OFFSET
0,4
COMMENTS
For a prime p we have sigma(p) = p + 1. Suppose p = x_1 + ... + x_k is a partition of p into distinct parts with k >= 2. If none of the parts equals 1, then sigma(x_i) >= x_i + 1 for all i, hence Sum_{i=1..k} sigma(x_i) >= p + k >= p + 2 > sigma(p). If 1 is a part, write p = 1 + y_1 + ... + y_{k-1}. Then Sum_{i=1..k} sigma(x_i) >= p + (k - 1), and equality with sigma(p) = p + 1 can occur only if k = 2 and y_1 is prime, which forces p = 3. Hence a(p) = 1 for all primes p other than 3, and a(3) = 2.
We define A000203(0) = 0.
LINKS
Felix Huber, Table of n, a(n) for n = 0..500
FORMULA
EXAMPLE
a(3) = 2: [3], [1, 2], since sigma(3) = 4 = sigma(1) + sigma(2).
a(26) = 9: [26], [5, 6, 15], [3, 4, 9, 10], [1, 7, 8, 10], [3, 6, 7, 10], [1, 2, 4, 9, 10], [1, 2, 6, 7, 10], [2, 3, 4, 8, 9], [2, 3, 6, 7, 8], since sigma(26) = 42 and each listed partition has sum of sigma values 42.
MAPLE
A394248 := proc(n)
local h, g, f;
h := proc(k) option remember; `if`(k <= 0, 0, NumberTheory:-sigma(k)) end proc:
g := proc(k) option remember; `if`(k <= 0, 0, g(k - 1) + h(k)) end proc:
f := proc(i, j, k) option remember;
`if`(i = 0 and j = 0, 1, `if`(i < 0 or j < 0 or k = 0 or j > g(k), 0,
f(i, j, k - 1) + f(i - k, j - h(k), k - 1)))
end proc:
`if`(n = 0, 1, f(n, h(n), n))
end proc:
seq(A394248(n), n = 0 .. 74);
MATHEMATICA
a[n_]:=Count[Total/@DivisorSigma[1, Select[IntegerPartitions[n], DuplicateFreeQ]], DivisorSigma[1, n]]; a[0]=1; Array[a, 70, 0] (* James C. McMahon, May 01 2026 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Felix Huber, Apr 27 2026
STATUS
approved
