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The run lengths transform of the balanced ternary expansion of n corresponds to the run lengths transform of the binary expansion of a(n).
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%I #11 Mar 17 2024 12:51:14

%S 0,1,2,2,3,4,5,5,5,4,5,6,6,7,8,9,9,10,11,10,10,10,11,11,10,10,9,8,9,

%T 10,10,11,12,13,13,13,12,13,14,14,15,16,17,17,18,19,18,18,18,19,20,21,

%U 21,22,23,22,21,21,20,20,21,21,21,20,21,22,22,23,23,22

%N The run lengths transform of the balanced ternary expansion of n corresponds to the run lengths transform of the binary expansion of a(n).

%C For any v > 0, the value v appears A225081(v-1) times in the sequence.

%H Rémy Sigrist, <a href="/A371263/b371263.txt">Table of n, a(n) for n = 0..9841</a>

%F abs(a(n+1) - a(n)) <= 1.

%e The first terms, alongside the balanced ternary expansion of n and the binary expansion of a(n), are:

%e n a(n) bter(n) bin(a(n))

%e -- ---- ------- ---------

%e 0 0 0 0

%e 1 1 1 1

%e 2 2 1T 10

%e 3 2 10 10

%e 4 3 11 11

%e 5 4 1TT 100

%e 6 5 1T0 101

%e 7 5 1T1 101

%e 8 5 10T 101

%e 9 4 100 100

%e 10 5 101 101

%e 11 6 11T 110

%e 12 6 110 110

%e 13 7 111 111

%e 14 8 1TTT 1000

%e 15 9 1TT0 1001

%o (PARI) a(n) = { my (r = [], d, l, v = 0); while (n, d = centerlift(Mod(n, 3)); l = 0; while (centerlift(Mod(n, 3))==d, n = (n-d)/3; l++;); r = concat(l, r);); for (k = 1, #r, v = (v+k%2)*2^r[k]-k%2); v }

%Y See A371256 for a similar sequence.

%Y Cf. A225081, A371265.

%K nonn,base

%O 0,3

%A _Rémy Sigrist_, Mar 16 2024