A307096 Positive integers m such that for any positive integer k the last k bits of the binary expansion of m is not a multiple of 3.
1, 5, 13, 17, 29, 37, 49, 61, 65, 77, 101, 113, 125, 133, 145, 157, 193, 205, 229, 241, 253, 257, 269, 293, 305, 317, 389, 401, 413, 449, 461, 485, 497, 509, 517, 529, 541, 577, 589, 613, 625, 637, 769, 781, 805, 817, 829, 901, 913, 925, 961, 973, 997, 1009
Offset: 1
Examples
29 is 11101_2 and none of 11101_2, 1101_2, 101_2, 1_2 are divisible by 3.
Programs
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Maple
f := n-> if(n != 0, add(2^(k-1)*`if`((n mod 2^k) mod 3 = 0, 1, 0), k = 1 .. ceil(log(n)/log(2))), 0); ker := []; for n from 1 to 1024 do if f(n) = 0 then ker := [op(ker), n] end if end do; ker; # Alternative: A1:= {1}: A2:= {}: for d from 1 to 12 do if d::odd then A1:= A1 union map(`+`,A2,2^d) else A2:= A2 union map(`+`,A1,2^d) fi od: sort(convert(A1 union A2,list)); # Robert Israel, Apr 25 2019 -
Mathematica
Select[Range[10^3], Function[s, NoneTrue[Array[FromDigits[Take[s, -#], 2] &, Length@ s], Mod[#, 3] == 0 &]]@ IntegerDigits[#, 2] &] (* Michael De Vlieger, Mar 24 2019 *)
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PARI
isok(n) = {if (n % 3, my(b=binary(n)); for (k=1, #b-1, b[k] = 0; if ((fromdigits(b, 2) % 3) == 0, return (0));); return (1);); return (0);} \\ Michel Marcus, Apr 24 2019
Comments