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Variant of the Chandra-sutra (A014701) using 3 instead of 2, and a mod argument using residues 1 and 2.
0

%I #18 Mar 24 2017 00:47:56

%S 0,1,1,2,2,2,3,3,2,3,3,3,4,4,3,4,4,3,4,4,4,5,5,4,5,5,3,4,4,4,5,5,4,5,

%T 5,4,5,5,5,6,6,5,6,6,4,5,5,5,6,6,5,6,6,4,5,5,5,6,6,5,6,6,5,6,6,6,7,7,

%U 6,7,7,5,6,6,6,7,7,6,7,7,4,5,5

%N Variant of the Chandra-sutra (A014701) using 3 instead of 2, and a mod argument using residues 1 and 2.

%C x :> x/3 if x == 0 mod 3, x :> x - x mod 3 otherwise. This sequence gives the number of steps needed to reach 0 or 1.

%C In base 3, number of 0's + (number of other digits - 1) * 2 + (1 if leading digit is 2).

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Ternary.html">Ternary.</a>

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Ternary_numeral_system">Ternary numeral system</a>

%e 8 -> 6 -> 2 -> 0.

%e 28 -> 27 -> 9 -> 3 -> 1.

%e In base 3 the process is more obvious, e.g., 19 is 201 and the sequence is 201 -> 200 -> 20 -> 2 ->0, so a(19)=4. The number of zeros is 1, other digits is 2 and the leading digit is a 2, so we also have a(19) = 1 + (2-1)*2 + 1 = 4.

%o (JavaScript)

%o for (i=1;i<300;i++) {

%o c=0;

%o n=i;

%o while (n>1) {c++;m=n%3;if (m==0) n/=3; else n-=m;}

%o document.write(c+", ");

%o }

%Y Cf. A014701.

%K nonn,base

%O 1,4

%A _Jon Perry_, Nov 26 2013