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A190508
a(n) = n+[n*s/r]+[n*t/r]+[n*u/r]; r=golden ratio, s=r^2, t=r^3, u=r^4.
4
8, 18, 26, 36, 47, 55, 65, 73, 84, 94, 102, 112, 123, 131, 141, 149, 160, 170, 178, 188, 196, 207, 217, 225, 235, 246, 254, 264, 272, 283, 293, 301, 311, 322, 330, 340, 348, 358, 369, 377, 387, 395, 406, 416, 424, 434, 445, 453, 463, 471, 482, 492, 500, 510, 518, 529, 539, 547, 557, 568, 576, 586, 594, 605, 615, 623, 633, 644
OFFSET
1,1
COMMENTS
This is one of four sequences that partition the positive integers. In general, suppose that r, s, t, u are positive real numbers for which the sets {i/r: i>=1}, {j/s: j>=1}, {k/t: k>=1}, {h/u: h>=1} are pairwise disjoint. Let a(n) be the rank of n/r when all the numbers in the four sets are jointly ranked. Define b(n), c(n), d(n) as the ranks of n/s, n/t, n/u, respectively.
It is easy to prove that
a(n)=n+[n*s/r]+[n*t/r]+[n*u/r],
b(n)=n+[n*r/s]+[n*t/s]+[n*u/s],
c(n)=n+[n*r/t]+[n*s/t]+[n*u/t],
d(n)=n+[n*r/u]+[n*s/u]+[n*t/u], where []=floor.
Taking r=golden ratio, s=r^2, t=r^3, u=r^4 gives a=A190508, b=A190509, c=A054770, d=A190511.
FORMULA
A190508: a(n)=n+[n*r]+[n*r^2]+[n*r^3]
A190509: b(n)=[n/r]+n+[n*r]+[n*r^2]
A054770: c(n)=[n/r^2]+[n/r]+n+[n*r]
A190511: d(n)=[n/r^3]+[n/r^2]+[n/r]+n
MATHEMATICA
r=GoldenRatio; s=r^2; t=r^3; u=r^4;
a[n_] := n + Floor[n*s/r] + Floor[n*t/r]+Floor[n*u/r];
b[n_] := n + Floor[n*r/s] + Floor[n*t/s]+Floor[n*u/s];
c[n_] := n + Floor[n*r/t] + Floor[n*s/t]+Floor[n*u/t];
d[n_] := n + Floor[n*r/u] + Floor[n*s/u]+Floor[n*t/u];
Table[a[n], {n, 1, 120}] (*A190508*)
Table[b[n], {n, 1, 120}] (*A190509*)
Table[c[n], {n, 1, 120}] (*A054770*)
Table[d[n], {n, 1, 120}] (*A190511*)
CROSSREFS
Cf. A190509, A054770, A190511 (the other three sequences in the partition of N).
Sequence in context: A201056 A352081 A257404 * A298161 A195419 A110188
KEYWORD
nonn,changed
AUTHOR
Clark Kimberling, May 11 2011
STATUS
approved