A308266 Sum of the middle parts in the partitions of n into 3 parts.
0, 0, 1, 1, 3, 5, 8, 11, 18, 22, 31, 40, 51, 62, 80, 93, 114, 135, 159, 183, 217, 244, 282, 320, 362, 404, 459, 505, 565, 625, 690, 755, 836, 906, 993, 1080, 1173, 1266, 1378, 1477, 1596, 1715, 1841, 1967, 2115, 2248, 2404, 2560, 2724, 2888, 3077, 3249, 3447
Offset: 1
Keywords
Examples
Figure 1: The partitions of n into 3 parts for n = 3, 4, ... 1+1+8 1+1+7 1+2+7 1+2+6 1+3+6 1+1+6 1+3+5 1+4+5 1+1+5 1+2+5 1+4+4 2+2+6 1+1+4 1+2+4 1+3+4 2+2+5 2+3+5 1+1+3 1+2+3 1+3+3 2+2+4 2+3+4 2+4+4 1+1+1 1+1+2 1+2+2 2+2+2 2+2+3 2+3+3 3+3+3 3+3+4 ... ----------------------------------------------------------------------- n | 3 4 5 6 7 8 9 10 ... ----------------------------------------------------------------------- a(n) | 1 1 3 5 8 11 18 22 ... -----------------------------------------------------------------------
Crossrefs
Cf. A308265.
Programs
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Mathematica
Table[Sum[Sum[i, {i, k, Floor[(n - k)/2]}], {k, Floor[n/3]}], {n, 100}] Table[Total[IntegerPartitions[n,{3}][[All,2]]],{n,60}] (* Harvey P. Dale, Jul 09 2020 *)
Formula
a(n) = Sum_{k=1..floor(n/3)} Sum_{i=k..floor((n-k)/2)} i.
Conjectures from Colin Barker, Jul 16 2019: (Start)
G.f.: x^3*(1 + x + x^2 + x^3 + x^4) / ((1 - x)^4*(1 + x)^2*(1 + x + x^2)^2).
a(n) = 2*a(n-2) + 2*a(n-3) - a(n-4) - 4*a(n-5) - a(n-6) + 2*a(n-7) + 2*a(n-8) - a(n-10) for n>10.
(End)