A014677 First differences of A001468.

1, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, 0, -1, 1, -1, 1

Offset: 0

Author

N. J. A. Sloane, Nov 07 2001

Keywords

• sign

Comments

A001468 is an infinite Fibonacci word with strings of 2's of length A001468(n) delimited by 1's. - Paul D. Hanna, Dec 17 2004

c(n):=a(n-1), n >= 1, is -1 if n is a Wythoff B-number from A001950, it is 0 if n=A(B(m)+1) for some m >= 1, with A(k):=A000201(k) (Wythoff A-numbers) and it is +1 if n=A(A(m)+1)=B(m)+1 for some m >= 0, with B(0):=0. - Wolfdieter Lang, Oct 13 2006

This sequence is a symbolic sequence as discussed in the paper "Morphisms, Symbolic Sequences, and Their Standard Forms". It can be derived directly from the 2-block morphism induced by the morphism generating A001468. Since A001468 is the Fibonacci word A003849, but on the alphabet {2,1}, with an extra 1 in front, this 2-block morphism has 3-symbol Fibonacci as a fixed point: A270788. The 2-blocks in A001468 are 12, 21, and 22, yielding the differences a(n) = 1, a(n) = -1, and a(n) = 0. In 3-symbol Fibonacci these correspond to the letters 2, 1, and 3. Expressing this coding with pi given by pi(1)=-1, pi(2)=1, pi(3)=0, we obtain the formula below. Wolfdieter Lang's Wythoff description of (a(n)) follows from the corresponding Wythoff description in A270788. - Michel Dekking, Dec 30 2019

Links

• F. Michel Dekking, Morphisms, Symbolic Sequences, and Their Standard Forms, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.

Crossrefs

Cf. A001468, A000045. Essentially equal to A270788.

Cf. A000201, A001950, A005614, A005713.

Programs

Python

from math import isqrt
def A014677(n): return (n+isqrt(m:=5*(n+2)**2)>>1)-(n+1+isqrt(m-10*n-15)&-2)+(n+isqrt(m-20*n-20)>>1)+1 # Chai Wah Wu, Aug 25 2022

Formula

abs(a(n)) = floor(f*ceiling(n/f)) - ceiling(f*floor(n/f)) where f=phi=(1+sqrt(5))/2; for n > 1, abs(a(n)) = A005713(n-1). - Benoit Cloitre, Apr 21 2003

G.f. equals the continued fraction: A(x) = [0;1, 1/x, 1/x, 1/x^2, 1/x^3, 1/x^5, 1/x^8, ..., 1/x^Fibonacci(n), ...]. - Paul D. Hanna, Dec 17 2004

a(n) = b(n) - b(n-1) with b(n):=A005614(n), n >= 1.

a(n) = pi(A270788(n)), n >= 1, where pi is the letter-to-letter map pi(1)=-1, pi(2)=1, pi(3)=0. - Michel Dekking, Dec 30 2019