Assuming the generating function is $\frac{1}{\prod\limits_{k=1}^{60}{1-x^k}}$ \frac{1}{\prod\limits_{k=1}^{60}{(1-x^k)}}$ less than two hours of gp/pari computations gave the 5738 digit answer:
EDIT: Here is how the number was computed. The generating function is $\frac{1}{\prod\limits_{k=1}^{60}{1-x^k}}$ \frac{1}{\prod\limits_{k=1}^{60}{(1-x^k)}}$ means the sequence satisfies linear recurrence with constant coefficients. These are known to be tractable efficiently computable assuming arithmetic operations in the range of the result are tractable. A good computational resource for recurrences is the free book "Matters Computational" was: "Algorithms for Programmers" by Jörg Arndt. Basically the method is fast binary exponentiation of a matrix or in $\mathbb{Z}[x]/poly(x)$. The book has code parts of which I used. Got a linear recurrence of width order 1830. My gp/pari code is here. A curiosity of the challenge is the result is so small - I wouldn't even try fibonacci(10^100).$fibonacci(10^{100})$.
EDIT2 A closed form possibly leading to faster approach (avoiding computing the recurrence) appears in the paperA GENERAL METHOD FOR DETERMINING A CLOSED FORMULA FOR THE NUMBER OF PARTITIONS OF THE INTEGER $n$ INTO $m$ POSITIVE INTEGERS FOR SMALL VALUES OF $m$, W. J. A. COLMAN

