, 205, 211, 217, 225, 231, 237, 244, 250, 257, 263, 269, 276, 282, 288, 295, 302, 308, 315, 321, 327, 334, 340, 346, 353, 359, 365, 372, 379, 386, 392, 398, 405, 411, 417, 424, 430, 436, 443, 450
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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+[ns/r]+[nt/r],
b(n)=n+[nr/s]+[nt/s],
c(n)=n+[nr/t]+[ns/t], where []=floor.
Taking r=1/2, s=sinh(1), t=cosh(1) gives
a=A190265, b=A190279, c=A190280.


LINKS

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


FORMULA

A190265: a(n)=n+[2n*sinh(1)]+[2n*cosh(1)].
A190279: b(n)=n+[(n/2)*csch(1)]+[n*coth(1)].
A190280: c(n)=n+[(n/2)*sech(1)]+[n*tanh(1)].


MATHEMATICA

r=1/2; s=Sinh[1]; t=Cosh[1];
a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
c[n_] := n + Floor[n*r/t] + Floor[n*s/t];
Table[a[n], {n, 1, 120}] (*A190265*)
Table[b[n], {n, 1, 120}] (*A190279*)
Table[c[n], {n, 1, 120}] (*A190280*)


CROSSREFS

Cf. A190279, A190280.
Sequence in context: A288794 A177708 A100357 * A135358 A187391 A081846
Adjacent sequences: A190262 A190263 A190264 * A190266 A190267 A190268


KEYWORD

nonn


AUTHOR

Clark Kimberling, May 07 2011


STATUS

approved



