Bu sayilarin icerisinde $3$ carpani icerenler $6k+3$ cinsinden olur.
$3$'e tam bolunen ($6$ ardil) $$\frac{99-3}{6}+1=17$$ sayi. $9$'a tam bolunen ($18$ ardil)$$\frac{99-9}{18}+1=6$$ sayi ve $27$'ye tam bolunen ($54$ ardil) $$\frac{81-27}{54}+1=2$$ ve $81$'e bolunen ($162$ ardil) $$\frac{81-81}{162}+1=1$$ sayi var.
Simdi bunu formulize edersek (
bu sorudaki gibi) toplamimiz, yani $$1\cdot 3\cdots (2n+1)$$ icersindeki $3$ carpani sayisi $$k+\sum\limits_{m =1}^k \bigg\lfloor \frac {2n+1-3^m}{2\cdot3^m} \bigg\rfloor=k+\sum\limits_{m =1}^k \bigg\lfloor \frac {2n+1}{2\cdot3^m}-\frac12 \bigg\rfloor$$ olur (burada $k$ dedigimiz $3^k \le 2n+1 <3^{k+1}$ sartini saglayan ilk tam sayi).
$p\ne2$ asal sayisi icin $$1\cdot 3\cdots (2n+1)$$ icersindeki $p$ carpani sayisi $$k+\sum\limits_{m =1}^k \bigg\lfloor \frac {2n+1-p^m}{2\cdot p^m} \bigg\rfloor=k+\sum\limits_{m =1}^k \bigg\lfloor \frac {2n+1}{2\cdot p^m}-\frac12 \bigg\rfloor$$ olur (burada $k$ dedigimiz $p^k \le 2n+1 <p^{k+1}$ sartini saglayan ilk tam sayi).
$l$ baska bir asal sayi olsun. ($p\ne l$). $$1\cdot(1+l)\cdots (1+n\cdot l)$$carpimi icerisndeki $p$ carpani sayisi $$k+\sum\limits_{m =1}^k \bigg\lfloor \frac {l\cdot n+1-p^m}{l\cdot p^m} \bigg\rfloor=k+\sum\limits_{m =1}^k \bigg\lfloor \frac {l\cdot n+1}{l\cdot p^m}-\frac1l \bigg\rfloor$$ olur (burada $k$ dedigimiz $p^k \le l\cdot n+1 <p^{k+1}$ sartini saglayan ilk tam sayi).
$l$ asal degil de $p$ ile aralarinda asal olsa bile yukaridaki esitlik yine saglanir: $$1\cdot(1+l)\cdots (1+n\cdot l)$$carpimi icerisndeki $p$ carpani sayisi $$k+\sum\limits_{m =1}^k \bigg\lfloor \frac {l\cdot n+1-p^m}{l\cdot p^m} \bigg\rfloor=k+\sum\limits_{m =1}^k \bigg\lfloor \frac {l\cdot n+1}{l\cdot p^m}-\frac1l \bigg\rfloor$$ olur (burada $k$ dedigimiz $p^k \le l\cdot n+1 <p^{k+1}$ sartini saglayan ilk tam sayi).