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a(n) = n+[ns/r]+[nt/r]; r=2, s=sqrt(2), t=1+sqrt(2).
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%I #13 Sep 08 2022 08:45:56

%S 2,5,8,10,14,17,19,22,25,29,31,34,37,39,43,46,49,51,54,58,60,63,66,68,

%T 72,75,78,80,84,87,89,92,95,99,101,104,107,109,113,116,118,121,124,

%U 128,130,133,136,138,142,145,148,150,153,157,159,162,165,169,171,174,177,179,183,186,188,191,194,198,200,203,206,208,212,215,218,220,223,227,229,232,235,237,241,244

%N a(n) = n+[ns/r]+[nt/r]; r=2, s=sqrt(2), t=1+sqrt(2).

%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 With r=2, s=sqrt(2), t=1+sqrt(2), a=A189457, b=A189458, c=A186222 (conjectured).

%H G. C. Greubel, <a href="/A189457/b189457.txt">Table of n, a(n) for n = 1..10000</a>

%t r=2; s=2^(1/2); t=1+2^(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}] (*A189457*)

%t Table[b[n], {n, 1, 120}] (*A189458*)

%t Table[c[n], {n, 1, 120}] (*A186222, conjectured*)

%o (PARI) vector(120, n, floor(n+floor(n*sqrt(2)/2)+floor(n*(1+sqrt(2))/2))) \\ _G. C. Greubel_, Aug 19 2018

%o (Magma) [Floor(n + Floor(n*Sqrt(2)/2) + Floor(n*(1+Sqrt(2))/2)): n in [1..120]]; // _G. C. Greubel_, Aug 19 2018

%Y Cf. A189458, A186222.

%K nonn

%O 1,1

%A _Clark Kimberling_, Apr 22 2011