A246159 Inverse function to the injection A048724.

0, 0, 0, 1, 0, 3, 2, 0, 0, 7, 6, 0, 4, 0, 0, 5, 0, 15, 14, 0, 12, 0, 0, 13, 8, 0, 0, 9, 0, 11, 10, 0, 0, 31, 30, 0, 28, 0, 0, 29, 24, 0, 0, 25, 0, 27, 26, 0, 16, 0, 0, 17, 0, 19, 18, 0, 0, 23, 22, 0, 20, 0, 0, 21, 0, 63, 62, 0, 60, 0, 0, 61, 56, 0, 0, 57, 0, 59, 58, 0, 48, 0, 0, 49, 0, 51, 50, 0, 0, 55, 54, 0, 52, 0, 0, 53, 32

Offset: 0

Author

Antti Karttunen, Aug 18 2014

Keywords

• nonn

Comments

After a(0)=0, sequence has nonzero values a(n) = k at those positions n for which A048724(k) = n and zeros at those positions n which are not present in A048724.

Equally, sequence is obtained when the positive terms of A065620 are replaced with zeros and the sign of the negative terms is reversed.

Links

• Antti Karttunen, Table of n, a(n) for n = 0..8191

Crossrefs

Cf. A246160, A006068, A010059, A048724, A065620, A245710.

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==0)*a065620(n)
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 07 2017

Formula

a(n) = (1/2) * A010059(n) * A006068(n).

a(n) = -1 * A010059(n) * A065620(n).

a(n) = A246160(n) - A065620(n).

a(n) = A010059(n) * A006068(A245710(n)).

For all n, a(A048724(n)) = n.