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# Letp be the partition counting function and let g be the function given by g(n

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Problem 11795

(American Mathematical Monthly, Vol.121, August-September 2014) Proposed by M. Merca (Romania).

Letp be the partition counting function and let g be the function given by g(n) = 12⌈n/2⌉⌈(3n+1)/2⌉.

LetA(n) be the set of non-negative integer triples (i, j, k) such that g(i) + j + k = n. Prove that for n ≥ 1,

p(n) = 1 n

X

(i,j,k)∈A(n)

(−1)⌈i/2⌉−1g(i)p(j)p(k).

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

By the pentagonal number theorem

G(x) :=

Y n=1

(1 − xn) = X n=0

(−1)⌈n/2⌉xg(n).

Let P be the generating function for p(n), then

P (x) = X n=1

p(n)xn = 1 G(x), Moreover,

xG(x) = X n=1

(−1)⌈n/2⌉g(n)xg(n)= −G(x) X n=1

nxn 1 − xn and

xP(x) = X n=1

np(n)xn= 1 G(x)

X n=1

nxn 1 − xn.

Hence xP(x) = −xG(x)(P (x))2 and by extracting the coefficient of xn for n ≥ 1, we obtain np(n) = − X

(i,j,k)∈A(n)

(−1)⌈i/2⌉g(i)p(j)p(k).

where

A(n) := {(i, j, k) ∈ N3 : g(i) + j + k = n}.

 Note that in the original statement

A(n) := {(i, j, k) ∈ (N+)3 : g(i) + j + k = n}

but in this case the formula does not work because j and k do not assume the value 0 (recall that p(0) = 1).

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