Problem 12041
(American Mathematical Monthly, Vol.125, May 2018) Proposed by R. Stanley (USA).
For a positive integer c, let νp(c) denote the largest integer d such that pd divides c. Let
Hm=
m
Y
i=0 m
Y
j=0
i + j i
.
For n ≥1, prove
νp(Hpn−1) = 1 2
n − 1 p −1
p2n+ pn p −1
.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. We have that
νp(Hm) =
m
X
i=0 m
X
j=0
(νp((i + j)!) − νp(i!) − νp(j!))
=
m
X
i=0
(i + 1)νp(i!) +
m−1
X
i=0
(m − i)νp((i + m + 1)!) − 2(m + 1)
m
X
i=0
νp(i!)
=
m
X
i=0
(m − i) (νp((i + m + 1)!) − νp(i!)) − (m + 1)
m
X
i=0
νp(i!).
Let p be a prime. Then by Legendre Theorem,
νp(N !) =X
k≥1
jN pk
k=N − sp(N ) p −1
where sp(N ) is the sum of all the digits in the expansion of N in base p.
For m = pn−1,
pn−1
X
i=0
(pn−1 − i) (νp((i + pn)!) − νp(i!)) =
n
X
k=1 pn−1
X
i=0
(pn−1 − i)
ji+ pn pk
k
− j i
pk k
=
n
X
k=1 pn−1
X
i=0
(pn−1 − i)pn−k= pn(pn−1)2 2(p − 1) .
Moreover, sincePpn−1
i=0 sp(i) = 12npn(p − 1), it follows that
pn−1
X
i=0
νp(i!) =
pn−1
X
i=0
i − sp(i)
p −1 = pn(pn−1) 2(p − 1) −npn
2 . Hence
νp(Hpn−1) =pn(pn−1)2
2(p − 1) −pn pn(pn−1) 2(p − 1) −npn
2
=1 2
n − 1 p −1
p2n+ pn p −1
.
