

A189377


a(n) = n + floor(ns/r) + floor(nt/r) with r=2, s=(1+sqrt(5))/2, t=(1+sqrt(5))/2.


4



1, 3, 5, 8, 10, 11, 14, 16, 18, 21, 22, 24, 27, 29, 31, 32, 35, 37, 39, 42, 43, 45, 48, 50, 52, 55, 56, 58, 60, 63, 65, 66, 69, 71, 73, 76, 77, 79, 82, 84, 86, 87, 90, 92, 94, 97, 99, 100, 103, 105, 107, 110, 111, 113, 115, 118, 120, 121, 124, 126, 128, 131, 132, 134, 137, 139, 141, 144, 145, 147, 149, 152, 154, 155, 158, 160, 162, 165, 166, 168, 171, 173, 175, 176
(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.
With r=2, s=(1+sqrt(5))/2, t=(1+sqrt(5))/2 gives a=A189377, b=A189378, c=A189379.
(Conjecture) These are the numbers n such that (n+1)sections of the Fibonacci word contain "000" (the commoner bit) but not "111" (the rarer bit).  Don Reble, Apr 07 2021
Conjecture proved April 8 2021, using the Walnut theorem prover.  Jeffrey Shallit, Apr 09 2021


LINKS

Table of n, a(n) for n=1..84.


MATHEMATICA

r=2; s=(1+5^(1/2))/2; t=(1+5^(1/2))/2;
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}] (*A189377*)
Table[b[n], {n, 1, 120}] (*A189378*)
Table[c[n], {n, 1, 120}] (*A189379*)


CROSSREFS

Cf. A189378, A189379.
Sequence in context: A189755 A141436 A282897 * A073608 A155945 A096985
Adjacent sequences: A189374 A189375 A189376 * A189378 A189379 A189380


KEYWORD

nonn


AUTHOR

Clark Kimberling, Apr 20 2011


STATUS

approved



