A385462 Numbers t which have a proper divisor d_i(t) such that (d_i(t) + sigma(t))/t is an integer k.

2, 4, 8, 10, 16, 24, 32, 44, 60, 64, 84, 128, 136, 152, 168, 184, 252, 256, 270, 336, 512, 630, 752, 756, 792, 864, 884, 924, 936, 1024, 1140, 1170, 1488, 1638, 2048, 2144, 2268, 2272, 2528, 2808, 2970, 3672, 4096, 4320, 4464, 4680, 5148, 5472, 6804, 7308, 7644, 8192, 8384

Offset: 1

Author

Lechoslaw Ratajczak, Jun 29 2025

Keywords

• nonn

Comments

Consecutive elements of this sequence for which k = 2 are consecutive deficient-perfect numbers (A271816) > 1.

Consecutive elements of this sequence for which k = 3 are consecutive non-perfect elements of A364977.

Let b_k(m) be the number of elements of this sequence with the same k and <= m.

--------------------------------------------

m | b_2(m) | b_3(m) | b_4(m) | b_5(m) |

--------------------------------------------

10^3 | 16 | 13 | - | - |

10^4 | 24 | 31 | 2 | - |

10^5 | 37 | 62 | 5 | - |

10^6 | 54 | 107 | 19 | - |

10^7 | 73 | 175 | 43 | 1 |

10^8 | 98 | 254 | 80 | 3 |

10^9 | 128 | 357 | 141 | 13 |

--------------------------------------------

Are there any odd terms in this sequence for which k > 2? If they exist, they are > 10^9.

Contains 2^k * (2^(k+1) + 2^j - 1) if 0 <= j <= k and 2^(k+1) + 2^j - 1 is prime. - Robert Israel, Jun 30 2025

Examples

			4 is in this sequence because sigma(4) + d_1(4) = 7 + 1 = 8 and 8/4 = 2.
24 is in this sequence because sigma(24) + d_7(24) = 60 + 12 = 72 and 72/24 = 3.
4320 is in this sequence because sigma(4320) + d_47(4320) = 15120 + 2160 = 17280 and 17280/4320 = 4.
		

Crossrefs

Cf. A000005, A000203, A007691, A054027, A271816, A364977.

Programs

Maple

filter:= proc(n) local s;
  s:= - numtheory:-sigma(n) mod n;
  ormap(d -> d mod n = s, numtheory:-divisors(n) minus {n})
end proc:
select(filter, [$1..10^4]); # Robert Israel, Jun 30 2025

Mathematica

Select[Range[8384],AnyTrue[(Drop[Divisors[#],-1]+DivisorSigma[1,#])/#,IntegerQ]&] (* James C. McMahon, Jul 05 2025 *)

Maxima

(n:1, for t:1 thru 10000 do (s:divsum(t), (A:args(divisors(t)),
              for i:1 thru length(A)-1 do (y:s+A[i],
                      if mod(y,t)=0 then (print(n,"",t), n:n+1)))));

PARI

isok(t) = my(s=sigma(t)); fordiv(t, d, if ((dMichel Marcus, Jun 30 2025