Problem 11782
(American Mathematical Monthly, Vol.121, June-July 2014)
Proposed by M. Merca (Romania).
Prove that
∞
X
k=1
(−1)k−1kpk(n − k(k + 1)/2) =
∞
X
k=−∞
(−1)kτ (n − k(3k − 1)/2).
Here, pk(n) denotes the number of partitions of n in which the greatest part is less than or equal to k, and n is the number of divisors of n.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
By the q-binomial theorem
n
X
k=0
n k
zkq(k+12 ) =
n
Y
k=1
(1 + zqk) where n k
= Qn
j=n−k+1(1 − qj) Qk
j=1(1 − qj) .
Let |z| < 1, then by taking the limit as n goes to infinity, we obtain
∞
X
k=0
zkq(k+12 ) Qk
j=1(1 − qj)=
∞
Y
k=1
(1 + zqk).
Now we take the derivative of both sides with respect to z,
∞
X
k=1
kzk−1q(k+12 ) Qk
j=1(1 − qj)=
∞
Y
k=1
(1 + zqk)
∞
X
k=1
qk 1 + zqk,
and, as z goes to −1, we have
∞
X
k=1
(−1)k−1kq(k+12 ) Qk
j=1(1 − qj) =
∞
Y
k=1
(1 − qk)
∞
X
k=1
qk 1 − qk.
By using the pentagonal number theorem, we can write the above identity in the following way
∞
X
k=1
(−1)k−1kq(k+12 )
∞
X
j=0
pk(j)qj =
∞
X
k=−∞
(−1)kxk(3k−1)2
!
·
∞
X
k=1
τ (k)qk
! .
Finally, by extracting the coefficient of xn of both sides, we get
∞
X
k=1
(−1)k−1kpk(n − k(k + 1)/2) =
∞
X
k=−∞
(−1)kτ (n − k(3k − 1)/2).
