Skip to main content
2 of 2
edited body
Alexey Ustinov
  • 12.3k
  • 7
  • 87
  • 119

Extended comment.

The relation $$0=P_0(n) S_n+ P_1(n) S_{n-1}+ P_2(n) S_{n-2}+ P_3(n) S_{n-3}+ P_4(n) S_{n-4}$$ for all $n\geqslant 4$ is equivalent to $$ 0=\sum_{n\geqslant 4} (P_0(n) S_n+ P_1(n) S_{n-1}+ P_2(n) S_{n-2}+ P_3(n) S_{n-3}+ P_4(n) S_{n-4})x^n. $$ Denote $\sum_{n=0}^\infty S_n x^n=f(x)$. Then $f'(x)=\sum_{n\geqslant 0} S_n\cdot nx^{n-1}$, $f''(x)=\sum_{n\geqslant 0} S_n\cdot n(n-1)x^{n-2}$. Thus for $i\in \{0,1,2,3,4\}$ we have $$\sum_{n\geqslant 4} P_i(n)S_{n-i}x^n= \sum_{n\geqslant i} P_i(n)S_{n-i}x^n-\sum_{n=i}^3P_i(n)S_{n-i}x^i=\\ \sum_{n\geqslant 0} P_i(n+i)S_{n}x^{n+i}-\sum_{n=i}^3P_i(n)S_{n-i}x^i.$$ Denote $P_i(n+i)=a_i+b_in+c_in(n-1)$ and $\sum_{n=i}^3P_i(n)S_{n-i}x^i=g_i(x)$. Then $$\sum_{n\geqslant 0} P_i(n+i)S_{n}x^{n+i}-\sum_{n=i}^3P_i(n)S_{n-i}x^i=x^i(a_if(x)+b_ixf'(x)+c_ix^2f''(x))-g_i(x).$$ Thus the differential equation for $f$ has a form $$ q_0(x)f(x)+xq_1(x)f'(x)+x^2q_2(x)f''(x)=A(x) $$ for some polynomials $q_0,q_1,q_2$ of degree at most 4 and $A$ of degree at most 3.

Fedor Petrov
  • 108.9k
  • 9
  • 264
  • 459