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Let be $a(k,m),k,m\geq 0$ an infinite matrix then the set $$T_k=\{(a(k,0),a(k,1),...,a(k,i),...),(a(k,0),a(k+1,1),...,a(k+i,i),...)\}$$is called angle of matrix

$a(k,0)$ is edge of $T_k$

$a(k,i),a(k+i,i),i>0$ are conjugate elements of $T_k$

$(a(k,0),a(k,1),...,a(k,i),...)$ is horizontal ray of $T_k$

$(a(k,0),a(k+1,1),...,a(k+i,i),...)$is diagonal ray of $T_k$

Elements of diagonal ray of $T_0$ are $1$

Elements above diagonal ray of $T_0$ are $0$

Elements of edge of $T_k,k>0$ are $0$

Each element of diagonal ray of $T_k,k>0$ is sum of his conjugate and elements of horizontal ray of $T_k$ that are placed on left.

Prove that sum of elements of row $k$ is partition function $p(k)$

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Where does this come from, please? –  Gerry Myerson Dec 12 '12 at 12:30
When you edit a question such that parts of existing answers no longer make sense, please mark the edit as such. –  joriki Dec 12 '12 at 18:12
I correct a typo (1 is replaced by 0) your observation is correct. –  Adi Dani Dec 12 '12 at 18:39

1 Answer 1

up vote 2 down vote accepted

This is a very unnecessarily complicated, ambiguous and partly erroneous reformulation of a simple recurrence relation for the number $a(k,m)$ of partitions of $k$ with greatest part $m$. It works out if the following changes and interpretations are made:

  • both instances of $k\gt1$ are replaced by $k\ge1$,
  • "his conjugate" is interpreted as "its upper conjugate" (each entry has two conjugates), and
  • "on the left" is interpreted as "to the left of its upper conjugate".

The resulting recurrence relation is

$$ a(k,m)=\sum_{i=1}^ma(k-m,i)\;, $$

which simply states that a partition of $k$ with greatest part $m$ arises by adding a part $m$ to any partition of $k-m$ with greatest part not greater than $m$.

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