### averaging five fun

Comment

What possible number of sets has each 'target N'
so for instance with a target of 4 there are 2 sets:
2-4-4-4-6
2-3-4-5-6
And with target 5 there are also 2 sets:
2-5-5-6-7
3-4-5-5-8

This is a fun poser. I looked at it and found some patterns. I wanted to express what I found as an single elegant formula but couldn't manage it, so here is a description instead.

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To calculate how many sets for an even number target:
(N divide by 2) add (N-6, + N-12, + N-18, ................for any positive number)

so for example
N4 =(4/2=2) (but nothing to add because 4 minus 6 is negative) answer= 2 sets
N8 =(8/2=4) + (8-6) (this time there is 2 to add because 8 minus 6 is positive) answer = 6 sets
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To calculate how many sets for an odd number target:
N-3 + N-9 + N-15........ for any positive number

So for example
N5 =(5-3=2) answer is 2 sets
N9 =(9-3=6) answer is 6 sets
N13=(13-3=10) + (13-9=4) answer is 14 sets
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I calculated 832 sets for a target of number of 100, but I could be wrong about it all. I'd love to have an elegant formula. Things to think about: modulus, partitions, triangles numbers, golden ratio ...... It levels at N=12. I kept looking at modulus tricks but couldn't quite nail it.
Thanks for another great thought provoking article!
3 :0
4 :2
5 :2
6 :3
7 :4
8 :6
9 :6
10 :9
11 :10
12 :12
13 :14
14 :17
15 :18
16 :22
17 :24
18 :27
19 :30
20 :34
21 :36
22 :41
23 :44
24 :48
25 :52
26 :57

100:832