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A214971
Integers k for which the base-phi 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
OFFSET
1,2
COMMENTS
Conjecture: L(2k-1) 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(2k-1). - Michel Dekking, Jun 25 2019
Conjecture: If D is the difference sequence, then D-3 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((n-1)*phi) + 2*(n-1) 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 beta-expansions'. - Michel Dekking, Jun 25 2019
a(n) equals A198270(n-1) for 0<n<15, and a(n) equals either A198270(n-1) or A198270(n-1)+1 for all n<90, after which the two sequences very slowly diverge from each other. - Greg Dresden, Aug 15 2020
LINKS
J.-P. Allouche and F. M. Dekking, Generalized Beatty sequences and complementary triples, arXiv:1809.03424 [math.NT], 2018.
Michel Dekking, Base phi representations and golden mean beta-expansions, arXiv:1906.08437 [math.NT], 2019.
FORMULA
a(n) = floor((n-1)*phi) + 2*n - 1. - Primoz Pirnat, Jun 09 2024
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
Clark Kimberling, Oct 17 2012
STATUS
approved