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

0,6

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

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.

Prog

(Scheme, three different implementations)
(define (A246159 n) (* -1 (A010059 n) (A065620 n)))
(define (A246159 n) (* (/ 1 2) (A010059 n) (A006068 n)))
(define (A246159 n) (* (A010059 n) (A006068 (A245710 n))))
(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

Crossrefs

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

Sequence in context: A373418 A293500 A240659 * A059033 A133209 A144553

Adjacent sequences: A246156 A246157 A246158 * A246160 A246161 A246162

Keyword

nonn

Author

Antti Karttunen, Aug 18 2014

Status

approved