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A343030
Number of 1-bits in the binary expansion of n which have an odd number of 0-bits at less significant bit positions.
5
0, 0, 1, 0, 0, 1, 2, 0, 1, 0, 1, 1, 0, 2, 3, 0, 0, 1, 2, 0, 1, 1, 2, 1, 2, 0, 1, 2, 0, 3, 4, 0, 1, 0, 1, 1, 0, 2, 3, 0, 1, 1, 2, 1, 1, 2, 3, 1, 0, 2, 3, 0, 2, 1, 2, 2, 3, 0, 1, 3, 0, 4, 5, 0, 0, 1, 2, 0, 1, 1, 2, 1, 2, 0, 1, 2, 0, 3, 4, 0, 1, 1, 2, 1, 1, 2, 3
OFFSET
0,7
COMMENTS
See A343029 for further notes.
FORMULA
a(n) = A343029(n) - A004718(n).
a(n) = A000120(n) - A343029(n), where A000120 is the number of 1-bits in n (binary weight).
a(2*n) = A000120(n) - a(n).
a(2*n+1) = a(n).
G.f. satisfies g(x) = (x-1)*g(x^2) + A000120(x^2).
G.f.: (1/2)* Sum_{k>=0} x^(2^k)*( (1-x^(2^k))/(1-x) - Prod_{j=0..k-1} x^(2^j)-1 )/( 1-x^(2*2^k ) ).
a(2(2^n - 1)) = n. - Michael S. Branicky, Apr 03 2021
EXAMPLE
n = 628 = binary 1001110100
^ ^^^ a(n) = 4
PROG
(PARI) a(n) = my(t=0, ret=0); for(i=0, if(n, logint(n, 2)), if(bittest(n, i), ret+=t, t=!t)); ret;
(Python)
def a(n):
b = bin(n)[2:]
return sum(bi=='1' and b[i:].count('0')%2==1 for i, bi in enumerate(b))
print([a(n) for n in range(87)]) # Michael S. Branicky, Apr 03 2021
CROSSREFS
Cf. A343029, A004718, A000120, A000918 (indices of new highs).
Sequence in context: A136745 A214157 A246720 * A246690 A317748 A090465
KEYWORD
nonn,easy
AUTHOR
Kevin Ryde, Apr 03 2021
STATUS
approved