Let us show that once a solution exist for (1,2,3,... n), creating a 3E-Partition for (1,2,3,...n+2, n+3) is trivial .
My proof:
Let the 3E-Partition for sum(n) be S1, S2, S3.
We have to add 3 additional numbers n+1 , n+2 ,n+3 i.e. to increase each subset by n+2.
We will replace number 1 (present in one of the subsets, say S2) by n+3; the number n+2 will go to another subset (e.g. S1) without replacement.
Two unassigned numbers 1 and n+1 go to the last subset
(S3 in our example).
Since existence of a solution for n=5 and 4=6 was shown de facto, the process of full induction is concluded.

Ans:5,6.8.9.11,12.....(3n-1),(3n),....

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