A326632 Sum of the sixth largest parts of the partitions of n into 10 squarefree parts.
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 2, 4, 5, 9, 11, 16, 21, 31, 37, 51, 61, 83, 99, 129, 150, 197, 229, 288, 332, 417, 479, 589, 676, 830, 950, 1150, 1310, 1588, 1802, 2148, 2431, 2897, 3264, 3843, 4322, 5067, 5684, 6604, 7380, 8557, 9538, 10961, 12198
Offset: 0
Keywords
Crossrefs
Programs
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Mathematica
Table[Sum[Sum[Sum[Sum[Sum[Sum[Sum[Sum[Sum[m * MoebiusMu[r]^2 * MoebiusMu[q]^2 * MoebiusMu[p]^2 * MoebiusMu[o]^2 * MoebiusMu[m]^2 * MoebiusMu[l]^2 * MoebiusMu[k]^2 * MoebiusMu[j]^2 * MoebiusMu[i]^2 * MoebiusMu[n - i - j - k - l - m - o - p - q - r]^2 , {i, j, Floor[(n - j - k - l - m - o - p - q - r)/2]}], {j, k, Floor[(n - k - l - m - o - p - q - r)/3]}], {k, l, Floor[(n - l - m - o - p - q - r)/4]}], {l, m, Floor[(n - m - o - p - q - r)/5]}], {m, o, Floor[(n - o - p - q - r)/6]}], {o, p, Floor[(n - p - q - r)/7]}], {p, q, Floor[(n - q - r)/8]}], {q, r, Floor[(n - r)/9]}], {r, Floor[n/10]}], {n, 0, 50}]
Formula
a(n) = Sum_{r=1..floor(n/10)} Sum_{q=r..floor((n-r)/9)} Sum_{p=q..floor((n-q-r)/8)} Sum_{o=p..floor((n-p-q-r)/7)} Sum_{m=o..floor((n-o-p-q-r)/6)} Sum_{l=m..floor((n-m-o-p-q-r)/5)} Sum_{k=l..floor((n-l-m-o-p-q-r)/4)} Sum_{j=k..floor((n-k-l-m-o-p-q-r)/3)} Sum_{i=j..floor((n-j-k-l-m-o-p-q-r)/2)} mu(r)^2 * mu(q)^2 * mu(p)^2 * mu(o)^2 * mu(m)^2 * mu(l)^2 * mu(k)^2 * mu(j)^2 * mu(i)^2 * mu(n-i-j-k-l-m-o-p-q-r)^2 * m, where mu is the Möbius function (A008683).