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A190323
n + [n*s/r] + [n*t/r]; r=1, s=sinh(Pi/2), t=cosh(Pi/2).
3
5, 11, 16, 23, 28, 34, 40, 46, 51, 58, 63, 69, 74, 81, 86, 92, 98, 104, 109, 116, 121, 127, 132, 139, 144, 150, 156, 162, 167, 174, 179, 185, 190, 197, 202, 208, 214, 220, 225, 232, 237, 243, 248, 255, 260, 266, 272, 278, 283, 290, 295, 301, 306, 313, 319, 324, 331, 336, 342, 348, 354, 359, 365, 371, 377, 382, 389, 394, 400
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, s=sinh(Pi/2), t=cosh(Pi/2) gives
LINKS
FORMULA
A190323: f(n) = n + [n*sinh(Pi/2)] + [n*cosh(Pi/2)].
A190324: g(n) = n + [n*csch(Pi/2)] + [n*coth(Pi/2)].
A190325: h(n) = n + [n*sech(Pi/2)] + [n*tanh(Pi/2)].
MATHEMATICA
r=1; s=Sinh[Pi/2]; t=Cosh[Pi/2];
f[n_] := n + Floor[n*s/r] + Floor[n*t/r];
g[n_] := n + Floor[n*r/s] + Floor[n*t/s];
h[n_] := n + Floor[n*r/t] + Floor[n*s/t];
Table[f[n], {n, 1, 120}] (* A190323 *)
Table[g[n], {n, 1, 120}] (* A190324 *)
Table[h[n], {n, 1, 120}] (* A190325 *)
PROG
(PARI) for(n=1, 100, print1(n + floor(n*sinh(Pi/2)) + floor(n*cosh(Pi/2)), ", ")) \\ G. C. Greubel, Apr 04 2018
(Magma) R:=RealField(); [n + Floor(n*Sinh(Pi(R)/2)) + Floor(n*Cosh(Pi(R)/2)): n in [1..100]]; // G. C. Greubel, Apr 04 2018
CROSSREFS
Sequence in context: A272666 A314174 A314175 * A314176 A314177 A314178
KEYWORD
nonn
AUTHOR
Clark Kimberling, May 08 2011
STATUS
approved