Binary word double runs

Probability for consecutive double runs (either 0s or 1s) of lenght k in binary words of length n, use p_binary_words_doub_runl(k,n)

julia>using AnalyticComb
julia>p_binary_words_doub_runl(6,200) #e.g. 100000011010... or 01111110101...
0.166...

Restricted summands problem

Solving Polya partitions with restricted summands (denumerants) problem^[1]. What is the number of ways of giving change of 99 cents using pennies (1 cent), nickels (5 cents), dimes (10 cents) and quarters (25 cents)?

That is, the number of ways to obtain 99 by summing 1s,5s,10s and 25s.

Or number of distinct sets $S = \{k_1,k_2,_k_3,k_4}\, \quad k \in \mathbb{N}$ satisfying the equation :

$1 k_1 + 5 k_2 + 10 k_3 + 25 k_4 = 99$.

The generating function is:$P(z) = SEQ(z)*SEQ(z^5)*SEQ(z^{10})*SEQ(z^{25})$ and the solution is the cofficient of $z^{99}$ in the expansion: $[z^{99}] T(P(z))$.

$T(P(z)) = ... + 213 z^{99} + ...$

Function restricted_sum_part_gf(r) returns the generating function for elements in r and restricted_sum_part(k,r) returns the coefficient in $z^k$ for the generating function with elements in r.

julia> restricted_sum_part_gf([1,5,10,25]) # examine the generating function SEQ(z)*SEQ(z^5)*SEQ(z^10)*SEQ(z^25)
                 1                  
────────────────────────────────────
        ⎛     5⎞ ⎛     10⎞ ⎛     25⎞
(1 - z)⋅⎝1 - z ⎠⋅⎝1 - z  ⎠⋅⎝1 - z  ⎠

julia>restricted_sum_part(99,[1,5,10,25]) # Counts for 99 as a sum of elements in (1,5,10,25).
213