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.

1, 2, 2, 6, 30, 330, 13530, 5019630, 69777876630, 351229105131280530, 24509789089304573335878465330, 8608552999157278550998626549630446732052243030

Offset

1,2

Comments

From Petros Hadjicostas, May 11 2020: (Start)

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.

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).

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)

Links

Petros Hadjicostas, Table of n, a(n) for n = 1..17

Formula

Conjecture: a(n) = A003686(n+1) - A003686(n) for n >= 3. - R. J. Mathar, Apr 24 2007

Mathematica

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 *)

Prog

(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

Crossrefs

See A064847 for another version.

Cf. A003686, A064183, A064526, A070231, A070233, A070234.

Sequence in context: A097801 A164347 A052584 * A117394 A267073 A003308

Adjacent sequences: A094300 A094301 A094302 * A094304 A094305 A094306

Keyword

nonn

Author

Amarnath Murthy, Apr 29 2004

Extensions

More terms from Gareth McCaughan, Jun 10 2004

Status

approved