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Let $r>4$ be a positive integer. Let $p_{n}$ be the sequence of prime numbers with $n≥3$. I wante to solve this difference equation:

$$H_{n+1}-(r^{2n+1}+1)H_{n}=2-r^{2n+1}p_{n}$$ where $H_{n}$ is the unknown and $H_{3}=5$. I have no idea to start.

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  • $\begingroup$ Do you mean $p_3 =2,p_4=3,p_5=5$ etc. ? $\endgroup$ – Jeb Sep 29 '15 at 18:54
  • $\begingroup$ @Jeb: Yes, this is the case. $\endgroup$ – DER Sep 29 '15 at 18:55
  • $\begingroup$ What are you assuming for $H_3$ ?, i.e.the initial data $\endgroup$ – Jeb Sep 29 '15 at 18:56
  • $\begingroup$ @Jeb: We can take it 5. $\endgroup$ – DER Sep 29 '15 at 18:57
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Notice you have $$H_{n+1} = (r^{2n+1}+1) H_n + (2 - r^{2n+1 } p_n) $$ The first few terms are $$ H_4 = ( r^7 +1) H_3 + (2 - p_3r^{7} )$$ $$H_5 = ( r^9 +1) H_4 + (2 - p_4 r^9 ) = ( r^9 +1) ( r^7 +1)H_3 + ( r^9 +1)(2 - p_3r^{7} ) + (2 - p_4 r^9) $$ Thus if you iterate this you'll find $$ H_n = H_3\prod_{k=3}^{n-1} ( r^{2k+1} +1) + \sum_{k=3}^{n-1}(2-r^{2k+1}p_k) \prod_{l=k}^{n-2} ( r^{2l+3} +1) $$

I have changed $n$ by $k$ in the indice of the prime number $p_n$ in the last equation.

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  • $\begingroup$ Changed the bounds. $\endgroup$ – Jeb Sep 29 '15 at 19:12
  • $\begingroup$ @ jeb: Ok I will verify that by direct calculations. $\endgroup$ – DER Sep 29 '15 at 19:17
  • $\begingroup$ @ jeb: It seems also the expression is wrong even for $n=5$.. $\endgroup$ – DER Sep 30 '15 at 9:09
  • $\begingroup$ It checks out... Do you understand the notation I'm using $\endgroup$ – Jeb Sep 30 '15 at 18:31

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