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).
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, -1, -2, -3, -4, -4, -4, -4, -3, -3, -3, -2, -3, -4, -4, -4, -4, -3, -2, -1, -1, 4, 4, 3, 4, 5, 5, 5, 5, 4, 3, 2, 2, 5, 6, 6, 4, 4, 5, 6, 7, 7, 7, 7, 6, 6, 6, 5, 6
Offset: 0
Links
- Rémy Sigrist, Table of n, a(n) for n = 0..12287
Crossrefs
Cf. A318722.
Programs
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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)
Comments