Numbers Expressed as Sums of Binomial Coefficients
== Definition == Let $n \in \Z_{> 0}$ be a (strictly) positive integer. Then for all $k \in \Z_{> 0}$, it is possible to express $k$ uniquely in the form: {{begin-eqn}} {{eqn | l = k       | r = \sum_{j \mathop = 1}^n \dbinom {k_j} j       | c =  }} {{eqn | r = \dbinom {k_1} 1 + \dbinom {k_2} 2 + \cdots + \dbinom {k_n} n       | c =  }} {{end-eqn}} such that $0 \le k_1 < k_2 < \cdots < k_n$. == Proof == === Existence of Representation === Proof by induction: For all $k \in \Z_{\ge 0}$, let $\map P k$ be the proposition that it is possible to express $k$ in the form: :$\displaystyle k = \sum_{j \mathop = 1}^n \dbinom {k_j} j$ such that $0 \le k_1 < k_2 < \cdots < k_n$. ==== Basis for the Induction ==== $\map P 0$ is true, as $\displaystyle 0 = \sum_{j \mathop = 1}^n \dbinom {j - 1} j$. This is our basis for the induction. ==== Induction Hypothesis ==== Now we need to show that, if $\map P m$ is true, where $r \ge 2$, then it logically follows that $\map P {m + 1}$ is true. So this is our induction hypothesis: $m$ can be expressed in the form: :$\displaystyle m = \sum_{j \mathop = 1}^n \dbinom {m_j} j$ where $0 \le m_1 < m_2 < \cdots < m_n$. Then we need to show: $m + 1$ can be expressed in the form: :$\displaystyle m = \sum_{j \mathop = 1}^n \dbinom {k_j} j$ where $0 \le k_1 < k_2 < \cdots < k_n$. ==== Induction Step ==== This is our induction step: Suppose the first $c$ $m_j$ are consecutive, that is: :$\forall \, j \in \N: 1 \le j < c: m_{j + 1} - m_j = 1$ and $m_{c + 1} - m_c > 1$ Then: {{begin-eqn}} {{eqn | l = m + 1       | r = \sum_{j \mathop = 1}^n \dbinom {m_j} j + 1       | c = from the induction hypothesis }} {{eqn | r = \sum_{j \mathop = c + 1}^n \dbinom {m_j} j + \sum_{j \mathop = 1}^c \dbinom {m_j} j + 1       | c =  }} {{eqn | r = \sum_{j \mathop = c + 1}^n \dbinom {m_j} j + \sum_{j \mathop = 1}^c \dbinom {m_1 + j - 1} j + \dbinom {m_1 + 0 - 1} 0       | c = Binomial Coefficient with Zero }} {{eqn | r = \sum_{j \mathop = c + 1}^n \dbinom {m_j} j + \dbinom {m_1 + c} c       | c = Sum of r+k Choose k up to n }} {{eqn | r = \sum_{j \mathop = c + 1}^n \dbinom {m_j} j + \dbinom {m_c - 1} c       | c =  }} {{eqn | r = \sum_{j \mathop = c + 1}^n \dbinom {m_j} j + \dbinom {m_c - 1} c + \sum_{j = 1}^{c - 1} \dbinom {j - 1} j       | c =  }} {{end-eqn}} Since $0 \le m_1 < m_2 < \cdots < m_n$, we must have $m_j \ge j - 1$ for each $j \le n$. Hence $0 \le 0 < 1 < \cdots < c - 2 < m_c - 1 < m_{c + 1} < \cdots < m_n$. Thus the expression above satisfy the conditions. So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 0}: \sum_{k \mathop = 0}^n \binom {r + k} k = \binom {r + n + 1} n$ {{qed|lemma}} === Uniqueness of Representation === Suppose $k$ can be expressed in the form: :$\displaystyle k = \sum_{j \mathop = 1}^n \dbinom {k_j} j = \sum_{j \mathop = 1}^n \dbinom {m_j} j$ where $0 \le k_1 < k_2 < \cdots < k_n$ and $0 \le m_1 < m_2 < \cdots < m_n$.  $k_i$ and $m_i$ are not all equal. Let $c$ be the largest integer such that $k_c \ne m_c$.  assume $k_c < m_c$. Then: {{begin-eqn}} {{eqn | l = k       | r = \sum_{j \mathop = 1}^n \dbinom {k_j} j }} {{eqn | r = \sum_{j \mathop = c + 1}^n \dbinom {m_j} j + \sum_{j \mathop = 1}^c \dbinom {k_j} j }} {{eqn | o = \le       | r = \sum_{j \mathop = c + 1}^n \dbinom {m_j} j + \sum_{j \mathop = 1}^c \dbinom {k_c - c + j} j       | c = $0 \le k_1 < k_2 < \cdots < k_n$ }} {{eqn | r = \sum_{j \mathop = c + 1}^n \dbinom {m_j} j + \dbinom {k_c - c + c + 1} c - \dbinom {k_c - c} 0       | c = Sum of r+k Choose k up to n }} {{eqn | o = \le       | r = \sum_{j \mathop = c + 1}^n \dbinom {m_j} j + \dbinom {m_c} c - 1       | c = $k_c + 1 \le m_c$ }} {{eqn | o = <       | r = \sum_{j \mathop = c + 1}^n \dbinom {m_j} j + \dbinom {m_c} c       | c =  }} {{eqn | o = \le       | r = \sum_{j \mathop = 1}^n \dbinom {m_j} j       | c = $\dbinom {m_j} j \ge 0$ }} {{eqn | r = k       | c =  }} {{end-eqn}} which is a contradiction. Thus the representation is unique.  == Examples == {{:Numbers Expressed as Sums of Binomial Coefficients/Examples}} == Also see == * Definition:Combinatorial Number System == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Inverse of Stirling's Triangle expressed as Matrix|next = Numbers Expressed as Sums of Binomial Coefficients/Examples/n = 3}}: $\S 1.2.6$: Binomial Coefficients: Exercise $56$