A162721 A positive integer k is included if, when k is represented in binary, it contains the binary representation of every distinct prime dividing k as substrings, without overlapping of the substrings.

2, 3, 4, 5, 7, 8, 11, 13, 16, 17, 19, 23, 27, 29, 31, 32, 37, 41, 43, 47, 53, 54, 59, 61, 63, 64, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 108, 109, 113, 125, 126, 127, 128, 131, 137, 139, 149, 151, 157, 163, 167, 173, 175, 179, 181, 191, 193, 197, 199, 211, 216, 223, 227, 229, 233, 239, 241, 243, 245, 251, 252, 256

Offset: 1

Author

Leroy Quet, Jul 11 2009

Keywords

• base
• nonn

Comments

Includes all primes and powers of 2, but no even semiprimes except 4. Contains p^2 for primes p in A018826. - Robert Israel, Jan 10 2023

Contains no squarefree numbers except primes. - Robert Israel, Jan 12 2023

Examples

			20 in binary is 10100. The distinct primes dividing 20 are 2 and 5, which are 10 and 101 in binary. Both 10 and 101 occur in 10100, but with overlapping. So 20 is not in this sequence. However, 54 in binary is 110110. 54 is divisible by 2 and 3, which are 10 and 11 in binary. Both 10 and 11 occur in 110110 without overlapping. (1{10}{11}0.) So 54 is in this sequence.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A018826, A162722.

Programs

Maple

# Requires Maple 2018 or later
satfilter:= proc(n) local n2, P, nP, X, P2, J, Cons, Clause, i,j,k, Ck;
  n2:= convert(n,base,2);
  P:= numtheory:-factorset(n);
  nP:= nops(P);
  P2:= map(convert,P,base,2);
  J:= map(t -> map(s -> [$s..s+nops(t)-1],select(i -> n2[i..i+nops(t)-1] = t, [$1..nops(n2)+1-nops(t)])), P2);
  if member([],J) then return false fi;
  Cons:= true;
  for i from 1 to nops(J) do
    Clause:= X[i,J[i][1]];
    for j from 2 to nops(J[i]) do
      Clause:= Clause &or X[i,J[i][j]]
    od;
    Cons:= Cons &and Clause;
  od;
  for k from 1 to nops(n2) do
    Ck:= {};
    for i from 1 to nP do
      for j from 1 to nops(J[i]) do if member(k,J[i,j]) then Ck:= Ck union {X[i,J[i,j]]} fi od od;
    if nops(Ck) >= 2 then for i from 2 to nops(Ck) do for j from 1 to i-1 do Cons:= Cons &and (¬(Ck[i]) &or ¬(Ck[j])) od od fi;
  od:
  Logic:-Satisfiable(Cons);
end proc:
select(satfilter, [$2..1000]); # Robert Israel, Jan 10 2023

Extensions

More terms from Sean A. Irvine, Dec 09 2010