A189999 - OEIS

%I #9 Sep 08 2022 08:45:56

%S 3,7,10,14,17,22,25,29,32,36,39,44,48,51,55,58,62,66,70,73,77,80,85,

%T 89,92,96,99,103,107,111,114,118,121,125,130,133,137,140,144,148,152,

%U 155,159,162,166,170,174,178,181,185,188,193,196,200,203,207,210,215,219,222,226,229,234,237,241,244,248,251,256,260,263,267,270

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

%C 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

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

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

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

%C Taking r=1, s=sinh(1), t=cosh(1) gives

%C a=A189999, b=A190000, c=A190001.

%H G. C. Greubel, <a href="/A189999/b189999.txt">Table of n, a(n) for n = 1..10000</a>

%F A189999:  a(n) = n + [n*sinh(1)] + [n*cosh(1)].

%F A190000:  b(n) = n + [n*csch(1)] + [n*coth(1)].

%F A190001:  c(n) = n + [n*sech(1)] + [n*tanh(1)].

%t r=1; s=Sinh[1]; t=Cosh[1];

%t a[n_] := n + Floor[n*s/r] + Floor[n*t/r];

%t b[n_] := n + Floor[n*r/s] + Floor[n*t/s];

%t c[n_] := n + Floor[n*r/t] + Floor[n*s/t];

%t Table[a[n], {n, 1, 120}]  (*A189999*)

%t Table[b[n], {n, 1, 120}]  (*A190000*)

%t Table[c[n], {n, 1, 120}]  (*A190001*)

%o (PARI) for(n=1,100, print1(n + floor(n*sinh(1)) + floor(n*cosh(1)), ", ")) \\ _G. C. Greubel_, Jan 11 2018

%o (Magma) [n + Floor(n*Sinh(1)) + Floor(n*Cosh(1)): n in [1..100]]; // _G. C. Greubel_, Jan 11 2018

%Y Cf. A190000, A190001, A190002.

%K nonn

%O 1,1

%A _Clark Kimberling_, May 03 2011