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If p(n) is the number of integer partitions of n then the generating function of this sequence is P(z

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Problem 11237

(American Mathematical Monthly, Vol.113, August-September 2006) Proposed by E. Deutsch (USA).

Prove that the number of2s occurring in all partitions of n is equal to the number of singletons occurring in all partitions of n −1, where a singleton in a partition is a part occurring once.

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

If p(n) is the number of integer partitions of n then the generating function of this sequence is

P(z) =

∞

X

n=0

p(n)zn=

∞

Y

n=1

1

(1 − zn) = 1 + z + 2z2+ 3z3+ 5z4+ 7z5+ 11z6+ 15z7+ 22z8+ o(z8).

In order to compute the number of 2s we introduce a parameter t 1 − z2

1 − tz2 ·P(z) then the generating function which counts the 2s is

T(z) =

∞

X

n=0

t(n)zn= d dt

 1 − z2 1 − tz2 ·P(z)

 t=1

= z2

1 − z2·P(z) = z2+z3+3z4+4z5+8z6+11z7+19z8+o(z8).

On the other hand, the generating function which counts the singletons that are equal to n ≥ 1 is zn(1 − zn) · P (z)

therefore the generating function which counts all the singletons is

S(z) =

∞

X

n=0

s(n)zn=

∞

X

n=1

zn(1 − zn)

!

·P(z) =

 z

1 − z− z2 1 − z2



·P(z) = z

1 − z2 ·P(z).

Since T (z) = zS(z) then

t(n) = s(n − 1).



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