The multinomial coefficients can be used to expand expressions of the form ${\left( {{x_1} + {x_2} + {x_3} + ...} \right)^n}$ in the basis of monomial symmetric polynomials (MSP). For example,
$$\begin{array}{c}{\left( {{x_1} + {x_2} + {x_3} + ...} \right)^5} = {M_{\left( 5 \right)}} + \frac{{5!}}{{4!}}{M_{\left( {4,1} \right)}} + \frac{{5!}}{{3!2!}}{M_{3,2}} + \frac{{5!}}{{3!}}{M_{3,1,1}} + \frac{{5!}}{{2!2!}}{M_{\left( {2,2,1} \right)}} + \frac{{5!}}{{2!}}{M_{\left( {2,1,1,1} \right)}} + 5!{M_{\left( {1,1,1,1,1} \right)}}\\ = {M_{\left( 5 \right)}} + 5{M_{\left( {4,1} \right)}} + 10{M_{3,2}} + 20{M_{3,1,1}} + 30{M_{\left( {2,2,1} \right)}} + 60{M_{\left( {2,1,1,1} \right)}} + 120{M_{\left( {1,1,1,1,1} \right)}}\end{array}$$
If each coefficient is multiplied by the number of permutations of the exponents of the MSP, we get the following sequence:
$$1,\;5\frac{{2!}}{{1!1!}},\;10\frac{{2!}}{{1!1!}},\;20\frac{{3!}}{{1!2!}}\;,30\frac{{3!}}{{2!1!}}\;,60\frac{{4!}}{{1!3!}}\;,120\frac{{5!}}{{5!}}$$
Which simplifies to $1,10,20,60,90,240,120$.
For values of $n$ different from 5 it follows the pattern:
- [1] $1;$
- [2] $1; 2;$
- [3] $1; 6; 6;$
- [4] $1; 8, 6; 36; 24;$
- [5] $1; 10, 20; 60, 90; 240; 120;$
These numbers appear to be generated by the polynomials in the coefficients of the function:
$$\frac{{\sum\limits_{k = 1}^\infty {{x_k}\frac{{{t^k}}}{{k!}}} }}{{1 - \sum\limits_{k = 1}^\infty {{x_k}\frac{{{t^k}}}{{k!}}} }} = {x_1}\frac{t}{{1!}} + \left( {{x_2} + 2x_1^2} \right)\frac{{{t^2}}}{{2!}} + \left( {{x_3} + 6{x_1}{x_2} + 6x_1^3} \right)\frac{{{t^3}}}{{3!}} + \left( {{x_4} + 6x_2^2 + 8{x_1}{x_3} + 36{x_2}x_1^2 + 24x_1^4} \right)\frac{{{t^4}}}{{4!}} + \left( {{x_5} + 10{x_1}{x_4} + 20{x_3}{x_2} + 60x_1^2{x_3} + 90{x_1}x_2^2 + 240x_1^3{x_2}} + 120x_1^5 \right)\frac{{{t^5}}}{{5!}} + ...$$
I have tried many things but have had no success in proving this relationship. I am wondering if anybody can offer a proof or further insight. Thank you in advance.
