A192002 Counting sequence for Wythoff AB-numbers smaller than n.

0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 18, 18, 18, 19, 19, 19, 19, 19, 20, 20, 20, 21, 21, 21

Offset: 1

Author

Wolfdieter Lang, Jun 28 2011

Keywords

• nonn
• easy

Comments

a(n) is the number of Wythoff AB-numbers from A003623 which are less than n.

a(n) is also the number of Wythoff A-pairs (two consecutive numbers which are both Wythoff A-numbers) not exceeding n.

a(n) is also the number of Wythoff BA-numbers (including 2 = B(A(1)) which however has Wythoff representation 0 for B(1)) not exceeding n-2. From the identity B(A(n)) = A(B(n)) - 1.

For the Wythoff representation of numbers see A189921 and A135817.

Examples

			a(9) = 2 = A(10) + A(9) - (3*9+1) = 16 + 14 - 28.
a(9) = 2 = z(9) - z(8) - 9 = 6 + 5 - 9.
There are a(9)=2 AB-numbers <9, namely 3=A(B(1)) and 8=A(B(B(1))).
There are a(9)=2 A-pairs <=9, namely 3,4 and 8,9.
There are a(9)=2 BA-numbers <=7, namely 2 (see the comment above) and 7 = B(A(B(1))).
		

Links

• Martin Griffiths, A formula for an infinite family of Fibonacci-word sequences, Fib. Q., 56 (2018), 75-80.

Crossrefs

Cf. A000201 (Wythoff A-numbers), A001950 (Wythoff B-numbers), A003623 (Wythoff AB-numbers), A123740.

Programs

Python

from math import isqrt
def A192002(n): return (n+isqrt(m:=5*n**2)>>1)+(n+1+isqrt(m+10*n+5)>>1)-3*n-1 # Chai Wah Wu, Aug 10 2022

Formula

a(n) = Sum_{j=0..n-1} A123740(j), with A123740(0)=0.

a(n) = A(n+1) + A(n) - (3*n+1), with the Wythoff A-numbers A000201.

a(n) = z(n) + z(n-1) - n, with z(n) = A005206(n) = A060143(n+1) which counts A-numbers <=n.

Note that no floor function definitions are necessary.

A(n) (which is as Beatty sequence also floor(n*phi), with phi=(1+sqrt(5))/2) can be defined from the rabbit sequence A005614(n-1), n>=1, which results from a substitution rule, via z(n) by A(n):= z(n-1) + n, B(n):= A(n) + n.

a(n) = floor(n/phi) - floor((1+n)/(1+phi)). - Frank Ruskey, Nov 30 2011