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Let en,k denote the number of ways to partition them in k affinity groups with no two members of a group seated next to each other

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Problem 11151

(American Mathematical Monthly, Vol.112, April 2005) Proposed by D. Knuth (USA).

Suppose n people are sitting at a circular table. Let en,k denote the number of ways to partition them in k affinity groups with no two members of a group seated next to each other. For example e4,3= 2, e5,3= 5, and e6,3= 10. For k ≥ 2 find the generating function fk(z) =P

n=0en,kzn. Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

We easily note that

f2(z) = z2+ z4+ z6+ z8+ · · · =

X

j=1

z2j= z2 1 − z2.

We consider the number of ways to partition n persons in k affinity groups at a linear table. It is the sum of the number of partitions with the 1st and the nth persons in different groups, that is just en,k, and the number of partitions with the 1st and the nth persons in the same group, say dn,k. Assume that n ≥ k ≥ 3, then in order to compute en,k we either put the nth person by itself or we distribute him in one of the affinity groups of n − 1 persons which does not contain the 1st person or the (n − 1)th person:

en,k= [en−1,k−1+ dn−1,k−1] + [(k − 2) · en−1,k+ (k − 1) · dn−1,k] .

In order to compute dn,k we put the nth person in the same group of the 1st one provided it does not contain the (n − 1)th person:

dn,k= 1 · en−1,k+ 0 · dn−1,k= en−1,k. Hence, eliminating dn,k, we find the following recurrence for n ≥ k ≥ 3

en,k= en−1,k−1+ en−2,k−1+ (k − 2)en−1,k+ (k − 1)en−2,k.

Since en,k = 0 for k > n, multiplying this equation by zn and summing up for n from 3 to ∞ we get fk(z) = (z + z2)fk−1(z) + (k − 2)z + (k − 1)z2 fk(z)

that is, for k ≥ 3,

fk(z) = (z + z2)fk−1(z)

1 − (k − 2)z − (k − 1)z2 = z(1 + z)fk−1(z)

(1 + z)2−kz(1 + z) = zfk−1(z) 1 − (k − 1)z. Finally

fk(z) =

k

Y

j=3

z 1 − (j − 1)z

f2(z) = z 1 + z

k−1

Y

j=1

z 1 − jz. For example

f3(z) = z3

(1 − z2)(1 − 2z)= z3+ 2z4+ 5z5+ 10z6+ 21z7+ 42z8+ 85z9+ · · · .

 Remark. It is interesting to note that

fk(z) = z

1 + zgk−1(z) k ≥ 2 where

gk(z) =

X

n=0

n k

 zn =

k

Y

j=1

z 1 − jz

is the generating function of the Stirling numbers of the second kind, i. e. the numbers of partitions of a set of n elements in k non-empty subsets (see Concrete Mathematics by Graham-Knuth-Patashnik).

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