A351599 -id:A351599 - cp's OEIS frontend

A143146 a(n) is the smallest positive multiple of n that has the same number of 0's as 1's in its binary representation.

2, 2, 9, 12, 10, 12, 35, 56, 9, 10, 44, 12, 52, 42, 135, 240, 153, 180, 38, 180, 42, 44, 184, 216, 50, 52, 135, 56, 232, 150, 527, 992, 165, 170, 35, 180, 37, 38, 156, 240, 41, 42, 172, 44, 135, 184, 141, 240, 49, 50, 153, 52, 212, 216, 165, 56, 228, 232, 177, 180

Offset: 1

Leroy Quet, Jul 27 2008

			For n = 7, checking: 7*1 = 7 = 111_2; 7*2 = 14 = 1110_2; 7*3 = 21 = 10101_2; 7*4 = 28 = 11100_2. All of these have two many 1's in binary. But 7*5 = 35 = 100011_2, which has both three 0's and three 1's. So a(7) = 35.

Michael S. Branicky, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)

Cf. A031443, A143147, A351599.

a:=proc(n) local b,k: b:=proc(m) convert(m,base,2) end proc: for k while add(b(k*n)[j],j=1..nops(b(k*n))) <> nops(b(k*n))-add(b(k*n)[j],j=1..nops(b(k*n))) do end do: k*n end proc: seq(a(n),n=1..60); # Emeric Deutsch, Aug 16 2008

spm[n_]:=Module[{k=1},While[DigitCount[k*n,2,0]!=DigitCount[k*n,2,1], k++]; k*n]; Array[spm,60] (* Harvey P. Dale, Apr 25 2014 *)

def a(n):
    m = n
    b = bin(m)[2:]
    while len(b) != 2*b.count("1"):
        m += n
        b = bin(m)[2:]
    return m
print([a(n) for n in range(1, 61)]) # Michael S. Branicky, May 15 2022

a(n) = n * A351599(n). - Rémy Sigrist, Jul 11 2022

More terms from Emeric Deutsch, Aug 16 2008

A355639 a(n) is the least k > 0 such that the balanced ternary expansion of k*n contains as many negative trits as positive trits.

Original entry on oeis.org

1, 2, 1, 2, 2, 4, 1, 8, 1, 2, 2, 14, 2, 2, 4, 4, 1, 8, 1, 14, 1, 8, 7, 2, 1, 16, 1, 2, 2, 8, 2, 2, 1, 14, 4, 2, 2, 2, 7, 2, 2, 4, 4, 2, 10, 4, 1, 4, 1, 2, 8, 8, 1, 8, 1, 8, 1, 14, 4, 4, 1, 8, 1, 8, 5, 2, 7, 14, 2, 2, 1, 2, 1, 2, 1, 16, 7, 2, 1, 8, 1, 2, 2, 8

Offset: 0

Rémy Sigrist, Jul 11 2022

The sequence is well defined: for n > 0, by the pigeonhole principle, there are necessarily two distinct integers i and j (say with i > j) such that 3^i == 3^j (mod n); the value 3^i - 3^j is a positive multiple of n containing exactly one positive trit and one negative trit, so a(n) <= (3^i - 3^j) / n.

			For n = 5:
- the first multiple of 5 (alongside their balanced ternary expansions) are:
      k  k*5  bter(k*5)  #1  #T
      -  ---  ---------  --  --
      1    5        1TT   1   2
      2   10        101   2   0
      3   15       1TT0   1   2
      4   20       1T1T   2   2
- negative and positive trits are first balanced for k = 4,
- so a(5) = 4.

See A351599 for a similar sequence.

Cf. A065363, A174658, A355640.

a(n) = { for (k=1, oo, my (m=k*n, s=0, d); while (m, m=(m-d=[0,1,-1][1+m%3])/3; s+=d); if (s==0, return (k))) }

a(n) = 1 iff n belongs to A174658.

cp's OEIS Frontend

A143146 a(n) is the smallest positive multiple of n that has the same number of 0's as 1's in its binary representation.

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