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 A157409 Minimum of { k > 0 : [2^n / 3^k] mod 6 = 3 } if such k exists, 0 otherwise. 0
 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 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified February 26 11:42 EST 2024. Contains 370352 sequences. (Running on oeis4.)