A326686 Sum of the third largest parts of the partitions of n into 10 primes.
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 2, 5, 5, 8, 10, 12, 14, 22, 24, 30, 37, 41, 50, 65, 72, 83, 104, 108, 136, 154, 175, 189, 234, 241, 298, 314, 371, 384, 487, 472, 596, 604, 732, 729, 928, 889, 1134, 1089, 1360, 1318, 1685
Offset: 0
Keywords
Crossrefs
Programs
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Mathematica
Table[Sum[Sum[Sum[Sum[Sum[Sum[Sum[Sum[Sum[j * (PrimePi[i] - PrimePi[i - 1]) (PrimePi[j] - PrimePi[j - 1]) (PrimePi[k] - PrimePi[k - 1]) (PrimePi[l] - PrimePi[l - 1]) (PrimePi[m] - PrimePi[m - 1]) (PrimePi[o] - PrimePi[o - 1]) (PrimePi[p] - PrimePi[p - 1]) (PrimePi[q] - PrimePi[q - 1]) (PrimePi[r] - PrimePi[r - 1]) (PrimePi[n - i - j - k - l - m - o - p - q - r] - PrimePi[n - i - j - k - l - m - o - p - q - r - 1]), {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)} c(r) * c(q) * c(p) * c(o) * c(m) * c(l) * c(k) * c(j) * c(i) * c(n-i-j-k-l-m-o-p-q-r) * j, where c = A010051.
Extensions
Name clarified by Sean A. Irvine, Jan 12 2025