A318723 - cp's OEIS frontend

A318723 Let f(0) = 0 and f(t4^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).

%I A318723 #9 Sep 04 2018 03:05:59
%S A318723 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 A318723 -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 A318723 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 A318723 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 A318723 See A318722 for the real part of g and additional comments.
%H A318723 Rémy Sigrist, <a href="/A318723/b318723.txt">Table of n, a(n) for n = 0..12287</a>
%o A318723 (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 A318723 Cf. A318722.
%K A318723 sign,base
%O A318723 0,8
%A A318723 _Rémy Sigrist_, Sep 02 2018

cp's OEIS Frontend

A318723 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).

A318723 Let f(0) = 0 and f(t4^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).