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A265263
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Change every other 1 bit in binary expansion of n to 0.
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1
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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
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internal format)
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OFFSET
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0,3
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COMMENTS
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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).
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LINKS
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EXAMPLE
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a(15) = 10, because if we delete the 2nd and 4th bits of 1111, we get 1010.
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PROG
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(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)
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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