A134863 Wythoff BAB numbers.

7, 20, 28, 41, 54, 62, 75, 83, 96, 109, 117, 130, 143, 151, 164, 172, 185, 198, 206, 219, 227, 240, 253, 261, 274, 287, 295, 308, 316, 329, 342, 350, 363, 376, 384, 397, 405, 418, 431, 439, 452, 460, 473, 486, 494, 507, 520, 528, 541, 549, 562, 575, 583, 596

Offset: 1

Author

Clark Kimberling, Nov 14 2007

Keywords

• nonn

Comments

The lower and upper Wythoff sequences, A and B, satisfy the complementary equation BAB = 2A+3B-1.

Also numbers with suffix string 1010, when written in Zeckendorf representation. - A.H.M. Smeets, Mar 24 2024

The asymptotic density of this sequence is 1/phi^5 = phi^5 - 11 = A244593 - 4 = 0.0901699... . - Amiram Eldar, Mar 24 2025

Links

• A.H.M. Smeets, Table of n, a(n) for n = 1..20000
• Jon Asier Bárcena-Petisco, Luis Martínez, María Merino, Juan Manuel Montoya, and Antonio Vera-López, Fibonacci-like partitions and their associated piecewise-defined permutations, arXiv:2503.19696 [math.CO], 2025. See p. 5.
• Clark Kimberling, Complementary equations and Wythoff Sequences, Journal of Integer Sequences 11 (2008), Article 08.3.3.

Crossrefs

Cf. A000201, A001950, A003622, A003623, A035336, A101864, A134859, A035337, A134860, A134861, A134862, A035338, A134864, A035513, A244593.

Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864. The eight triples AAA, AAB, ..., BBB are A134859, A134860, A035337, A134862, A134861, A134863, A035338, A134864, resp.

Programs

Mathematica

A[n_] := Floor[n * GoldenRatio]; B[n_] := Floor[n * GoldenRatio^2]; a[n_] := B[A[B[n]]]; Array[a, 100] (* Amiram Eldar, Mar 24 2025 *)

Python

from sympy import floor
from mpmath import phi
def A(n): return floor(n*phi)
def B(n): return floor(n*phi**2)
def a(n): return B(A(B(n))) # Indranil Ghosh, Jun 10 2017

Python

from math import isqrt
def A134863(n): return 5*(n+isqrt(5*n**2)>>1)+3*n-1 # Chai Wah Wu, Aug 11 2022

Formula

a(n) = B(A(B(n))), n>=1, with A=A000201, the lower Wythoff sequence and B=A001950, the upper Wythoff sequence.

From A.H.M. Smeets, Mar 24 2024: (Start)

a(n) = 2*A(n) + 3*B(n) - 1 (see Clark Kimberling 2008), with A=A000201, B=A001950, the lower and upper Wythoff sequences, respectively.

Equals {A035336}\{A134861} (= Wythoff BA \ Wythoff BAA). (End)