A265263 Change every other 1 bit in binary expansion of n to 0.

0, 1, 2, 2, 4, 4, 4, 5, 8, 8, 8, 9, 8, 9, 10, 10, 16, 16, 16, 17, 16, 17, 18, 18, 16, 17, 18, 18, 20, 20, 20, 21, 32, 32, 32, 33, 32, 33, 34, 34, 32, 33, 34, 34, 36, 36, 36, 37, 32, 33, 34, 34, 36, 36, 36, 37, 40, 40, 40, 41, 40, 41, 42, 42, 64, 64, 64, 65, 64, 65, 66, 66, 64, 65, 66, 66, 68, 68, 68, 69, 64, 65, 66, 66, 68, 68, 68, 69, 72, 72, 72

Offset: 0

Author

Jeffrey Shallit, Dec 06 2015

Keywords

• nonn
• base

Comments

a(n) obeys the recurrences a(2n) = a(n), a(4n+1) = 2a(n) + a(2n+1), a(8n+3) = 4a(n) + a(4n+3), and a(8n+7) = -2a(n) + a(2n+1) + 2a(4n+3).

Examples

			a(15) = 10, because if we delete the 2nd and 4th bits of 1111, we get 1010.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..8192

Programs

PARI

a(n) = my (b=binary(n), o=0); for (k=1, #b, if (b[k], b[k]=o++%2)); fromdigits(b, 2) \\ Rémy Sigrist, Feb 19 2020

Python

def a(n):
    switch, b, out = False, bin(n)[2:], ""
    for bi in b:
        if bi == "0": out += "0"
        else: out += "0" if switch else "1"; switch = not switch
    return int(out, 2)
print([a(n) for n in range(91)]) # Michael S. Branicky, Jul 15 2022