A125225 Numbers n such that n-1 can be represented as a sum of a subset of divisors of n.

1, 2, 4, 6, 8, 12, 16, 18, 20, 24, 28, 30, 32, 36, 40, 42, 48, 54, 56, 60, 64, 66, 70, 72, 78, 80, 84, 88, 90, 96, 100, 104, 108, 112, 120, 126, 128, 132, 140, 144, 150, 156, 160, 162, 168, 176, 180, 192, 196, 198, 200, 204, 208, 210, 216, 220, 224, 228, 234, 240, 252

Offset: 1

Author

David Eppstein, Jan 13 2007

Keywords

• nonn

Comments

The definition is related to that for semiperfect numbers (A005835). Every practical number (A005153) belongs to this sequence but not necessarily vice versa; e.g., 70 is in this sequence but not practical. Every number n in this sequence has sigma(n) >= 2n-1 (A103288) but, despite being abundant, 102 is not in this sequence.

Such numbers can be used to construct inheritance puzzles of the type described by Premchand Anne (see link).

Does the sequence contain A005231 (the odd abundant numbers)? - Robert Israel, Aug 05 2016

Examples

			70 is in this sequence because 70-1=69=35+14+10+7+2+1 and all numbers in the sum are divisors of 70.
		

Links

• T. D. Noe, Table of n, a(n) for n=1..1000
• Premchand Anne, Egyptian fractions and the inheritance problem, The College Mathematics Journal 29 (4) (1998) 296-300.

Crossrefs

Cf. A005835, A000225, A005153, A103288.

Programs

Maple

ss:= proc(n, S) local s, Sp;  option remember;
  if n = 0 then return true
  elif S = {} then return false
  fi;
  s:= max(S);
  if s > n then return procname(n, select(`<=`,S,n))
  elif s = n then return true
  fi;
  s:= min(S);
  Sp:= subs(s=NULL,S);
  if s > n then false
  else procname(n-s,Sp) or procname(n,Sp)
  fi
end proc:
select(n -> ss(n-1, numtheory:-divisors(n)), [$1..1000]); # Robert Israel, Aug 05 2016

Mathematica

okQ[n_] := With[{dd = Divisors[n]}, AnyTrue[Range[Length[dd], 1, -1], AnyTrue[Subsets[dd, {#}], Total[#] == n-1&]&]]; okQ[1] = True;
Select[Range[1000], okQ] (* Jean-François Alcover, Jul 23 2020 *)

PARI

padbin(n, len) = {b = binary(n); while(length(b) < len, b = concat(0, b);); b;}
isok(n) = {if (n == 1, return (1)); d = divisors(n); nbd = #d; for (i = 1, 2^nbd-1, b = padbin(i, nbd); s = sum(j = 1, nbd, d[j]*b[j]); if (s == (n - 1), return (1));); return (0);} \\ Michel Marcus, Aug 30 2013