Number theory expression
In mathematics, Legendre's formula gives an expression for the exponent of the largest power of a prime p that divides the factorial n !. It is named after Adrien-Marie Legendre . It is also sometimes known as de Polignac's formula , after Alphonse de Polignac .
For any prime number p and any positive integer n , let
ν
p
(
n
)
{\displaystyle \nu _{p}(n)}
p that divides n (that is, the p -adic valuation of n ). Then
ν
p
(
n
!
)
=
∑
i
=
1
∞
⌊
n
p
i
⌋
,
{\displaystyle \nu _{p}(n!)=\sum _{i=1}^{\infty }\left\lfloor {\frac {n}{p^{i}}}\right\rfloor ,}
where
⌊
x
⌋
{\displaystyle \lfloor x\rfloor }
floor function . While the sum on the right side is an infinite sum, for any particular values of n and p it has only finitely many nonzero terms: for every i large enough that
p
i
>
n
{\displaystyle p^{i}>n}
⌊
n
p
i
⌋
=
0
{\displaystyle \textstyle \left\lfloor {\frac {n}{p^{i}}}\right\rfloor =0}
ν
p
(
n
!
)
=
∑
i
=
1
L
⌊
n
p
i
⌋
,
{\displaystyle \nu _{p}(n!)=\sum _{i=1}^{L}\left\lfloor {\frac {n}{p^{i}}}\right\rfloor \,,}
where
L
=
⌊
log
p
n
⌋
{\displaystyle L=\lfloor \log _{p}n\rfloor }
For n = 6, one has
6
!
=
720
=
2
4
⋅
3
2
⋅
5
1
{\displaystyle 6!=720=2^{4}\cdot 3^{2}\cdot 5^{1}}
ν
2
(
6
!
)
=
4
,
ν
3
(
6
!
)
=
2
{\displaystyle \nu _{2}(6!)=4,\nu _{3}(6!)=2}
ν
5
(
6
!
)
=
1
{\displaystyle \nu _{5}(6!)=1}
ν
2
(
6
!
)
=
∑
i
=
1
∞
⌊
6
2
i
⌋
=
⌊
6
2
⌋
+
⌊
6
4
⌋
=
3
+
1
=
4
,
ν
3
(
6
!
)
=
∑
i
=
1
∞
⌊
6
3
i
⌋
=
⌊
6
3
⌋
=
2
,
ν
5
(
6
!
)
=
∑
i
=
1
∞
⌊
6
5
i
⌋
=
⌊
6
5
⌋
=
1.
{\displaystyle {\begin{aligned}\nu _{2}(6!)&=\sum _{i=1}^{\infty }\left\lfloor {\frac {6}{2^{i}}}\right\rfloor =\left\lfloor {\frac {6}{2}}\right\rfloor +\left\lfloor {\frac {6}{4}}\right\rfloor =3+1=4,\\[3pt]\nu _{3}(6!)&=\sum _{i=1}^{\infty }\left\lfloor {\frac {6}{3^{i}}}\right\rfloor =\left\lfloor {\frac {6}{3}}\right\rfloor =2,\\[3pt]\nu _{5}(6!)&=\sum _{i=1}^{\infty }\left\lfloor {\frac {6}{5^{i}}}\right\rfloor =\left\lfloor {\frac {6}{5}}\right\rfloor =1.\end{aligned}}}
Since
n
!
{\displaystyle n!}
n , we obtain at least one factor of p in
n
!
{\displaystyle n!}
p in
{
1
,
2
,
…
,
n
}
{\displaystyle \{1,2,\dots ,n\}}
⌊
n
p
⌋
{\displaystyle \textstyle \left\lfloor {\frac {n}{p}}\right\rfloor }
p
2
{\displaystyle p^{2}}
p , each multiple of
p
3
{\displaystyle p^{3}}
p , etc. Adding up the number of these factors gives the infinite sum for
ν
p
(
n
!
)
{\displaystyle \nu _{p}(n!)}
One may also reformulate Legendre's formula in terms of the base-p expansion of n . Let
s
p
(
n
)
{\displaystyle s_{p}(n)}
p expansion of n ; then
ν
p
(
n
!
)
=
n
−
s
p
(
n
)
p
−
1
.
{\displaystyle \nu _{p}(n!)={\frac {n-s_{p}(n)}{p-1}}.}
For example, writing n = 6 in binary as 610 = 1102 , we have that
s
2
(
6
)
=
1
+
1
+
0
=
2
{\displaystyle s_{2}(6)=1+1+0=2}
ν
2
(
6
!
)
=
6
−
2
2
−
1
=
4.
{\displaystyle \nu _{2}(6!)={\frac {6-2}{2-1}}=4.}
Similarly, writing 6 in ternary as 610 = 203 , we have that
s
3
(
6
)
=
2
+
0
=
2
{\displaystyle s_{3}(6)=2+0=2}
ν
3
(
6
!
)
=
6
−
2
3
−
1
=
2.
{\displaystyle \nu _{3}(6!)={\frac {6-2}{3-1}}=2.}
Write
n
=
n
ℓ
p
ℓ
+
⋯
+
n
1
p
+
n
0
{\displaystyle n=n_{\ell }p^{\ell }+\cdots +n_{1}p+n_{0}}
p . Then
⌊
n
p
i
⌋
=
n
ℓ
p
ℓ
−
i
+
⋯
+
n
i
+
1
p
+
n
i
{\displaystyle \textstyle \left\lfloor {\frac {n}{p^{i}}}\right\rfloor =n_{\ell }p^{\ell -i}+\cdots +n_{i+1}p+n_{i}}
ν
p
(
n
!
)
=
∑
i
=
1
ℓ
⌊
n
p
i
⌋
=
∑
i
=
1
ℓ
(
n
ℓ
p
ℓ
−
i
+
⋯
+
n
i
+
1
p
+
n
i
)
=
∑
i
=
1
ℓ
∑
j
=
i
ℓ
n
j
p
j
−
i
=
∑
j
=
1
ℓ
∑
i
=
1
j
n
j
p
j
−
i
=
∑
j
=
1
ℓ
n
j
⋅
p
j
−
1
p
−
1
=
∑
j
=
0
ℓ
n
j
⋅
p
j
−
1
p
−
1
=
1
p
−
1
∑
j
=
0
ℓ
(
n
j
p
j
−
n
j
)
=
1
p
−
1
(
n
−
s
p
(
n
)
)
.
{\displaystyle {\begin{aligned}\nu _{p}(n!)&=\sum _{i=1}^{\ell }\left\lfloor {\frac {n}{p^{i}}}\right\rfloor \\&=\sum _{i=1}^{\ell }\left(n_{\ell }p^{\ell -i}+\cdots +n_{i+1}p+n_{i}\right)\\&=\sum _{i=1}^{\ell }\sum _{j=i}^{\ell }n_{j}p^{j-i}\\&=\sum _{j=1}^{\ell }\sum _{i=1}^{j}n_{j}p^{j-i}\\&=\sum _{j=1}^{\ell }n_{j}\cdot {\frac {p^{j}-1}{p-1}}\\&=\sum _{j=0}^{\ell }n_{j}\cdot {\frac {p^{j}-1}{p-1}}\\&={\frac {1}{p-1}}\sum _{j=0}^{\ell }\left(n_{j}p^{j}-n_{j}\right)\\&={\frac {1}{p-1}}\left(n-s_{p}(n)\right).\end{aligned}}}
Legendre's formula can be used to prove Kummer's theorem . As one special case, it can be used to prove that if n is a positive integer then 4 divides
(
2
n
n
)
{\displaystyle {\binom {2n}{n}}}
n is not a power of 2.
It follows from Legendre's formula that the p -adic exponential function
p
−
1
/
(
p
−
1
)
{\displaystyle p^{-1/(p-1)}}
Legendre, A. M. (1830), Théorie des Nombres , Paris: Firmin Didot Frères Moll, Victor H. (2012), Numbers and Functions , American Mathematical Society , ISBN 978-0821887950 MR 2963308 Leonard Eugene Dickson , History of the Theory of Numbers 
![{\displaystyle {\begin{aligned}\nu _{2}(6!)&=\sum _{i=1}^{\infty }\left\lfloor {\frac {6}{2^{i}}}\right\rfloor =\left\lfloor {\frac {6}{2}}\right\rfloor +\left\lfloor {\frac {6}{4}}\right\rfloor =3+1=4,\\[3pt]\nu _{3}(6!)&=\sum _{i=1}^{\infty }\left\lfloor {\frac {6}{3^{i}}}\right\rfloor =\left\lfloor {\frac {6}{3}}\right\rfloor =2,\\[3pt]\nu _{5}(6!)&=\sum _{i=1}^{\infty }\left\lfloor {\frac {6}{5^{i}}}\right\rfloor =\left\lfloor {\frac {6}{5}}\right\rfloor =1.\end{aligned}}}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/62facfa3f93f01476aa9253ab337a2c3ae823e15)
