A157409 Minimum of { k > 0 : [2^n / 3^k] mod 6 = 3 } if such k exists, 0 otherwise.

0, 0, 0, 0, 0, 2, 1, 0, 3, 0, 0, 3, 1, 3, 0, 0, 2, 0, 1, 5, 4, 12, 7, 2, 1, 11, 0, 15, 10, 4, 1, 4, 10, 3, 2, 9, 1, 4, 11, 15, 10, 2, 1, 7, 4, 7, 3, 7, 1, 21, 12, 4, 2, 4, 1, 6, 5, 8, 7, 2, 1, 4, 3

Offset

0,6

Comments

It is conjectured that a(n) = 0 only for n in {0,1,2,3,4,7,9,10,14,15,17,26}.

Let N, K defined by: K = max {a(n) for all n <= N}. The following pairs (N : K) for N > 26 mark points where K increases.

(27 : 15), (49 : 21), (110 : 29), (118 : 34), (165 : 58), (2769 : 61), (2837 : 65), (3661 : 70), (14354 : 74), (59913 : 103), (1786453 : 112), (2702893 : 117), (2712849 : 121).

Links

Table of n, a(n) for n=0..62.

Peter Luschny, An arithmetic conjecture.

Example

a(20) = 4 because MOD([2^20 / 3^4], 6) = 3.

Maple

a := proc(m) local l, i, u, A; A := convert(2^m, base, 3); u := 0;
for i from 0 to nops(A)-1 do if A[i+1] = 1 then u := u + 1 ;
elif A[i+1] = 0 then if type(u, odd) then RETURN(i) fi fi od;
0 end: seq(a(i), i=0..62);

Mathematica

a[n_] := Module[{k}, For[k = 1, k <= n, k++, If[Mod[Floor[2^n/3^k], 6] == 3, Return[k]]]; 0]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Jun 17 2019 *)

Crossrefs

Sequence in context: A194812 A305320 A159813 * A245960 A340867 A178616

Adjacent sequences: A157406 A157407 A157408 * A157410 A157411 A157412

Keyword

easy,nonn

Author

Peter Luschny, Mar 06 2009

Status

approved