In this paper, Hamilton showed that $n\ge 47$ suffices for removing the $n-1,n-2,n-3,n-4,n-5$ degree term from the degree $n$ polynomial without solving equation higher than the quintic by a $n-1$ degree Tschirnhaus transformation, $$y=x^{n-1}+c_1x^{n-2}+c_2x^{n-3}+\dots+c_{n-1}$$ For the equation $$x^n+a_1x^{n-1}+a_2x^{n-2}+a_3x^{n-3}+\dots+a_n=0$$ to yields $$y^n+C_1y^{n-1}+C_2y^{n-2}+C_3y^{n-3}+\dots+C_n=0$$ By determining the coefficients $a_i$, one can solve, $$C_1=C_2=C_3=C_4=C_5=0$$ By equations not higher than quintic, or not higher than degree $n-1$.
One can prove that the $C_i$ will have the form, $$C_1=\alpha_1\\ C_2=\alpha_2c_{n-2}+\alpha_3\\ C_3=\alpha_4c_{n-2}^2+\alpha_5c_{n-2}+\alpha_6\\ C_4=\alpha_7c_{n-2}^3+\alpha_8c_{n-2}^2+\alpha_9c_{n-2}+\alpha_{10}$$ If $a_1=a_2=a_3=a_4=0$ (this is doable in radicals) with the first 4 $C_i$ missing their leading term $(c_{n-2},c_{n-2}^2,c_{n-2}^3,c_{n-2}^4)$ respectively, and the $\alpha_i$ are polynomials in other parameters.
Given that $\alpha_1,\alpha_2,\alpha_4,\alpha_7$ are linears, $\alpha_3=\alpha_5=\alpha_8$ are quadratics, $\alpha_6,\alpha_9$ are cubics and $\alpha_{10}$ are a quartic, one use the formula described in the paper: $$m(4,3,2,1)\le1+m(9,5,2)\le2+m(15,6,1)\le3+m(21,6)\color{red}{\le4+m(26,5)\le5+m(30,4)\le6+m(33,3)\le7+m(35,2)\le8+m(36,1)\le9+m(36)=9+36=45}$$ We ignore the red part and focus on the $m(21,6)$ which is $21$ linears and $6$ quadratics.
To achieve $m(4,3,2,1)\le42$ by solving equations not higher than quintic, then $m(21,6)\le39$, and one needs to solve for the $6$ quadratics and $21$ linears by radicals in $39$ variables. After that, the system is solved by equations not higher than quintic. It can be shown possible by splitting variables into $3$ parts, focus on a quadratic and split it into: $$A=A_{200}+A_{020}+A_{002}+A_{110}+A_{101}+A_{011}+A_{100}+A_{010}+A_{001}+A_{000}$$ With $A_{ijk}$ have $i,j,k$ being the total degree of the first part, second part and third part of the splitted variables respectively. The first part is used to solve and leave a free parameter: $$A_{200}=0$$ The second part is used to solve and also leave a free parameter: $$A_{020}=A_{110}=0$$ The third part is used to solve the most quadratics and linears: $$A_{002}+A_{001}+A_{000}=A_{101}+A_{100}=A_{011}+A_{010}=0$$ And the $2$ free parameters is used to solve $2$ quadratics. In particular $$m(a,b+2)\le 2+m(a+2b,b)$$ $$\implies m(21,6)\le 2+m(29,4)\le 4+m(33,2)=4+35=39$$ Therefore it is possible to reduce equations of degree $44$ in radicals, however a numerical example seems to be way too complex for this post.
To achieve $m(4,3,2,1)\le 19$ by solving equations not higher than degree $20$, we first free a parameter from the linears, quadratics, cubics and eliminate the quartic $C_4$ along with $C_5$ using extra help from the parameter $c_{19}$. One can easily prove that after splitting the variables into $2$ parts and split the equations $\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5,\alpha_6$ into parts, the remaining equations will be: $$m(3+2+1,2+1,1)=m(6,3,1)\le 1+m(9,3)=1+9+3=13$$ The few last steps is solving equations as worse as degree $2^3=8$,and thus $$m(4,3,2,1)\le 14$$ Which implies a Tschirnhaus transformation of degree $15$ also suffices for reducing equations of degree $21$, but we will stick with transformation of degree $20$.
In particular, I want to reduce the sample equation: $$x^{21}+3x^{16}-x^{15}+x^{14}+2x^{13}+2x^{12}+x^{11}\\-x^{10}-x^9+x^8+4x^7-3x^6-2x^5-x^4+2x^3+x^2+x+1=0$$ Into: $$y^{21}+C_6y^{15}+C_7y^{14}+\dots=0$$ With all the $c_i$'s determined without solving equations higher than degree $20$.
Question: Will there be a numerical example to reduce the sample equation with constrains above?
