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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 imaginary part of g(n).
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%I #9 Sep 04 2018 03:05:59

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

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

%U 5,5,5,5,4,3,2,2,5,6,6,4,4,5,6,7,7,7,7,6,6,6,5,6

%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 imaginary part of g(n).

%C See A318722 for the real part of g and additional comments.

%H Rémy Sigrist, <a href="/A318723/b318723.txt">Table of n, a(n) for n = 0..12287</a>

%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)))); imag((z-1-I)/2)

%Y Cf. A318722.

%K sign,base

%O 0,8

%A _Rémy Sigrist_, Sep 02 2018