

A285099


a(n) is the zerobased index of the second least significant 1bit in the base2 representation of n, or 0 if there are fewer than two 1bits in n.


5



0, 0, 0, 1, 0, 2, 2, 1, 0, 3, 3, 1, 3, 2, 2, 1, 0, 4, 4, 1, 4, 2, 2, 1, 4, 3, 3, 1, 3, 2, 2, 1, 0, 5, 5, 1, 5, 2, 2, 1, 5, 3, 3, 1, 3, 2, 2, 1, 5, 4, 4, 1, 4, 2, 2, 1, 4, 3, 3, 1, 3, 2, 2, 1, 0, 6, 6, 1, 6, 2, 2, 1, 6, 3, 3, 1, 3, 2, 2, 1, 6, 4, 4, 1, 4, 2, 2, 1, 4, 3, 3, 1, 3, 2, 2, 1, 6, 5, 5, 1, 5, 2, 2, 1, 5, 3, 3, 1, 3, 2, 2, 1, 5, 4, 4, 1, 4, 2, 2, 1, 4
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OFFSET

0,6


LINKS



FORMULA

This is a 2regular sequence, satisfying the identities
a(4n) = a(n) + a(2n),
a(4n+2) = a(4n+1),
a(8n+1) = a(2n+1) + 2a(4n+1),
a(8n+3) = a(4n+3),
a(8n+5) = 2a(4n+3),
a(8n+7) = a(4n+3). (End)


EXAMPLE

For n = 3, "11" in binary, the second least significant 1bit (the second 1bit from the right) is at position 1 (when the rightmost position is position 0), thus a(3) = 1.
For n = 4, "100" in binary, there is just one 1bit present, thus a(4) = 0.
For n = 5, "101" in binary, the second 1bit from the right is at position 2, thus a(5) = 2.
For n = 25, "11001" in binary, the second 1bit from the right is at position 3, thus a(25) = 3.


MATHEMATICA

a007814[n_]:=IntegerExponent[n, 2]; a[n_]:=If[DigitCount[n, 2, 1]<2, 0, a007814[BitAnd[n, n  1]]]; Table[a[n], {n, 0, 150}] (* Indranil Ghosh, Apr 20 2017 *)


PROG

(Scheme) (define (A285099 n) (if (<= (A000120 n) 1) 0 (A007814 (A004198bi n ( n 1))))) ;; A004198bi implements bitwiseand.
(Python)
import math
def a007814(n): return int(math.log(n  (n & n  1), 2))
def a(n): return 0 if bin(n)[2:].count("1") < 2 else a007814(n & (n  1)) # Indranil Ghosh, Apr 20 2017


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



