Let $N(n)$ be a set of integers, which can be presented using first $n$ Egyptian fractions:

$$ N(n):=\{m\in\mathbb{Z}:\ \ m=\sum_{i=1}^n\frac{\epsilon_i}{i},\ \epsilon_i=0\ \text{or}\ 1\} $$

I want to write a code in Mathematica that gives $N(n)$, but I think A[e,n] does not define what I need above: for example

A[e_,n_]:=Sum[e/i,{e,0,1},{i,1,n}];
n=1000;    
Sum[If[IntegerQ[A[e,n]]==True,m,0],{m,1,10}]

Thanks.

P.S. example

\begin{align} 1=&1\\ 2=&1+\frac12+\frac13+\frac04+\frac05+\frac16 \end{align} therefore $N(6)=\{1,2\}$

Let $N(n)$ be a set of integers, which can be presented using first $n$ Egyptian fractions:

$$ N(n):=\{m\in\mathbb{Z}:\ \ m=\sum_{i=1}^n\frac{\epsilon_i}{i},\ \epsilon_i=0\ \text{or}\ 1\} $$

I want to write a code in Mathematica that gives $N(n)$, but I think A[e,n] does not define what I need above: for example

A[e_,n_]:=Sum[e/i,{e,0,1},{i,1,n}];
n=1000;    
Sum[If[IntegerQ[A[e,n]]==True,m,0],{m,1,10}]

Thanks.

P.S. example

\begin{align} 0=&0\\ 1=&1\\ 2=&1+\frac12+\frac13+\frac04+\frac05+\frac16 \end{align} therefore $N(6)=\{0,1,2\}$

Let

$$ N(n):=\{m\in\mathbb{Z}:\ \ m=\sum_{i=1}^n\frac{\epsilon_i}{i},\ \epsilon_i=0\ \text{or}\ 1\} $$

I want to write a code in Mathematica that gives the above result, but I think A[e,n] does not define what I need above: for example

A[e_,n_]:=Sum[e/i,{e,0,1},{i,1,n}];
n=1000;    
Sum[If[IntegerQ[A[e,n]]==True,m,0],{m,1,10}]

Thanks.

P.S. example

\begin{align} 1=&1\\ 2=&1+\frac12+\frac13+\frac04+\frac05+\frac16 \end{align} therefore $N(6)=\{1,2\}$

Let $N(n)$ be a set of integers, which can be presented using first $n$ Egyptian fractions:

$$ N(n):=\{m\in\mathbb{Z}:\ \ m=\sum_{i=1}^n\frac{\epsilon_i}{i},\ \epsilon_i=0\ \text{or}\ 1\} $$

I want to write a code in Mathematica that gives $N(n)$, but I think A[e,n] does not define what I need above: for example

A[e_,n_]:=Sum[e/i,{e,0,1},{i,1,n}];
n=1000;    
Sum[If[IntegerQ[A[e,n]]==True,m,0],{m,1,10}]

Thanks.

P.S. example

\begin{align} 1=&1\\ 2=&1+\frac12+\frac13+\frac04+\frac05+\frac16 \end{align} therefore $N(6)=\{1,2\}$