4. Let $b$ take odd numbers between 2 and 16, and $c$ take any natural number. Then the number of quadratic equations $3 x^{2}+$ $(b+1) x+c=0$ that can be formed with two distinct real roots is ( ).
(A) 64
(B) 66
(C) 107
(D) infinitely many
4. (A).
$\because$ The equation $3 x^{2}+(b+1) x+c=0$ has two distinct real roots, $\therefore \Delta=(b+1)^{2}-12 c>0 \Rightarrow c<\frac{1}{12}(b+1)^{2}$.
Given $b=3,5,7,9,11,13,15$.
When $b=3$, $c<\frac{4}{3}$, there is 1 value of $c$ that meets the condition; when $b=5$, $c<3$, there are 2 values of $c$ that meet the condition; when $b=7$, $c<\frac{16}{3}$, there are 5 values of $c$ that meet the condition; when $b=9$, $c<\frac{25}{3}$, there are 8 values of $c$ that meet the condition; when $b=11$, $c<12$, there are 11 values of $c$ that meet the condition; when $b=13$, $c<\frac{49}{3}$, there are 16 values of $c$ that meet the condition; when $b=15$, $c<\frac{64}{3}$, there are 21 values of $c$ that meet the condition. Therefore, a total of $1+2+5+8+11+16+21=64$ (quadratic equations) are formed.