A342932 The unique sequence {a(1), a(2), a(3), a(4), ...} of digits 1, 2, or 3 such that the number a(n)a(n-1)...a(2)a(1), read in base 6, is divisible by 3^n.

3, 1, 2, 3, 3, 1, 1, 3, 1, 2, 2, 1, 1, 3, 1, 3, 1, 3, 3, 3, 2, 3, 3, 2, 1, 1, 2, 1, 3, 1, 2, 1, 1, 3, 3, 2, 2, 3, 1, 3, 2, 2, 2, 3, 3, 1, 2, 2, 2, 2, 2, 1, 3, 2, 2, 3, 2, 2, 1, 2, 1, 3, 2, 2, 3, 1, 1, 1, 1, 1, 3, 2, 2, 2, 3, 3, 2, 1, 3, 1, 1, 2, 2, 3, 1, 3, 2, 3, 2, 3, 1, 1, 3, 1, 2, 3, 3, 2, 3, 2, 3, 1, 1, 2, 3

Offset

1,1

Comments

The distribution seems to be uniform but random (empirical observation).

To prove that such a digit sequence exists and is unique is a good (but uncommon) example of a proof by induction.

Links

Table of n, a(n) for n=1..105.

Rémy Sigrist, Colored scatterplot of (n, 3*o(n)-n) for = 1..1000000 (where o corresponds to the ordinal transform of the sequence)

Eugen Ionascu, Mathematica File

Example

3 is divisible by 3^1;
13_6 = 1*6 + 3 = 9, which is divisible by 3^2,
213_6 = 2*6^2 + 1*6 + 3 = 81, which is divisible by 3^3.

Mathematica

nd[n_] := Module[{k, i, s, ss, L, a}, L = Array[f, n]; f[1] = 3;
  Do[s = Sum[6^(k - 1)*f[k], {k, 1, i - 1}];
   ss = Mod[2^(i - 1)*s/3^(i - 1), 3];
   If[ss == 0, f[i] = 3, If[ss == 1, f[i] = 2, f[i] = 1]], {i, 2, n}];
  s = Sum[6^(k - 1)*f[k], {k, 1, n}];
  {L, s/3^n}]

Prog

(Python)
n, div, divnum = 0, 1, 0
while n < 87:
    div, a = 3*div, 1
    while (a*6**n+divnum)%div != 0:
        a = a+1
    divnum, n = divnum+a*6**n, n+1
    print(a, end=', ') # A.H.M. Smeets, Apr 13 2021
(PARI) { q=0; t=1; for (n=1, 105, print1 (d=[3, 1, 2][1+lift(-q/Mod(t, 3))]", "); q=(t*d+q)/3; t*=2) } \\ Rémy Sigrist, Apr 15 2021

Crossrefs

Cf. A023396, A053312, A126933.

Sequence in context: A029279 A260869 A274514 * A026181 A327239 A322849

Adjacent sequences: A342929 A342930 A342931 * A342933 A342934 A342935

Keyword

nonn,base

Author

Eugen Ionascu, Mar 29 2021

Status

approved