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%I #11 Dec 27 2017 01:41:19
%S 6,12,19,25,31,38,44,50,57,63,69,77,83,89,96,102,108,115,121,128,134,
%T 140,147,154,160,167,173,179,186,192,198,205,211,217,225,231,237,244,
%U 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
%N a(n) = n + [ns/r] + [nt/r]; r=1/2, 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+[ns/r]+[nt/r],
%C b(n)=n+[nr/s]+[nt/s],
%C c(n)=n+[nr/t]+[ns/t], where []=floor.
%C Taking r=1/2, s=sinh(1), t=cosh(1) gives
%C a=A190265, b=A190279, c=A190280.
%H G. C. Greubel, <a href="/A190265/b190265.txt">Table of n, a(n) for n = 1..10000</a>
%F A190265: a(n)=n+[2n*sinh(1)]+[2n*cosh(1)].
%F A190279: b(n)=n+[(n/2)*csch(1)]+[n*coth(1)].
%F A190280: c(n)=n+[(n/2)*sech(1)]+[n*tanh(1)].
%t r=1/2; 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}] (*A190265*)
%t Table[b[n], {n, 1, 120}] (*A190279*)
%t Table[c[n], {n, 1, 120}] (*A190280*)
%Y Cf. A190279, A190280.
%K nonn
%O 1,1
%A _Clark Kimberling_, May 07 2011
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