login
A290885
Let S be the sequence generated by these rules: 0 is in S, and if z is in S, then z * (1+i) and (z-1) * (1+i) + 1 are in S (where i denotes the imaginary unit), and duplicates are deleted as they occur; a(n) = the imaginary part of the n-th term of S.
3
0, -1, -1, -2, 0, -1, -1, -2, 2, 1, 1, 0, 2, 1, 1, 0, 4, 3, 3, 2, 4, 3, 3, 2, 6, 5, 5, 4, 6, 5, 5, 4, 4, 3, 3, 2, 4, 3, 3, 2, 6, 5, 5, 4, 6, 5, 5, 4, 8, 7, 7, 6, 8, 7, 7, 6, 10, 9, 9, 8, 10, 9, 9, 8, 0, -1, -1, -2, 0, -1, -1, -2, 2, 1, 1, 0, 2, 1, 1, 0, 4, 3
OFFSET
1,4
COMMENTS
See A290884 for the real part of the n-th term of S, and additional comments.
See A290886 for the square of the norm of the n-th term of S.
LINKS
EXAMPLE
Let f be the function z -> z * (1+i), and g the function z -> (z-1) * (1+i) + 1.
S(1) = 0 by definition; so a(1) = 0.
f(S(1)) = 0 has already occurred.
g(S(1)) = -i has not yet occurred; so S(2) = -i and a(2) = -1.
f(S(2)) = 1 - i has not yet occurred; so S(3) = 1 - i and a(3) = -1.
g(S(2)) = 1 - 2*i has not yet occurred; so S(4) = 1 - 2*i and a(4) = -2.
f(S(3)) = 2 has not yet occurred; so S(5) = 2 and a(5) = 0.
g(S(3)) = 2 - i has not yet occurred; so S(6) = 2 - i and a(6) = -1.
f(S(4)) = 3 - i has not yet occurred; so S(7) = 3 - i and a(7) = -1.
g(S(4)) = 3 - 2*i has not yet occurred; so S(8) = 3 - 2*i and a(8) = -2.
PROG
(PARI) See Links section.
(PARI) a(n) = -real(subst(Pol(binary(n-1)), 'x, I+1)); \\ Kevin Ryde, Apr 08 2020
CROSSREFS
KEYWORD
sign,look
AUTHOR
Rémy Sigrist, Aug 13 2017
STATUS
approved