

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(n1)...a(2)a(1), read in base 6, is divisible by 3^n.


1



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
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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



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
(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



KEYWORD

nonn,base


AUTHOR



STATUS

approved



