A341054 For any number n with balanced ternary expansion (d_1, ..., d_k), the balanced ternary expansion of a(n), say (t_1, ..., t_k), satisfies t_m = d_1 + ... + d_m mod 3 for m = 1..k.
0, 1, 3, 4, 2, 8, 9, 10, 12, 13, 11, 7, 5, 6, 25, 23, 24, 26, 27, 28, 30, 31, 29, 35, 36, 37, 39, 40, 38, 34, 32, 33, 21, 22, 20, 16, 14, 15, 17, 18, 19, 75, 76, 74, 70, 68, 69, 71, 72, 73, 79, 77, 78, 80, 81, 82, 84, 85, 83, 89, 90, 91, 93, 94, 92, 88, 86, 87
Offset: 0
Examples
The first terms, alongside their balanced ternary expansion (with T's standing for -1's), are: n a(n) bter(n) bter(a(n)) -- ---- ------- ---------- 0 0 0 0 1 1 1 1 2 3 1T 10 3 4 10 11 4 2 11 1T 5 8 1TT 10T 6 9 1T0 100 7 10 1T1 101 8 12 10T 110 9 13 100 111 10 11 101 11T 11 7 11T 1T1 12 5 110 1TT 13 6 111 1T0 14 25 1TTT 10T1 15 23 1TT0 10TT 16 24 1TT1 10T0
Links
Programs
-
PARI
a(n) = { my (d=[], s=Mod(0, 3)); while (n, my (t=centerlift(Mod(n, 3))); n=(n-t)\3; d=concat(t, d)); for (k=1, #d, d[k] = centerlift(s+=d[k])); fromdigits(d, 3) }
Comments