Conjecture and prove $a_n$ for $n\ge 0$.
$a_n=\sum_{i=0}^{n-1}{{n-1}\choose {i}}a_ia_{(n-1)-i},n\ge 1; a_0 $ a fixed constant.
Conjecture and prove $a_n$ for $n\ge 0$.
$a_n=\sum_{i=0}^{n-1}{{n-1}\choose {i}}a_ia_{(n-1)-i},n\ge 1; a_0 $ a fixed constant.
On
Conjecture and prove $a_n$ for $n\ge 0$.
$a_n=\sum_{i=0}^{n-1}{{n-1}\choose {i}}a_ia_{(n-1)-i},n\ge 1; a_0 $ a fixed constant.
$a_n=na_{n-1}a_0=n!a_0^{n+1},n\ge 1;$
$a_{n-1}=(n-1)!a_0^n,n\ge 2$
$a_0=a_0,a_1=1(1!)a_0a_0=a_0^2,a_2=2(1!)a_0^2a_0=2a_0^3$
$n_0=1,n_1=2$(strong induction)
$a_0,a_1,a_2,a_{n-1},a_n$ are true.
$a_n=na_{n-1}a_0=n(n-1)!a_0^na_0$
$a_{n+1}=(n+1)n!a_0^{n+1}a_0=(n+1)!a_0^{n+2}$
Is this correct? Comments/suggestion? Thanks.
Hint: Let's calculate more of the sequence: $$ a_0, a_0^2, 2a_0^3, 6a_0^4, 24a_0^5, 120a_0^6, \ldots $$ Perhaps you recognize the sequence $1,1,2,6,24,120$. If you don't, you can try plugging it in the Online Encyclopedia of Integer Sequences. Once you recognize the sequence, you can come up with a formula for $a_i$, which will be very easy to prove by induction.