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A094303 a(1) = 1, a(2) = 2, and a(n+1) = a(n) * sum of all previous terms up to a(n-1) for n >= 2.

%I A094303 #29 Sep 02 2024 01:18:51
%S A094303 1,2,2,6,30,330,13530,5019630,69777876630,351229105131280530,
%T A094303 24509789089304573335878465330,
%U A094303 8608552999157278550998626549630446732052243030
%N A094303 a(1) = 1, a(2) = 2, and a(n+1) = a(n) * sum of all previous terms up to a(n-1) for n >= 2.
%C A094303 From _Petros Hadjicostas_, May 11 2020: (Start)
%C A094303 _R. J. Mathar_'s conjecture is correct and this is identical to A064847 starting at n = 3. To see why this is the case, consider the sequences u(n) and v(n) defined by u(1) = v(1) = 1, and u(k+1) = u(k) + v(k), v(k+1) = u(k)*v(k) for k >= 1. Then u(n) = A003686(n) and v(n) = A064847(n) for n >= 1.
%C A094303 Then v(n) = u(n+1) - u(n), and thus Sum_{k=1..n-1} v(k) = u(n) - u(1) = u(n) - 1 for n >= 2. Then v(n-1) + ... + v(3) + (v(2) + 1) + v(1) = u(n) for n >= 3, and hence v(n)*(v(n-1) + ... + v(3) + (v(2) + 1) + v(1)) = u(n)*v(n) = v(n+1).
%C A094303 Since v(1) = 1 = a(1) and v(2) + 1 = 2 = a(2), the sequence (v(1), v(2) + 1, v(3), ..., v(n), ...) is identical to the current sequence. Hence, a(n) = v(n) = u(n+1) - u(n) = A003686(n+1) - A003686(n) for n >= 3. (End)
%H A094303 Petros Hadjicostas, <a href="/A094303/b094303.txt">Table of n, a(n) for n = 1..17</a>
%F A094303 Conjecture: a(n) = A003686(n+1) - A003686(n) for n >= 3. - _R. J. Mathar_, Apr 24 2007
%t A094303 nxt[{t1_,t2_,a_}]:=Module[{c=t1*a},{t1+t2,c,c}]; Join[{1},NestList[nxt,{1,2,2},10][[All,2]]] (* _Harvey P. Dale_, Aug 30 2020 *)
%o A094303 (PARI) lista(nn) = { my(va = vector(nn)); va[1] = 1; va[2] = 2; for(n=3, nn, va[n] = va[n-1]*sum(k=1, n-2, va[k]);); va; } \\ _Petros Hadjicostas_, May 11 2020
%Y A094303 See A064847 for another version.
%Y A094303 Cf. A003686, A064183, A064526, A070231, A070233, A070234.
%K A094303 nonn
%O A094303 1,2
%A A094303 _Amarnath Murthy_, Apr 29 2004
%E A094303 More terms from _Gareth McCaughan_, Jun 10 2004