n such that there is one sequence of length n having equal sum and product

[Don Reble]

> I think it is also worth mentioning the minimal constraint that terms > 2
> in A033179 must be of the form p+1 (since [ a+1, b+1, <1> x (ab - 1) ]
> is always a solution for k = ab+1, and non-trivial when a >= b >= 2).

Also, 2L-1 must not be composite; if it has proper factorization a*b, then
    1*(L-3), 2, (a+1)/2, (b+1)/2
is a second solution for length L.

Combining those ideas with those from Louis Marmet, we see that:

For L-1 and 2L-1 to be non-composite, L == 0,12,24 mod 30; or L=2,3,4,6.

But if L = 30n+12, then
    1*(30n+7), 2, 2, 2, 2, 2n+1
is a second solution.

So all entries in the sequence are 2,3,4,6, or 0,24 mod 30.

One can eliminate some of the multiples of 5 with such congruences;
for example,
    1*(35n+25),2,2,3,3,n+1	eliminates L=35n+30,
    1*(35n+22),4,9,n+1		eliminates L=35n+25
    1*(55n+37),4,14,n+1		eliminates L=55n+40
But it may be hard to find a complete set of congruences.

-- 
Don Reble  djr at nk.ca