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A180938
Smallest k such that k*n has an even number of 1's in its base-2 expansion.
3
3, 3, 1, 3, 1, 1, 9, 3, 1, 1, 3, 1, 3, 9, 1, 3, 1, 1, 3, 1, 3, 3, 1, 1, 3, 3, 1, 9, 1, 1, 33, 3, 1, 1, 3, 1, 3, 3, 1, 1, 3, 3, 1, 3, 1, 1, 3, 1, 3, 3, 1, 3, 1, 1, 3, 9, 1, 1, 3, 1, 3, 33, 1, 3, 1, 1, 3, 1, 3, 3, 1, 1, 3, 3, 1, 3, 1, 1, 3, 1, 3, 3, 1, 3, 1, 1, 5, 3, 1, 1, 5, 1, 11, 3, 1, 1, 3, 3, 1, 3, 1, 1, 9, 3, 1
OFFSET
1,1
COMMENTS
From Robert G. Wilson v, Sep 29 2010: (Start)
k must always be odd.
First occurrence of odd k: 3, 1, 87, 109, 7, 93, 457, 1143, 5501, 7921, 889, 12775, 11753, 635, 111209, 6093, 31, 33823, 7665, ..., .
(End)
LINKS
EXAMPLE
For n = 7, a(n) = 9, since the smallest multiple of 7 with an even number of 1's in its base-2 expansion is 9*7 = 63.
MATHEMATICA
a[n_] := Block[{k = 1}, While[OddQ@ DigitCount[k*n, 2, 1], k++ ]; k]; Array[a, 100] (* Robert G. Wilson v, Sep 29 2010 *)
PROG
(PARI) A180938(n) = my(k=1); while(hammingweight(k*n)%2, k += 2); k; \\ Antti Karttunen, Jul 09 2017
(Python)
def a(n):
k=1
while True:
if not bin(k*n)[2:].count('1')%2: return k
k+=1
print([a(n) for n in range(1, 61)]) # Indranil Ghosh, Jul 11 2017
CROSSREFS
Cf. A083420 (where records occur). - Alois P. Heinz, Oct 16 2011
Sequence in context: A215596 A268676 A354762 * A123561 A323626 A133695
KEYWORD
nonn,base
AUTHOR
Jeffrey Shallit, Sep 26 2010
EXTENSIONS
More terms from Robert G. Wilson v, Sep 29 2010
STATUS
approved