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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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`%I #7 Jul 13 2022 14:43:22
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`%S 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,
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`%T 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,
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`%U 7,14,2,2,1,2,1,2,1,16,7,2,1,8,1,2,2,8
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`%N 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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`%C 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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`%F a(n) = 1 iff n belongs to A174658.
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`%e For n = 5:
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`%e - the first multiple of 5 (alongside their balanced ternary expansions) are:
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`%e k k*5 bter(k*5) #1 #T
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`%e - --- --------- -- --
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`%e 1 5 1TT 1 2
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`%e 2 10 101 2 0
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`%e 3 15 1TT0 1 2
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`%e 4 20 1T1T 2 2
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`%e - negative and positive trits are first balanced for k = 4,
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`%e - so a(5) = 4.
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`%o (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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`%Y See A351599 for a similar sequence.
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`%Y Cf. A065363, A174658, A355640.
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`%K nonn,base
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`%O 0,2
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`%A _Rémy Sigrist_, Jul 11 2022
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