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

0,3

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

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.

Prog

(Scheme, two different implementations)
(define (A246160 n) (* (A010060 n) (A065620 n)))
(define (A246160 n) (if (zero? n) n (* (A010060 n) (1+ (A006068 (-1+ (A115384 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)*a065620(n)
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 07 2017

Crossrefs

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

Sequence in context: A048298 A123565 A258701 * A081120 A200038 A249093

Adjacent sequences: A246157 A246158 A246159 * A246161 A246162 A246163

Keyword

nonn

Author

Antti Karttunen, Aug 18 2014

Status

approved