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A393359
a(n) is the number of ways to successively choose subsets of sizes equal to the proper divisors of n from [n], each choice taken from the remaining elements.
1
1, 2, 3, 12, 5, 60, 7, 840, 504, 7560, 11, 0, 13, 360360, 2522520, 10810800, 17, 0, 19, 0, 465585120, 349188840, 23, 0, 1062600, 8923714800, 57366738000, 14456417976000, 29, 0, 31, 6498159880212000, 5662884427200, 4760556688800, 15200373988800, 0, 37, 102748681866600, 485721041551200, 0
OFFSET
1,2
COMMENTS
If the sum of the proper divisors of n exceeds n (i.e., n is abundant), the process cannot be completed and a(n) = 0.
Equivalently, for sigma(n) <= 2n, a(n) is the number of ordered set partitions of [n] into parts of sizes equal to the proper divisors of n together with one additional part of size n minus their sum.
For perfect n, the remainder part has size 0 and a(n) = n!/Product_{d|n, d<n} d!. Thus perfect numbers are exactly those n for which the selection process exhausts [n] without remainder.
FORMULA
a(n) = 0 if sigma(n) > 2*n, otherwise:
a(n) = Product_{i=1..tau(n)-1} binomial(n - Sum_{j=1..i-1} d_j, d_i), where d_1 > ... > d_{tau(n)-1} are the proper divisors of n.
a(n) = n!/((2*n - sigma(n))!*Product_{d|n, d<n} d!).
a(p) = p for primes p.
EXAMPLE
a(6) = binomial(6, 3)*binomial(3, 2)*binomial(1, 1) = 60 (perfect).
a(10) = binomial(10, 5)*binomial(5, 2)*binomial(3, 1) = 7560 (deficient).
a(11) = binomial(11, 1) = 11 (prime).
a(12) = 0 since sigma(12) = 28 > 24 (abundant).
MAPLE
A393359 := proc(n)
local d, k, m, a;
d := NumberTheory:-Divisors(n);
m := n;
a := 1;
for k from nops(d) - 1 by -1 to 1 do
m := m - d[k];
if m < 0 then return 0 end if;
a := a*binomial(m + d[k], d[k])
end do;
a
end proc:
seq(A393359(n), n = 1 .. 40);
KEYWORD
nonn
AUTHOR
Felix Huber, Mar 28 2026
STATUS
approved