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A132076
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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).
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1
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1, 2, -6, -12, -240, -65280, -4294901760, -18446744069414584320, -340282366920938463444927863358058659840, -115792089237316195423570985008687907852929702298719625575994209400481361428480
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OFFSET
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1,2
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COMMENTS
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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
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LINKS
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FORMULA
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For n >= 4, a(n) = -2^(2^(n-3)) * (2^(2^(n-3)) - 1).
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EXAMPLE
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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.
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PROG
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(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
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CROSSREFS
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KEYWORD
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easy,sign
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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