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Let

$f(1)=M_1$,

$f(2)=M_1+2M_2$,

$f(3)=2M_1+4M_2+3M_3$,

$f(4)=6M_1+12M_2+9M_3+4M_4$,

$f(5)=24M_1+48M_2+36M_3+16M_4+5M_5$

$....$

How to find $f(p)$ for $p>0?$

Is there a method to find such recurrence relation?

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  • $\begingroup$ How is $f(n)$ defined? i.e. what's the pattern on the coefficients? $\endgroup$
    – coffeemath
    Jan 18, 2018 at 12:58
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    $\begingroup$ Shouldn't it be: $$f(5)=24M_1+48M_2+36M_3+\color{red}{16}M_4+5M_5$$ If so, the pattern is clear enough. $\endgroup$ Jan 18, 2018 at 13:00
  • $\begingroup$ Yes, its 16. A typo. $\endgroup$
    – Harish
    Jan 18, 2018 at 13:02
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    $\begingroup$ The recurrence is thus: $$f(1)=M_1$$ $$f(n+1)=nf(n)+(n+1)M_{n+1}$$ $\endgroup$ Jan 18, 2018 at 13:05
  • $\begingroup$ I'll risk to say, that no such method exists, and you are just guessing what they want. With some guesses being a bit more "natural" or expected. $\endgroup$
    – dEmigOd
    Jan 18, 2018 at 13:07

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