cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A214970 Integers n that divide f(n), where f is the change-of-base function defined at A214969, using b = (golden ratio) and c=b^2.

Original entry on oeis.org

1, 4, 8, 11, 19, 22, 29, 48, 55, 58, 76, 124, 152, 192, 199, 323, 341, 348, 377, 398, 496, 521, 844, 912, 969, 992, 1042, 1292, 1364, 2208, 2255, 2356, 2388, 2584, 2728, 3072, 3376, 3553, 3571, 5779, 5952, 6119, 6169, 6252, 6752, 6820, 7142
Offset: 1

Views

Author

Clark Kimberling, Oct 17 2012

Keywords

Comments

Conjecture: L(2k-1) and L(2k)+1 are terms of the sequence, for all positive integers k, where L=A000032 (Lucas numbers).

Examples

			8 = r^4 + 1/r^4, where r = (1 + sqrt(5))/2 (the golden ratio), so f(8) = r^8 + 1/r^8 = 48, so 8 divides f(8).

Links

Crossrefs

Cf. A214969, A214971.

Programs

  • Mathematica
    r = GoldenRatio; f[x_] := Floor[Log[r, x]]; t[n_] := RealDigits[n, r, 1000]; p[n_] := Flatten[Position[t[n][[1]], 1]]; m[n_] := If[MemberQ[f[n] + 1 - p[n], 0], 1, 0]; t = Table[m[n], {n, 1, 900}]; Flatten[Position[t,1]]

A214971 Integers k for which the base-phi representation of k includes 1.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Oct 17 2012

Keywords

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 0A198270(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

Examples

			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.

Links

Crossrefs

Cf. A055778, A214969, A214970, A000032, A096270.

Programs

  • 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 *)
  • Python
    from math import isqrt
def A214971(n): return (n<<1)-1+(n-1+isqrt(5*(n-1)**2)>>1) # Chai Wah Wu, May 22 2025

Formula

a(n) = floor((n-1)*phi) + 2*n - 1. - Primoz Pirnat, Jun 09 2024

A178482 Phi-antipalindromic numbers.

Original entry on oeis.org

1, 3, 4, 7, 8, 10, 11, 18, 19, 21, 22, 25, 26, 28, 29, 47, 48, 50, 51, 54, 55, 57, 58, 65, 66, 68, 69, 72, 73, 75, 76, 123, 124, 126, 127, 130, 131, 133, 134, 141, 142, 144, 145, 148, 149, 151, 152, 170, 171, 173, 174
Offset: 1

Views

Author

Vladimir Shevelev, May 28 2010

Keywords

Comments

We call m a phi-antipalindromic number if for the vector (a,...,b) (a<...=2, either a(n)+1 or a(n)-1 is in the sequence; also either a(n)+3 or a(n)-3 is in the sequence.
Conjecture: this is the sequence of numbers k for which f(k) is an integer, where f(x) is the change-of-base function defined at A214969 using b=phi and c=b^2. - Clark Kimberling, Oct 17 2012
There is a 21-state automaton accepting the Zeckendorf representations of those n in this sequence. - Jeffrey Shallit, May 03 2023
Kimberling's conjecture has been proven by Ingrid Vukusic and myself. Along the way we prove an alternate characterization of the sequence: they are the positive integers whose base-phi expansion consists only of even exponents of phi. - Jeffrey Shallit, Aug 28 2025
Alternatively, this sequence consists of those numbers k such that either k or k-1 can be written as the (possibly empty) sum of distinct Lucas numbers L_i where i>=2 and i is even. - Jeffrey Shallit, Aug 28 2025

Examples

			The vectors of exponents of 4 and 5 are (-2,0,2) and (-4,-1,3) correspondingly (cf.A104605). Therefore by definition 4 is a phi-antipalindromic number, while 5 is not. Let n=38. Then k=5. Thus a(38)=A005248(5)+a(6)=123+10=133. The vector of exponents of phi in the base-phi expansion of 133 is (-10,-4,-2,2,4,10).

Links

Crossrefs

Cf. A005248, A055778, A104605, A104626, A104627, A104628.
For bisections see A171070, A171071.

Programs

  • Mathematica
    phiAPQ[1] = True; phiAPQ[n_] := Module[{d = RealDigits[n, GoldenRatio, 2*Ceiling[Log[GoldenRatio, n]]]}, e = d[[2]] - Flatten @ Position[d[[1]], 1]; Reverse[e] == -e]; Select[Range[200], phiAPQ] (* Amiram Eldar, Apr 23 2020 *)

Formula

For k>=1, a(2^k)=A005248(k); if 2^k
Showing 1-3 of 3 results.

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