# Combinatorial Theory Seminar Eindhoven University of by Jacobus H. van Lint (auth.)

By Jacobus H. van Lint (auth.)

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The proof we may always assume no rowsum or columnsum case). in both cases. 7) ~(r i) H _ i=l~ ~(r i - I) < i= 1 ~(r i We first remark = 2. Now (all that it is easily checked = 1. Then, taking ~ {~(2)} e all > 2) . 7) may possibly be proved by induction We a s s u m e ~ ( 1 ) r. I) r . 7) for ! on ~ or on the r i s or both. ,~). ,r %) ~ + 1. If r I = r 2 = ... 7) holds. 8) holds = 3 I/3 ~(r I - l)~(r 2 - I) ... 8) for ~ + i instead side is increased by I) < ~(rl)~(r2) of % and replace ... ~(r~) .

1). 6) is d e r i v e d t) -I. 2 u -I + n l o g ( l + u ) < "6 + log ~ u < -6172 u-I + ( n - l)u + log u . 3. 2) takes no more work than the proof of itself[ Partitions, s e r i e s cmd products. 43 Problem 2 that p(n) ment using series partitions and products of n into parts n 2 2. It follows (cf. is convex. We give a proof of this state- [3]). Let c(0) := I and c(n) := the number of > 3. Then c(1) = c(2) = 0, c(3) = I and c(n) that for n k 2 tbe coefficient of x n in the expansion k c(n-I) for of co (I + x) -I is positive.

The r e s u l t log p(n) t I-t < ~1 value 72 u-1 + nu for u = ~(6n) -! 2. S u b s t i t u t i n g result. is much w e a k e r of the proof n. T h e r e f o r e that f(t) its m i n i m a l the r e q u i r e d inequality slight m o d i f i c a t i o n I ~2 < -6- Hence (0 < u < ~) and find expression of u we find this > p(n)t n for 0 < t < I, n ( ~. 1). 1). 6) is d e r i v e d t) -I. 2 u -I + n l o g ( l + u ) < "6 + log ~ u < -6172 u-I + ( n - l)u + log u . 3. 2) takes no more work than the proof of itself[ Partitions, s e r i e s cmd products.