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a(n) = Product_{k = 0..n-1} (a(k) + a(n-1-k)), with a(0) = 1.
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%I #16 Jan 02 2020 16:13:27

%S 1,2,9,400,19456921,1101216948902114953248,

%T 76796373204229717290826972582321984854855228022915711475735049

%N a(n) = Product_{k = 0..n-1} (a(k) + a(n-1-k)), with a(0) = 1.

%H Andrew Howroyd, <a href="/A049299/b049299.txt">Table of n, a(n) for n = 0..8</a>

%F lim_{m -> oo} log(a(m+1))/log(a(m)) exists and equals 3. - _Roland Bacher_, Sep 06 2004.

%e a(3)=400 because 400=(1+9)*(2+2)*(9+1).

%o (PARI) a(n)={my(v=vector(n+1)); for(n=1, #v, v[n]=prod(k=1, n-1, v[k]+v[n-k])); v[#v]} \\ _Andrew Howroyd_, Jan 02 2020

%Y Cf. A000108 (Catalan numbers) where a(0) = 1, a(n) = Sum_{k=0..n-1} a(k)*a(n-k), A000012 (constant 1) where a(0) = 1, a(n) = Product_{k=0..n-1} a(k)*a(n-k) and A025192 (2*3^(n-1)) where a(0) = 1, a(n) = Sum_{k=0..n-1} a(k)+a(n-k). - _Henry Bottomley_, May 16 2000

%K easy,nonn

%O 0,2

%A _Leroy Quet_

%E Offset corrected and terms a(6) and beyond from _Andrew Howroyd_, Jan 02 2020