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A355639
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a(n) is the least k > 0 such that the balanced ternary expansion of k*n contains as many negative trits as positive trits.
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2
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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
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OFFSET
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0,2
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COMMENTS
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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.
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LINKS
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FORMULA
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EXAMPLE
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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.
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PROG
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(PARI) 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))) }
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CROSSREFS
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See A351599 for a similar sequence.
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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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