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

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

r=1/2; 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}]  (*A190326*)
Table[g[n], {n, 1, 120}]  (*A190327*)
Table[h[n], {n, 1, 120}]  (*A190328*)

Prog

(PARI) for(n=1, 100, print1(n + floor(2*n*sinh(Pi/2)) + floor(2*n*cosh(Pi/2)), ", ")) \\ G. C. Greubel, Apr 04 2018
(Magma) R:=RealField(); [n + Floor(2*n*Sinh(Pi(R)/2)) + Floor(2*n*Cosh(Pi(R)/2)): n in [1..100]]; // G. C. Greubel, Apr 04 2018

Crossrefs

Cf. A190327, A190328.

Sequence in context: A265415 A245071 A256825 * A185691 A042291 A041194

Adjacent sequences: A190323 A190324 A190325 * A190327 A190328 A190329

Keyword

nonn

Author

Clark Kimberling, May 08 2011

Status

approved