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 A189518 n+[ns/r]+[nt/r]; r=1, s=1/e, t=1/(e+1). 3
 1, 2, 4, 6, 7, 9, 10, 12, 14, 15, 17, 19, 20, 22, 24, 25, 27, 28, 30, 32, 33, 35, 37, 38, 40, 41, 43, 45, 46, 49, 50, 51, 53, 55, 56, 58, 59, 61, 63, 64, 67, 68, 69, 71, 73, 74, 76, 77, 80, 81, 82, 84, 86, 87, 89, 91, 92, 94, 95, 98, 99, 100, 102, 104, 105, 107, 109, 111, 112, 113, 116, 117, 118, 120, 122, 123, 125, 126, 129, 130, 131, 134, 135, 136, 138 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS 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 a(n)=n+[ns/r]+[nt/r], b(n)=n+[nr/s]+[nt/s], c(n)=n+[nr/t]+[ns/t], where []=floor. Taking r=1, s=1/e, t=1/(e+1) gives a=A189518, b=A189519, c=A189520. LINKS FORMULA The three sequences a=A189518, b=A189519, c=A189520 are given by a(n)=n+[n/e]+[n/(e+1)], b(n)=n+[ne]+[ne/(e+1)], c(n)=3n+[ne]+[n/e]. Is there a simple formula for the complement of a? MATHEMATICA r=1; s=1/E; t=1/(E+1); a[n_] := n + Floor[n*s/r] + Floor[n*t/r]; b[n_] := n + Floor[n*r/s] + Floor[n*t/s]; c[n_] := n + Floor[n*r/t] + Floor[n*s/t]; Table[a[n], {n, 1, 120}]  (*A189518*) Table[b[n], {n, 1, 120}]  (*A189519*) Table[c[n], {n, 1, 120}]  (*A189520*) CROSSREFS Cf. A189519, A189520. Sequence in context: A184658 A247910 A189407 * A213733 A218608 A288475 Adjacent sequences:  A189515 A189516 A189517 * A189519 A189520 A189521 KEYWORD nonn AUTHOR Clark Kimberling, Apr 23 2011 STATUS approved

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Last modified October 18 22:26 EDT 2019. Contains 328211 sequences. (Running on oeis4.)