A132076 a(1)=1, a(2)=2. a(n), for every positive integer n, is such that Product_{k=1..n} (Sum_{j=1..k} a(j)) = Sum_{k=1..n} Product_{j=1..k} a(j).

1, 2, -6, -12, -240, -65280, -4294901760, -18446744069414584320, -340282366920938463444927863358058659840, -115792089237316195423570985008687907852929702298719625575994209400481361428480

Offset

1,2

Comments

There are an infinite number of sequences {a(k)}, with different values for a(1) and a(2) (a(1) must be 0 or 1; a(2) can be anything), where Product_{k=1..n} (Sum_{j=1..k} a(j)) = Sum_{k=1..n} Product_{j=1..k} a(j), for all positive integers n. Setting a(1) to 1 and a(2) to 2 results in the sequence here.

All sequences (not necessarily integer sequences) with a(1) = 0 trivially have the property in the sequence name because each product is zero. For a general sequence in this family with a(1) = 1 and a(2) any integer, then a(3) = -a(2)^2 - a(2) and, for n >= 4, a(n) = -a(2)^(2^(n-3))*(a(2)^(2^(n-3))-1), so that all terms after a(2) are negatives of oblong (or promic) numbers (A002378). - Rick L. Shepherd, Aug 10 2014

Links

Rick L. Shepherd, Table of n, a(n) for n = 1..13

Formula

For n >= 4, a(n) = -2^(2^(n-3)) * (2^(2^(n-3)) - 1).

For n >= 4, a(n) = -A002378(A051179(n-3)). - Rick L. Shepherd, Aug 10 2014

Example

For n = 4, we have a(1) * (a(1)+a(2)) * (a(1)+a(2)+a(3)) * (a(1)+a(2)+a(3)+a(4)) = a(1) + a(1)*a(2) + a(1)*a(2)*a(3) + a(1)*a(2)*a(3)*a(4) =
1 * (1+2) * (1+2-6) * (1+2-6-12) = 1 + 1*2 + 1*2*(-6) + 1*2*(-6)*(-12) = 135.

Prog

(PARI)
a(n) = if(n<1, , if(n<3, n, if(n==3, -6, -2^(2^(n-3))*(2^(2^(n-3))-1)))) \\ Rick L. Shepherd, Aug 10 2014

Crossrefs

Cf. A002378, A051179.

Sequence in context: A179201 A105122 A351940 * A309743 A291062 A058046

Adjacent sequences: A132073 A132074 A132075 * A132077 A132078 A132079

Keyword

easy,sign

Author

Leroy Quet, Oct 30 2007

Extensions

More terms from Max Alekseyev, Apr 29 2010

Status

approved