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a(n) = (1 + Sum_{j=1..K-2} a(n-j)*a(n-j-1))/a(n-K) with a(1),...,a(K)=1, where K=9.
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%I #15 Mar 17 2017 22:53:03

%S 1,1,1,1,1,1,1,1,1,8,15,134,2143,289304,620267775,179446568646374,

%T 111304924045117731444223,19973286693341774343274474438923641617,

%U 277890644791720860549000282813744291843989379146578350083761

%N a(n) = (1 + Sum_{j=1..K-2} a(n-j)*a(n-j-1))/a(n-K) with a(1),...,a(K)=1, where K=9.

%H Seiichi Manyama, <a href="/A283821/b283821.txt">Table of n, a(n) for n = 1..24</a>

%H Matthew Christopher Russell, <a href="http://www.math.rutgers.edu/~zeilberg/Theses/MatthewRussellThesis.pdf">Using experimental mathematics to conjecture and prove theorems in the theory of partitions and commutative and non-commutative recurrences</a>, PhD Dissertation, Mathematics Department, Rutgers University, May 2016; see <a href="https://pdfs.semanticscholar.org/fdeb/e20954dacb7ec7a24afe2cf491b951c5a28d.pdf">also</a>.

%t a[n_]:=If[n<10, 1, (1 + Sum[a[n - j] * a[n - j - 1], {j, 7}])/a[n - 9]]; Table[a[n], {n, 1, 20}] (* _Indranil Ghosh_, Mar 17 2017 *)

%o (PARI) a(n) = if(n<10, 1, (1 + sum(j=1, 7, a(n - j) * a(n - j - 1)))/a(n - 9));

%o for(n=1, 20, print1(a(n),", ")) \\ _Indranil Ghosh_, Mar 17 2017

%K nonn

%O 1,10

%A _N. J. A. Sloane_, Mar 17 2017

%E More terms from _Indranil Ghosh_, Mar 17 2017