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Let f(0) = 0 and f(t*4^k + u) = i^t * ((1+i) * 2^k - f(u)) for any t in {1, 2, 3} and k >= 0 and u such that 0 <= u < 4^k (i denoting the imaginary unit); for any n >= 0, let g(n) = (f(A042968(n)) - 1 - i) / 2; a(n) is the real part of g(n).
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%I #27 Sep 04 2018 03:35:42

%S -1,-1,0,-1,-2,-2,-2,-2,-1,0,1,1,-2,-3,-3,-1,-1,-2,-3,-4,-4,-4,-4,-3,

%T -3,-3,-2,-3,-4,-4,-4,-4,-3,-2,-1,-1,1,2,2,0,0,1,2,3,3,3,3,2,-4,-5,-5,

%U -3,-3,-4,-5,-6,-6,-6,-6,-5,-2,-2,-3,-2,-1,-1,-1,-1,-2

%N Let f(0) = 0 and f(t*4^k + u) = i^t * ((1+i) * 2^k - f(u)) for any t in {1, 2, 3} and k >= 0 and u such that 0 <= u < 4^k (i denoting the imaginary unit); for any n >= 0, let g(n) = (f(A042968(n)) - 1 - i) / 2; a(n) is the real part of g(n).

%C See A318723 for the imaginary part of g.

%C See A318724 for the square of the modulus of g.

%C This sequence can be computed by considering the base 4 representation of A042968, hence the keyword base.

%C The function g runs uniquely through the set of Gaussian integers z such that Re(z) < 0 or Im(z) < 0.

%C The function g is related to the numbering of the cells in a Chair tiling (see representation of g(n) in Links section).

%C This sequence has similarities with A316657.

%H Rémy Sigrist, <a href="/A318722/b318722.txt">Table of n, a(n) for n = 0..12287</a>

%H Rémy Sigrist, <a href="/A318722/a318722.png">Colored scatterplot of (a(n), A318723(n)) for n = 0..3*4^9-1</a> (where the hue is function of n)

%H Rémy Sigrist, <a href="/A318722/a318722_2.png">Colored scatterplot of (a(n), A318723(n)) for n = 0..3*4^9-1</a> (where the color is function of the sum of digits of A042968(n) in base 4)

%H Rémy Sigrist, <a href="/A318722/a318722_1.png">Representation of g(n) for n = 0..3*4^2-1 in the complex plane, alongside the base 4 representation of A042968(n)</a>

%H Tilings Encyclopedia, <a href="https://tilings.math.uni-bielefeld.de/substitution/chair/">Chair</a>

%F a(n) = A318723(n) iff the base 4 representation of A042968(n) contains only 0's and 2's.

%F If A048647(A042968(m)) = A042968(n), then a(m) = A318723(n) and A318723(m) = a(n).

%o (PARI) a(n) = my (d=Vecrev(digits(1+n+n\3,4)), z=0); for (k=1, #d, if (d[k], z = I^d[k] * (-z + (1+I) * 2^(k-1)))); real((z-1-I)/2)

%Y Cf. A042968, A048647, A318723, A318724.

%K sign,base

%O 0,5

%A _Rémy Sigrist_, Sep 02 2018