A246160 Inverse function to the injection A065621.

0, 1, 2, 0, 4, 0, 0, 3, 8, 0, 0, 7, 0, 5, 6, 0, 16, 0, 0, 15, 0, 13, 14, 0, 0, 9, 10, 0, 12, 0, 0, 11, 32, 0, 0, 31, 0, 29, 30, 0, 0, 25, 26, 0, 28, 0, 0, 27, 0, 17, 18, 0, 20, 0, 0, 19, 24, 0, 0, 23, 0, 21, 22, 0, 64, 0, 0, 63, 0, 61, 62, 0, 0, 57, 58, 0, 60, 0, 0, 59, 0, 49, 50, 0, 52

Offset: 0

Author

Antti Karttunen, Aug 18 2014

Keywords

• nonn

Comments

Sequence has nonzero values a(n) = k at those points n for which A065621(k) = n and zeros at those positions n which are not present in A065621.

Equally, sequence is obtained when the negative terms of A065620 are replaced with zeros

Links

• Antti Karttunen, Table of n, a(n) for n = 0..8192

Crossrefs

Cf. A246159, A006068, A010060, A115384, A065620, A065621.

Programs

PARI

a065620(n) = if(n<3, n, if(n%2, -2*a065620((n - 1)/2) + 1, 2*a065620(n/2)));
a(n) = (hammingweight(n)%2)*a065620(n);
for(n=0, 100, print1(a(n),", ")) \\ Indranil Ghosh, Jun 07 2017

Python

def a065620(n): return n if n<3 else 2*a065620(n//2) if n%2==0 else -2*a065620((n - 1)//2) + 1
def a(n): return (bin(n)[2:].count("1")%2)*a065620(n)
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 07 2017

Formula

a(n) = A010060(n) * A065620(n).

a(n) = A246159(n) + A065620(n).

a(0) = 0, and for n >= 1, a(n) = A010060(n) * (1 + A006068(A115384(n)-1)).

For all n, a(A065621(n)) = n.