A136690 Final nonzero digit of n! in base 3.
1, 1, 2, 2, 2, 1, 2, 2, 1, 1, 1, 2, 2, 2, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2, 1, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 2, 2, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2, 1, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 2, 2, 1, 2, 2, 1, 1, 1, 2, 2, 2, 1, 2, 2, 1, 1, 1, 2, 2, 2, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2
Offset: 0
Examples
6! = 720 decimal = 222200 ternary, so a(6) = 2.
Links
Crossrefs
Programs
-
Mathematica
f[n_] := Mod[6 Times @@ (Rest[ FoldList[{1 + #1[[1]], #2! 2^(#1[[1]] #2)} &, {0, 0}, Reverse[ IntegerDigits[n, 3]]]]), 10][[2]]; # /. {0 -> 1} & /@ Mod[Table[f@n, {n, 0, 104}], 3] (* Robert G. Wilson v, Apr 17 2010 *) fnzd[n_]:=Module[{sidn3=Split[IntegerDigits[n!,3]]},If[MemberQ[ Last[ sidn3],0], sidn3[[-2,1]], sidn3[[-1,1]]]]; Array[fnzd,110,0] (* Harvey P. Dale, May 03 2018 *) -
PARI
a(n) = vecsum([bittest(220,b) |b<-digits(n,9)])%2 + 1; \\ Kevin Ryde, Dec 03 2022
Formula
From David Radcliffe, Sep 03 2021: (Start)
a(n) = (n! / A060828(n)) mod 3;
a(n) = 1 + (A189672(n) mod 2);
a(6*n) = a(6*n+1) = a(2*n);
a(6*n+2) = 3 - a(2*n);
a(6*n+3) = a(6*n+4) = 3 - a(2*n+1);
a(6*n+5) = a(2*n+1).
(End)
From Kevin Ryde, Dec 03 2022: (Start)
a(n) = 1 if n written in base 9 has an even number of digits {2,3,4,6,7}; and otherwise a(n) = 2.
Fixed point of the morphism 1 -> 1,1,2,2,2,1,2,2,1; 2 -> 2,2,1,1,1,2,1,1,2; starting from 1.
(End)
a(n) = A212307(n) mod 3. - Ridouane Oudra, Sep 25 2024
Extensions
More terms from Robert G. Wilson v, Apr 17 2010
Comments