A101642 a(n) = Knuth's Fibonacci (or circle) product "3 o n".

8, 13, 21, 29, 34, 42, 47, 55, 63, 68, 76, 84, 89, 97, 102, 110, 118, 123, 131, 136, 144, 152, 157, 165, 173, 178, 186, 191, 199, 207, 212, 220, 228, 233, 241, 246, 254, 262, 267, 275, 280, 288, 296, 301, 309, 317, 322, 330, 335, 343, 351, 356, 364, 369, 377

Offset: 1

Author

N. J. A. Sloane, Jan 26 2005

Keywords

• nonn

Comments

Let phi be the golden ratio. Using Fred Lunnon's formula in A101330 for Knuth's circle product, and the fact that phi^{-2} = 2-phi, plus [-x] = -[x]-1 for non-integer x, one obtains the formula below, expressing this sequence in terms of the lower Wythoff sequence. It follows in particular that the sequence of first differences 5,8,8,5,8,5,8,8,5,8,... of this sequence is the Fibonacci word A003849 on the alphabet {8,5}, shifted by 1. - Michel Dekking, Dec 23 2019

Also numbers with suffix string 0000, when written in Zeckendorf representation. - A.H.M. Smeets, Mar 20 2024

The asymptotic density of this sequence is 1/phi^4 = A094214^4 = 0.145898... . - 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.
• W. F. Lunnon, Proof of formula, 2008.

Crossrefs

Third row of array in A101330.

Cf. A101345 = Knuth's Fibonacci (or circle) product "2 o n".

Cf. A000201, A003849, A094214.

Programs

Mathematica

zeck[n_Integer] := Block[{k = Ceiling[ Log[ GoldenRatio, n*Sqrt[5]]], t = n, fr = {}}, While[k > 1, If[t >= Fibonacci[k], AppendTo[ fr, 1]; t = t - Fibonacci[k], AppendTo[fr, 0]]; k-- ]; FromDigits[fr]]; kfp[n_, m_] := Block[{y = Reverse[ IntegerDigits[ zeck[ n]]], z = Reverse[ IntegerDigits[ zeck[ m]]]}, Sum[ y[[i]]*z[[j]]*Fibonacci[i + j + 2], {i, Length[z1]}, {j, Length[z2]}]]; (* Robert G. Wilson v, Feb 04 2005 *)
Table[ kfp[3, n], {n, 50}] (* Robert G. Wilson v, Feb 04 2005 *)
Array[3*Floor[(# + 1)*GoldenRatio] + 2*# - 3 &, 100] (* Paolo Xausa, Mar 23 2024 *)

Python

from math import isqrt
def A101642(n): return 3*(n+1+isqrt(5*(n+1)**2)>>1)+(n<<1)-3 # Chai Wah Wu, Aug 29 2022

Formula

From Michel Dekking, Dec 23 2019: (Start)

a(n) = 3*A000201(n+1) + 2n - 3.

a(n) = A101345(n) + A000201(n+1) + n + 1. (End)

Extensions

More terms from David Applegate, Jan 26 2005

More terms from Robert G. Wilson v, Feb 04 2005