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

0,3

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].

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]

Prog

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

Crossrefs

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

Cf. A000120, A193231.

Sequence in context: A344779 A317586 A303780 * A261273 A097454 A139803

Adjacent sequences: A234019 A234020 A234021 * A234023 A234024 A234025

Keyword

nonn

Author

Antti Karttunen, Dec 28 2013

Status

approved