Talk:Legendre's formula

"Alternate form"

The article contained (until I just removed it) a long section titled "Alternate form"

The section I just removed
The following discussion has been closed. Please do not modify it.
One may also reformulate Legendre's formula in terms of the base-p expansion of n. Let $s_{p}(n)$ denote the sum of the digits in the base-p expansion of n; then $\nu _{p}(n!)={\frac {n-s_{p}(n)}{p-1}}.$ For example, writing n = 6 in binary as 610 = 1102, we have that $s_{2}(6)=1+1+0=2$ and so $\nu _{2}(6!)={\frac {6-2}{2-1}}=4.$ Similarly, writing 6 in ternary as 610 = 203, we have that $s_{3}(6)=2+0=2$ and so $\nu _{3}(6!)={\frac {6-2}{3-1}}=2.$ Proof Write $n=n_{\ell }p^{\ell }+\cdots +n_{1}p+n_{0}$ in base p. Then $\textstyle \left\lfloor {\frac {n}{p^{i}}}\right\rfloor =n_{\ell }p^{\ell -i}+\cdots +n_{i+1}p+n_{i}$, and therefore ${\begin{aligned}\nu _{p}(n!)&=\sum _{i=1}^{\ell }\left\lfloor {\frac {n}{p^{i}}}\right\rfloor \\&=\sum _{i=1}^{\ell }\left(n_{\ell }p^{\ell -i}+\cdots +n_{i+1}p+n_{i}\right)\\&=\sum _{i=1}^{\ell }\sum _{j=i}^{\ell }n_{j}p^{j-i}\\&=\sum _{j=1}^{\ell }\sum _{i=1}^{j}n_{j}p^{j-i}\\&=\sum _{j=1}^{\ell }n_{j}\cdot {\frac {p^{j}-1}{p-1}}\\&=\sum _{j=0}^{\ell }n_{j}\cdot {\frac {p^{j}-1}{p-1}}\\&={\frac {1}{p-1}}\sum _{j=0}^{\ell }\left(n_{j}p^{j}-n_{j}\right)\\&={\frac {1}{p-1}}\left(n-s_{p}(n)\right).\end{aligned}}$

I have removed it because (1) it was completely uncited, and (2) I do not believe that this formula is referred to by anyone as "Legendre's formula", or as a form of Legendre's formula -- I think it's a totally separate formula for the same quantity. I would not object to restoring it, provided sources can be found that support the idea that this version is also known as Legendre's formula -- but if so, the proof should probably be removed (or at least abbreviated). --JBL (talk) 16:41, 12 March 2022 (UTC)

"Alternate form" might not be the right section title, but I disagree that it doesn't belong in the article. The two formulas are not totally separate; they both show how $\nu _{p}(n!)$ depends on the base-$p$ representation of $n$. They are equivalent and can be derived from each other. The section was only long because of the proof; if the proof is not necessary, let's leave it out. The citation should be the Moll reference, which says "An explicit expression for the error term $\nu _{p}(n!)\sim n/(p-1)$ was discovered by A. M. Legendre" and then immediately states a theorem that $\nu _{p}(n!)={\frac {n-s_{p}(n)}{p-1}}$. Eric Rowland (talk) 21:39, 15 March 2022 (UTC)