

A293207


Lexicographically earliest sequence of positive terms such that the function f defined by f(n) = Sum_{k=1..n} (i^k * a(k)) for any n >= 0 is injective (where i denotes the imaginary unit), and a(n) != a(n+1) and a(n) != a(n+2) for any n > 0.


4



1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 4, 1, 2, 4, 1, 2, 3, 1, 4, 5, 1, 7, 2, 1, 4, 2, 1, 4, 5, 1, 2, 3, 1, 2, 6, 3, 1, 4, 2, 1, 9, 2, 1, 3, 2, 1, 11, 2, 1, 6, 2, 1, 3, 2, 1, 6, 4, 7, 1, 3, 2, 1, 3, 9, 1, 2, 3, 1, 4, 7, 5, 1, 2, 4, 1, 11, 4, 10, 1, 9, 2, 6, 7, 1, 9
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OFFSET

1,2


COMMENTS

See A293208 for the real part of f(n).
See A293209 for the imaginary part of f(n).
For any m > 0, let b_m be the lexicographically earliest sequence of positive terms such that the function f_m defined by f_m(n) = Sum_{k=1..n} (i^k * b_m(k)) for any n >= 0 is injective, and #{b_m(n), ..., b_m(n+m)} = m+1 for any n > 0:
 in particular, b_2 = a (this sequence) and f_2 = f,
 the representation of f shows an apparently chaotic initial phase followed by the emergence of a regular oscillating escape (see representations in Links section),
 the cases m=8 and m=12 have similarities with Langton's ant: the representation of f_m shows an apparently chaotic initial phase followed by the emergence of a regular escape (see representations in Links section),
 for the cases m=3, m=7, m=11 and m=15: b_m is m+1 periodic and b_m(n) = n for any n <= m+1,
 for the cases m=4, m=5, m=6, m=9, m=10, m=13, m=14 and m=16: the representation of f_m shows an apparently chaotic initial phase; it is unknown whether a regular escape emerges (see representation in Links section).
More informally, the sequence can be obtained with the following procedure:
 start at the origin, looking to the north,
 repeatedly: jump forward to the nearest nonvisited square (provided that the jump length is distinct from the two previous jump lengths) and turn 90° to the left,
 a(n) = length of nth jump and f(n1) = position before nth jump as a Gaussian integer.


LINKS

Rémy Sigrist, Table of n, a(n) for n = 1..60000
Wikipedia, Langton's ant
Rémy Sigrist, Representation of f(n) for n=0..60000
Rémy Sigrist, Representation of f_8(n) for n=0..121346
Rémy Sigrist, Representation of f_12(n) for n=0..4463502
Rémy Sigrist, Representation of f_6(n) for n=0..10000000
Rémy Sigrist, PARI program for A293207


FORMULA

a(56948 + 8*k + i) = (1k) * a(56948 + i) + k * a(56948 + i + 8) for any k >= 0 and i in 0..7.


EXAMPLE

f(0) = 0
i^1 = i.
f(0) + 1*i has not yet been visited; hence a(1) = 1 and f(1) = i.
i^2 = 1.
f(1) + 1*1 has not yet been visited, but a(1) = 1.
f(1) + 2*1 has not yet been visited; hence a(2) = 2 and f(2) = 2 + i.
i^3 = i.
f(2) + 1*i has not yet been visited, but a(1) = 1.
f(2) + 2*i has not yet been visited, but a(2) = 2.
f(2) + 3*i has not yet been visited; hence a(3) = 3 and f(3) = 2  2*i.
i^4 = 1.
f(3) + 1*1 has not yet been visited; hence a(4) = 1 and f(4) = 1  2*i.
i^5 = i.
f(4) + 1*i has not yet been visited, but a(4) = 1.
f(4) + 2*i has not yet been visited; hence a(5) = 2 and f(5) = 1.
i^6 = 1.
f(5) + 1*1 has not yet been visited, but a(4) = 1.
f(5) + 2*1 has not yet been visited, but a(5) = 2.
f(5) + 3*1 has not yet been visited; hence a(6) = 3 and f(6) = 4.
i^7 = i.
f(6) + 1*i has not yet been visited; hence a(7) = 1 and f(7) = 4  i.
i^8 = 1.
f(7) + 1*1 has not yet been visited, but a(7) = 1.
f(7) + 2*1 has not yet been visited; hence a(8) = 2 and f(8) = 2  i.
i^9 = i.
f(8) + 1*i has not yet been visited, but a(7) = 1.
f(8) + 2*i has not yet been visited, but a(8) = 2.
f(8) + 3*i has not yet been visited; hence a(9) = 3 and f(9) = 2 + 2*i.
i^10 = 1.
f(9) + 1*1 has not yet been visited; hence a(10) = 1 and f(10) = 3 + 2*i.


PROG

(PARI) See Links section.


CROSSREFS

See A293208, A293209.
Sequence in context: A117373 A132677 A010882 * A106590 A194074 A175469
Adjacent sequences: A293204 A293205 A293206 * A293208 A293209 A293210


KEYWORD

nonn


AUTHOR

Rémy Sigrist, Oct 02 2017


STATUS

approved



