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I'm reading Andrews' and Eriksson's text on Integer Partitions. On page 36, it is stated that P(n: parts = +1 or -1 mod6) = P(n: distinct parts = +1 or -1 mod3) (call this last partition P). Then the goal is to show that there is a bijection between P and P(n: 3-distinct parts, no consecutive multiples of 3)(call this last partition P1). The suggestion is to merge pairs of parts in P differing by at most two, starting from the smallest part. The authors suggest that consecutive multiples of 3 in P1 cannot appear in this way, but I don't see why this is so. For example, couldn't you map {1,2,5,7,10,11} in P to {1+2,5+7,10+11} in P1, giving three consecutive parts that are multiples of 3? Any help would be much appreciated.

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up vote 2 down vote accepted

No consecutive multiples of $3$ mean that any two multiples of $3$ that appear in the partition must differ by at least $6$. This is trivial because if you sort the values in the first partition and assume $a < b < c < d$ are some numbers in the partition then it is easy to see that $$(c+d)-(a+b) = (c-a) + (d-b) \geq 3+3 = 6$$ So you just misunderstood what the book meant about "consecutive multiples of 3". In your example $$(1+2,5+7,10+11) = (3,12,21)$$ aren't consecutive multiples of 3.

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Yes, I see now. Thank you very much – Adam Jun 7 '13 at 1:48

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