## Special partitions

### Problem 333

Published on Saturday, 16th April 2011, 01:00 pm; Solved by 615All positive integers can be partitioned in such a way that each and every term of the partition can be expressed as 2^{i}x3^{j}, where i,j ≥ 0.

Let's consider only those such partitions where none of the terms can divide any of the other terms.

For example, the partition of 17 = 2 + 6 + 9 = (2^{1}x3^{0} + 2^{1}x3^{1} + 2^{0}x3^{2}) would not be valid since 2 can divide 6. Neither would the partition 17 = 16 + 1 = (2^{4}x3^{0} + 2^{0}x3^{0}) since 1 can divide 16. The only valid partition of 17 would be 8 + 9 = (2^{3}x3^{0} + 2^{0}x3^{2}).

Many integers have more than one valid partition, the first being 11 having the following two partitions.

11 = 2 + 9 = (2^{1}x3^{0} + 2^{0}x3^{2})

11 = 8 + 3 = (2^{3}x3^{0} + 2^{0}x3^{1})

Let's define P(`n`) as the number of valid partitions of `n`. For example, P(11) = 2.

Let's consider only the prime integers `q` which would have a single valid partition such as P(17).

The sum of the primes `q` <100 such that P(`q`)=1 equals 233.

Find the sum of the primes `q` <1000000 such that P(`q`)=1.