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%I #19 Jul 01 2017 02:54:47
%S 3,7,10,14,18,22,26,29,34,37,41,44,48,53,56,60,63,68,72,75,79,82,87,
%T 90,94,98,102,106,109,113,117,121,125,128,132,136,140,144,147,151,155,
%U 159,162,166,171,174,178,181,186,190,193,197,200,205,208,212,216
%N n+floor(ns/r)+floor(nt/r), where r=sqrt(2), s=sqrt(3), t=sqrt(5).
%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
%C {i/r: i>=1}, {j/s: j>=1}, {k/t: k>=1} are disjoint.
%C 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=sqrt(2), s=sqrt(3), t=sqrt(5) yields
%C a=A184812, b=A184813, c=A184815.
%H G. C. Greubel, <a href="/A184812/b184812.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n)=n+floor(ns/r)+floor(nt/r), r=sqrt(2), s=sqrt(3), t=sqrt(5).
%t r=2^(1/2); s=3^(1/2); t=5^(1/2);
%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}] (* A184812 *)
%t Table[b[n],{n,1,120}] (* A184813 *)
%t Table[c[n],{n,1,120}] (* A184814 *)
%o (PARI) sr=sqrt(3/2);tr=sqrt(5/2);for(n=1,100,print1(n+floor(n*sr)+floor(n*tr)", ")) \\ _Charles R Greathouse IV_, Jul 15 2011
%o (Maxima) r:sqrt(2)$ s:sqrt(3)$ t:sqrt(5)$
%o makelist(n+floor(n*s/r)+floor(n*t/r),n,1,50); /* _Martin Ettl_, Oct 18 2012 */
%Y Cf. A184813, A184814. Associated partition of the primes: A184815, A184816, A184817.
%K nonn
%O 1,1
%A _Clark Kimberling_, Jan 22 2011
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