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A192002
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Counting sequence for Wythoff AB-numbers smaller than n.
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1
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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
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OFFSET
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1,9
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COMMENTS
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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 Wyhoff 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.
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REFERENCES
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Martin Griffiths, A formula for an infinite family of Fibonacci-word sequences, Fib. Q., 56 (2018), 75-80.
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LINKS
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FORMULA
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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
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EXAMPLE
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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))).
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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