A234022 a(n) = A000120(A193231(n)); number of 1-bits in blue code for n.

0, 1, 2, 1, 2, 1, 2, 3, 4, 3, 2, 3, 2, 3, 2, 1, 2, 1, 2, 3, 2, 3, 4, 3, 4, 5, 4, 3, 4, 3, 2, 3, 4, 3, 2, 3, 4, 5, 4, 3, 4, 5, 6, 5, 4, 3, 4, 5, 2, 3, 2, 1, 4, 3, 2, 3, 4, 3, 4, 5, 2, 3, 4, 3, 4, 3, 4, 5, 2, 3, 4, 3, 4, 5, 4, 3, 6, 5, 4, 5, 2, 3, 4, 3, 2, 1, 2

Offset: 0

Author

Antti Karttunen, Dec 28 2013

Keywords

• nonn

Links

• Antti Karttunen, Table of n, a(n) for n = 0..8191
• Joerg Arndt, Matters Computational (The Fxtbook), section 1.19 "Invertible transforms on words", pp. 49--55. [Cf. especially pages 50 & 51].

Crossrefs

A234023 gives the positions where abs(a(n)-a(n+1)) > 1.

Cf. A000120, A193231.

Programs

Python

def a065621(n): return n^(2*(n - (n&-n)))
def a048724(n): return n^(2*n)
l=[0, 1]
z=[0, 1]
for n in range(2, 101):
    if n%2==0: l.append(a048724(l[n//2]))
    else: l.append(a065621(1 + l[(n - 1)//2]))
    z.append(bin(l[-1])[2:].count("1"))
print(z) # Indranil Ghosh, Jun 05 2017

Scheme

(define (A234022 n) (A000120 (A193231 n)))

Formula

a(n) = A000120(A193231(n)).

A000035(a(n)) = A000035(n) = (n mod 2) for all n. [Even terms occur only on even indices and odd terms only on odd indices, respectively]