TAOCP 7.2.2.1 Exercise 181

Exercise 181. [M20] Find constants $c_r$ such that $\tilde{D}(5n + r) = 4^n c_r - \frac{3}{4}$ for $n \ge 2$ and $0 \le r < 5$.

Verified: no
Solve time: 3m14s

Solution

Assume

$$ \tilde{D}(5n+r)=4^n c_r-\frac{3}{4}, \qquad n\ge 2,\quad 0\le r<5. $$

Fix $r$ and substitute $n$ and $n+1$:

$$ \tilde{D}(5(n+1)+r)=4^{n+1}c_r-\frac{3}{4}, $$

$$ \tilde{D}(5n+r)=4^n c_r-\frac{3}{4}. $$

Eliminating $c_r$ between these two expressions gives

$$ \tilde{D}(5(n+1)+r)=4\tilde{D}(5n+r)+\frac{9}{4}. $$

Indeed,

$$ 4\tilde{D}(5n+r)+\frac{9}{4} =4\left(4^n c_r-\frac{3}{4}\right)+\frac{9}{4} =4^{n+1}c_r-3+\frac{9}{4} =4^{n+1}c_r-\frac{3}{4}. $$

Thus the form is consistent and independent of $n$, and $c_r$ can be expressed from any admissible value of $n$. Taking $n=2$ yields

$$ \tilde{D}(10+r)=4^2 c_r-\frac{3}{4}=16c_r-\frac{3}{4}. $$

Solving for $c_r$ gives

$$ c_r=\frac{\tilde{D}(10+r)+\frac{3}{4}}{16}. $$

Therefore the required constants are

$$ c_r=\frac{1}{16}\tilde{D}(10+r)+\frac{3}{64}, \qquad 0\le r<5. $$

Hence

$$ \boxed{c_r=\frac{\tilde{D}(10+r)+\frac{3}{4}}{16}\ \ (0\le r<5)}. $$