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# The nth Bernoulli number Bn is a rational number

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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)jn + 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)k2m 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)j2m + 2 j



(k + 1 − j)2m+1

k

X

j=0

(−1)j2m + 2 j



(k + 1 − j) (mod p)

= (k + 1)

k

X

j=0

(−1)j2m + 2 j



+ (2m + 2)

k

X

j=1

(−1)j−12m + 1 j − 1



= (k + 1)(−1)k2m + 1 k



+ (2m + 2)(−1)k−1

 2m k − 1



= (−1)k2m k

 .



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