A318724 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 square of the modulus of g(n).

1, 2, 1, 2, 5, 4, 5, 8, 5, 4, 5, 2, 8, 13, 10, 5, 10, 13, 18, 25, 20, 17, 16, 9, 13, 18, 13, 10, 17, 20, 25, 32, 25, 20, 17, 10, 10, 13, 8, 9, 16, 17, 20, 25, 18, 13, 10, 5, 32, 41, 34, 25, 34, 41, 50, 61, 52, 45, 40, 29, 29, 40, 45, 20, 17, 26, 37, 50, 53, 58

Offset

0,2

Comments

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

Links

Rémy Sigrist, Table of n, a(n) for n = 0..12287

Formula

a(n) = A318722(n)^2 + A318723(n)^2.

If A048647(A042968(m)) = A042968(n), then a(m) = a(n).

Prog

(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)))); norm((z-1-I)/2)

Crossrefs

Cf. A042968, A048647, A318722, A318723.

Sequence in context: A352479 A084600 A216760 * A167482 A328601 A137327

Adjacent sequences: A318721 A318722 A318723 * A318725 A318726 A318727

Keyword

nonn,base,look

Author

Rémy Sigrist, Sep 02 2018

Status

approved