## Maximum Integer Partition Product

### Problem 374

Published on Saturday, 3rd March 2012, 07:00 pm; Solved by 376An integer partition of a number `n` is a way of writing `n` as a sum of positive integers.

Partitions that differ only in the order of their summands are considered the same.
A partition of `n` into **distinct parts** is a partition of `n` in which every part occurs at most once.

The partitions of 5 into distinct parts are:

5, 4+1 and 3+2.

Let f(`n`) be the maximum product of the parts of any such partition of `n` into distinct parts and let m(`n`) be the number of elements of any such partition of `n` with that product.

So f(5)=6 and m(5)=2.

For `n`=10 the partition with the largest product is 10=2+3+5, which gives f(10)=30 and m(10)=3.

And their product, f(10)·m(10) = 30·3 = 90

It can be verified that

∑f(`n`)·m(`n`) for 1 ≤ `n` ≤ 100 = 1683550844462.

Find ∑f(`n`)·m(`n`) for 1 ≤ `n` ≤ 10^{14}.

Give your answer modulo 982451653, the 50 millionth prime.