A064771 Let S(n) = set of divisors of n, excluding n; sequence gives n such that there is a unique subset of S(n) that sums to n.

6, 20, 28, 78, 88, 102, 104, 114, 138, 174, 186, 222, 246, 258, 272, 282, 304, 318, 354, 366, 368, 402, 426, 438, 464, 474, 490, 496, 498, 534, 572, 582, 606, 618, 642, 650, 654, 678, 748, 762, 786, 822, 834, 860, 894, 906, 940, 942, 978, 1002, 1014, 1038

Offset: 1

Author

Jonathan Ayres (jonathan.ayres(AT)btinternet.com), Oct 19 2001

Keywords

• nonn
• nice

Comments

Perfect numbers (A000396) are a proper subset of this sequence. Weird numbers (A006037) are numbers whose proper divisors sum to more than the number, but no subset sums to the number.

Odd elements are rare: the first few are 8925, 32445, 351351, 442365; there are no more below 100 million. See A065235 for more details.

A065205(a(n)) = 1. - Reinhard Zumkeller, Jan 21 2013

Examples

			Proper divisors of 20 are 1, 2, 4, 5 and 10. {1,4,5,10} is the only subset that sums to 20, so 20 is in the sequence.
		

Links

• Giovanni Resta, Table of n, a(n) for n = 1..10000 (terms 1..200 from T. D. Noe, terms 201..5000 from Amiram Eldar)

Crossrefs

A005835 gives n such that some subset of S(n) sums to n. Cf. A065205.

Cf. A006037, A065205, A378448 (characteristic function).

Subsequences: A000396, A065235 (odd terms), A378519, A378530.

Cf. A027751.

Programs

Haskell

a064771 n = a064771_list !! (n-1)
a064771_list = map (+ 1) $ elemIndices 1 a065205_list
-- Reinhard Zumkeller, Jan 21 2013

Maple

filter:= proc(n)
  local P,x,d;
  P:= mul(x^d+1, d = numtheory:-divisors(n) minus {n});
  coeff(P,x,n) = 1
end proc:
select(filter, [$1..2000]); # Robert Israel, Sep 25 2024

Mathematica

okQ[n_]:= Module[{d=Most[Divisors[n]]}, SeriesCoefficient[Series[ Product[ 1+x^i, {i, d}], {x, 0, n}], n] == 1];Select[ Range[ 1100],okQ] (* Harvey P. Dale, Dec 13 2010 *)

Python

from sympy import divisors
def isok(n):
    dp = {0: 1}
    for d in divisors(n)[:-1]:
        u = {}
        for k in dp.keys():
            if (s := (d + k)) <= n:
                u[s] = dp.get(s, 0) + dp[k]
                if s == n and u[s] > 1:
                    return False
        for k,v in u.items():
            dp[k] = v
    return dp.get(n, 0) == 1
print([n for n in range(1, 1039) if isok(n)]) # Darío Clavijo, Sep 17 2024

Extensions

More terms from Don Reble, Jud McCranie and Naohiro Nomoto, Oct 22 2001