A157409 - OEIS

%I #9 May 09 2024 07:44:57

%S 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,

%T 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

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

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

%C 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.

%C (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).

%H Peter Luschny, <a href="http://www.luschny.de/math/vermutung.html">An arithmetic conjecture.</a>

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

%p a := proc(m) local l,i,u,A; A := convert(2^m, base, 3); u := 0;

%p for i from 0 to nops(A)-1 do if A[i+1] = 1 then u := u + 1 ;

%p elif A[i+1] = 0 then if type(u,odd) then RETURN(i) fi fi od;

%p 0 end: seq(a(i),i=0..62);

%t 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 *)

%o (PARI) a(n) = my(v2=2^n); for(k=1, logint(v2, 3)-1, (3==v2\(3^k)%6) && return(k)); 0; \\ _Ruud H.G. van Tol_, May 09 2024

%K easy,nonn

%O 0,6

%A _Peter Luschny_, Mar 06 2009