Problem 11720
(American Mathematical Monthly, Vol.120, August-September 2013]) Proposed by Ira Gessel (USA).
Let En(t) be the Eulerian polynomial defined by
∞
X
k=0
(k + 1)ntk= En(t) (1 − t)n+1, and let Bn be the nth Bernoulli number. Show that
(En+1(t) − (1 − t)n)Bn
is a polynomial with integer coefficients.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
It is known that for n ≥ 1,
En(t) =
n
X
k=0
n k
tk
where
n k
=
k
X
j=0
(−1)jn + 1 j
(k + 1 − j)n
is the so-called Eulerian number which gives the number of permutations of {1, 2, . . . , n} having k permutation ascents.
The nth Bernoulli number Bn is a rational number. If n is odd then B1 = −1/2 and Bn = 0 for n > 1. If n is even then B0= 1 and for n > 0 the denominator is given by the product of the primes p such that p − 1 divides n.
Therefore for n ∈ {0, 1},
(E1(t) − 1)B0= 0, and (E2(t) − (1 − t))B1= −t.
Moreover, the polynomial (En+1(t) − (1 − t)n)Bn ∈ Z[x] for all n ≥ 2 iff for any prime p such that p − 1 divides n = 2m
2m + 1 k
− (−1)k2m k
≡ 0 (mod p).
Indeed, by Fermat’s little theorem, for any integer x, x2m+1= xq(p−1)+1≡ x (mod p) which implies
2m + 1 k
=
k
X
j=0
(−1)j2m + 2 j
(k + 1 − j)2m+1
≡
k
X
j=0
(−1)j2m + 2 j
(k + 1 − j) (mod p)
= (k + 1)
k
X
j=0
(−1)j2m + 2 j
+ (2m + 2)
k
X
j=1
(−1)j−12m + 1 j − 1
= (k + 1)(−1)k2m + 1 k
+ (2m + 2)(−1)k−1
2m k − 1
= (−1)k2m k
.
