A006339

[David Wilson]

Yes thank you.  The method for computing A006339(n) is as follows:

Let k = 2n+1.  Then form every factorization k = PROD(x_i) with x_0 >= x_1 >= 
... >= x_m > 1.  For each such product, form the product PROD(p_i^((x_i-1)/2), 
where p_i is the ith prime == 0 (mod 4).  Choose the smallest such product, this 
will be a(n).

For example, to find A006339(4):

n = 4, giving k = 2n+1 = 9.

The factorizations of 9 into nondecreasing integers are 9 and 3*3.

These factorizations correspond to 5^4 and 5^1*13^1.

The least of these is 5^1*13^1 = 65, which is A006339(4).

I'm sure an algorithm could be written to avoid many unsuitable factorizations 
of k.