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A011934
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a(n) = |1^3 - 2^3 + 3^3 - 4^3 + ... + (-1)^(n+1)*n^3|.
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13
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0, 1, 7, 20, 44, 81, 135, 208, 304, 425, 575, 756, 972, 1225, 1519, 1856, 2240, 2673, 3159, 3700, 4300, 4961, 5687, 6480, 7344, 8281, 9295, 10388, 11564, 12825, 14175, 15616, 17152, 18785, 20519, 22356, 24300, 26353, 28519, 30800, 33200, 35721, 38367, 41140
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OFFSET
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0,3
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COMMENTS
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From the formula a(n) = n^3 - a(n-1) it follows that a(n-1) + a(n) = n^3. Thus the sum of two consecutive terms (call them the "former" and "latter" terms) is a cube of the index of the "latter" term. - Alexander R. Povolotsky, Jan 09 2008
The general formula for alternating sums of powers is in terms of the Swiss-Knife polynomials P(n,x) A153641 2^(-n-1)(P(n,1)-(-1)^k P(n,2k+1)). Thus we get expression a(k) = |2^(-4)(P(3,1)-(-1)^k P(3,2k+1))|. - Peter Luschny, Jul 12 2009
a(n) is the number of (w,x,y} having all terms in {0,...,n} and w<floor((x+y)/2). Also, the number of (w,x,y} having all terms in {0,...,n} and w>=floor((x+y)/2). [Clark Kimberling, Jun 02 2012]
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REFERENCES
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Eldon Hansen's _A Table of Series and Products_ (Prentice-Hall, 1975) gives the sum in Formula 6.2.2 in terms of Euler polynomials.
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LINKS
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Kenneth B. Davenport, Problem 913, Pi Mu Epsilon Journal, Vol. 10, No. 6, Spring 1997, p. 492.
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FORMULA
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a(n) = |(1/8)*(-1 + (-1)^n - 6*(-1)^n*n^2 - 4*(-1)^n*n^3)|. - Henry Bottomley, Nov 13 2000
{-a(n)-a(n+1)+n^3+3*n^2+3*n+1, a(0)=0, a(1)=1, a(2)=7, a(3)=20}. - Robert Israel, May 14 2008
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MAPLE
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a := n -> ((2*n+3)*n^2-(n mod 2))/4; # Peter Luschny, Jul 12 2009
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MATHEMATICA
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Abs[Accumulate[Times@@@Partition[Riffle[Range[0, 50]^3, {1, -1}, {1, -1, 2}], 2]]] (* Harvey P. Dale, May 20 2019 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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David Penney (david(AT)math.uga.edu)
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EXTENSIONS
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STATUS
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approved
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