A350534 Sum of the largest parts of the partitions of n into 3 parts whose largest part is equal to the product of the other two.
0, 0, 0, 1, 0, 2, 0, 3, 4, 4, 0, 11, 0, 6, 8, 16, 0, 18, 0, 21, 12, 10, 0, 40, 16, 12, 16, 31, 0, 52, 0, 36, 20, 16, 24, 88, 0, 18, 24, 74, 0, 76, 0, 51, 60, 22, 0, 121, 36, 60, 32, 61, 0, 100, 40, 108, 36, 28, 0, 198, 0, 30, 88, 125, 48, 124, 0, 81, 44, 140, 0, 243, 0, 36, 104
Offset: 0
Keywords
Examples
a(13) = 6 since we have 13 = 1+6+6, whose largest part is 6. Partitions not counted: 1+1+11, 1+2+10, 1+3+9, 1+4+8, 1+5+7, 2+2+9, 2+3+8, 2+4+7, 2+5+6, 3+3+7, 3+4+6, 3+5+5, 4+4+5.
Links
Programs
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Mathematica
Table[Sum[Sum[(n - i - k) KroneckerDelta[(n - i - k), (i*k)], {i, k, Floor[(n - k)/2]}], {k, Floor[n/3]}], {n, 0, 100}] -
PARI
first(n) = {my(res = vector(n)); for(i = 1, n \ 2, for(j = i, n\i, c = i + j + i*j; if(c <= n, res[c] += i*j))); concat(0, res)} \\ David A. Corneth, Jan 07 2022
Formula
a(n) = Sum_{k=1..floor(n/3)} Sum_{i=k..floor((n-k)/2)} [n-i-k = i*k] * (n-i-k), where [ ] is the Iverson bracket.