

A214971


Integers k for which the basephi representation of k includes 1.


9



1, 4, 8, 11, 15, 19, 22, 26, 29, 33, 37, 40, 44, 48, 51, 55, 58, 62, 66, 69, 73, 76, 80, 84, 87, 91, 95, 98, 102, 105, 109, 113, 116, 120, 124, 127, 131, 134, 138, 142, 145, 149, 152, 156, 160, 163, 167, 171, 174, 178, 181, 185, 189, 192, 196, 199, 203
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OFFSET

1,2


COMMENTS

Conjecture: L(2k1) and L(2k)+1 are terms of this sequence for all positive integers k, where L=A000032 (Lucas numbers).
Proof of this conjecture: this follows directly from the well known formula L(2k)=phi^{2k}+phi^{2k}, and the recursion L(2k+1)=L(2k)+L(2k1).  Michel Dekking, Jun 25 2019
Conjecture: If D is the difference sequence, then D3 is the infinite Fibonacci word A096270. If so, then A214971 can be generated as in Program 3 of the Mathematica section.  Peter J. C. Moses, Oct 19 2012
Conjecture: A very simple formula for this sequence seems to be a(n) = ceiling((n1)*phi) + 2*(n1) for n>1; thus, see the related sequence A004956.  Thomas Baruchel, May 14 2018
Moses' conjecture is equivalent to Baruchel's conjecture: Baruchel's conjecture expresses that this sequence is a generalized Beatty sequence, and since A096270 equals the Fibonacci word A005614 with an initial zero, this follows directly from Lemma 8 in Allouche and Dekking.  Michel Dekking, May 04 2019
The conjectures by Baruchel and Moses are proved in my paper 'Base phi representations and golden mean betaexpansions'.  Michel Dekking, Jun 25 2019
a(n) equals A198270(n1) for 0<n<15, and a(n) equals either A198270(n1) or A198270(n1)+1 for all n<90, after which the two sequences very slowly diverge from each other.  Greg Dresden, Aug 15 2020


LINKS



FORMULA



EXAMPLE

1 = 1,
4 = r^2 + 1 + 1/r^2,
8 = r^4 + 1 + 1/r^4,
11 = r^4 + r^1 + 1 + 1/r^2 + 1/r^4.
where r = phi = (1 + sqrt(5))/2 = the golden ratio.


MATHEMATICA

(* 1st program *)
r = GoldenRatio; f[x_] := Floor[Log[r, x]];
t[n_] := RealDigits[n, r, 1000]
p[n_] := Flatten[Position[t[n][[1]], 1]]
Table[{n, f[n] + 1  p[n]}, {n, 1, 47}] (* {n, exponents of r in base phi repr of n} *)
m[n_] := If[MemberQ[f[n] + 1  p[n], 0], 1, 0]
u = Table[m[n], {n, 1, 900}]
Flatten[Position[u, 1]] (* A214971 *)
(* 2nd program *)
A214971 = Map[#[[1]] &, Cases[Table[{n, Last[#]  Flatten[Position[First[#], 1]] &[RealDigits[n, GoldenRatio, 1000]]}, {n, 1, 5000}], {_, {___, 0, ___}}]] (* Peter J. C. Moses, Oct 19 2012 *)
(* 3rd program; see Comments *)
Accumulate[Flatten[{1, Nest[Flatten[# /. {0 > {0, 1}, 1 > {0, 1, 1}}] &, {0}, 8] + 3}]] (* Peter J. C. Moses, Oct 19 2012 *)


CROSSREFS



KEYWORD

nonn,base,easy


AUTHOR



STATUS

approved



