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A290536
Let S be the sequence generated by these rules: 0 is in S, and if z is in S, then z + 1 and z * (1+i) are in S (where i denotes the imaginary unit), and duplicates are deleted as they occur; a(n) = the real part of the n-th term of S.
4
0, 1, 2, 1, 3, 2, 2, 0, 4, 3, 3, 0, 3, 1, 1, -2, 5, 4, 4, 0, 4, 1, 1, -4, 4, 2, 2, -2, -1, -1, -4, 6, 5, 5, 0, 5, 1, 1, -6, 5, 2, 2, -4, -3, -3, -8, 5, 3, 3, -2, -1, -1, -6, 0, -4, -3, -3, -4, 7, 6, 6, 0, 6, 1, 1, -8, 6, 2, 2, -6, -5, -5, -12, 6, 3, 3, -4, -3
OFFSET
1,3
COMMENTS
See A290537 for the imaginary part of the n-th term of S.
See A290538 for the square of the norm of the n-th term of S.
The representation of the first terms of S in the complex plane has nice fractal features (see also Links section).
The sequence S is a "complex" variant of A232559.
The sequence S is a permutation of the Gaussian integers (Z[i]):
- let u be the function defined over Z[i] by z -> z+1,
- let v be the function defined over Z[i] by z -> z*(1+i),
- for m, n, o, p and q >= 0,
let f(m,n,o,p,q) = u^m(v(u^n(v(u^o(v(v(u^p(v(v(u^q(0)))))))))))
(where w^k denotes the k-th iterate of w),
- f(m,0,0,0,0) = m, and any nonnegative integer x can be represented in this way for some m >= 0,
- f(m,n,0,0,0) = m+n + n*i, and any Gaussian integer x+y*i with 0 <= x and 0 <= y <= x can be represented in this way for some m and n >= 0,
- f(m,n,o,0,0) = f(m,n,0,0,0) + 2*o*i, and any Gaussian integer x+y*i with 0 < x and 0 <= y can be represented in this way for some m, n and o >= 0,
- f(m,n,o,p,0) = f(m,n,o,0,0) - 4*p, and any Gaussian integer x+y*i with 0 <= y can be represented in this way for some m, n, o and p >= 0,
- f(m,n,o,p,q) = f(m,n,o,p,0) - 8*q*i, and any Gaussian integer x+y*i can be represented in this way for some m, n, o, p and q >= 0,
- in other words, any Gaussian integer can be reached from 0 after a finite number of steps chosen in { u, v }, QED.
EXAMPLE
S(1) = 0 by definition; so a(1) = 0.
S(1)+1 = 1 has not yet occurred; so S(2) = 1 and a(2) = 1.
S(1)*(i+i) = 0 has already occurred.
S(2)+1 = 2 has not yet occurred; so S(3) = 2 and a(3) = 2.
S(2)*(1+i) = 1+i has not yet occurred; so S(4) = 1+i and a(4) = 1.
S(3)+1 = 3 has not yet occurred; so S(5) = 3 and a(5) = 3.
PROG
(PARI) See Links section.
CROSSREFS
KEYWORD
sign,look
AUTHOR
Rémy Sigrist, Aug 05 2017
STATUS
approved