

A178757


Smallest k such that k*n has an odd number of 1's in its base2 expansion.


4



1, 1, 7, 1, 5, 7, 1, 1, 9, 5, 1, 7, 1, 1, 19, 1, 17, 9, 1, 5, 1, 1, 3, 7, 1, 1, 3, 1, 3, 19, 1, 1, 33, 17, 1, 9, 1, 1, 3, 5, 1, 1, 7, 1, 9, 3, 1, 7, 1, 1, 7, 1, 5, 3, 1, 1, 3, 3, 1, 19, 1, 1, 67, 1, 65, 33, 1, 17, 1, 1, 3, 9, 1, 1, 5, 1, 5, 3, 1, 5, 1, 1, 5, 1, 5, 7, 1, 1, 5, 9, 1, 3, 1, 1, 3, 7, 1, 1, 11, 1
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OFFSET

1,3


COMMENTS

a(n) is the least integer k such that t(k*n) = 1, where t(i) is the ThueMorse sequence.
k must always be odd.
First occurrence of odd ks: 1, 23, 5, 3, 9, 99, 325, 315, 17, 15, 3021, 195, 189, 645, 2313, 2295, 33, 7695, 1785, 1799, 105, 387, 23529, 5643, ..., .
Records occur at n: 1, 3, 9, 15, 33, 63, 129, 255, 513, 1023, 2049, 4095, 8193, 16383, 32769, 65535, 131073, 262143, ..., . (End)
In their paper Morgenbesser et al. prove a(n) <= n+4. They also prove that, for all n>=1, it is possible to find a k such that A010060(k*n) = 1, with k <= n+4 and Hamming weight of k <= 3. That sequence differs from the current sequence at n=195, 315, 387, 390, 630, 645, 765, 771, 774, 780, ...  Michel Marcus, Jan 24 2016


LINKS



EXAMPLE

For n = 3 the sequence has value 7, since 21 is 10101 in base 2, with an odd number of 1's, and no smaller multiple works.


MATHEMATICA

f[n_] := Block[{k = 1}, While[ EvenQ@ DigitCount[k*n, 2, 1], k++ ]; k]; Array[f, 105] (* Robert G. Wilson v, Sep 29 2010 *)


PROG

(PARI) a(n) = {my(k = 1); while (hammingweight(k*n) % 2 != 1, k++); k; } \\ Michel Marcus, Jan 23 2016
(Python)
def a(n):
k = 1
while not bin(k*n).count("1")%2 == 1: k += 1
return k


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



