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Search: id:A011934
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| A011934 |
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|1^3 - 2^3 + 3^3 - 4^3 + ... + (-1)^(n+1)*n^3|. |
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+0
5
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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
(list; graph; listen)
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OFFSET
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0,3
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COMMENT
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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 (pevnev(AT)juno.com), Jan 09 2008
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REFERENCES
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Problem 913 of the Spring 1997 issue of the Pi Mu Epsilon Journal.
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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Skidmore College Problem Group, Solution to Problem #913 from the Pi Mu Epsilon Journal
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FORMULA
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(1/8)[ -1 + (-1)^n - 6*(-1)^n*n^2 - 4*(-1)^n*n^3 ]. - Henry Bottomley (se16(AT)btinternet.com), Nov 13 2000
a(n) = n^3-a(n-1) = a(n-1)+A032528(n) = ceiling(A015238(n+1)/4) = ceiling[(n+1)^2*(2n-1)/4] - Henry Bottomley (se16(AT)btinternet.com), Nov 13 2000
G.f. = (x^3 + 4*x^2 + x)/(x^5 - 3*x^4 + 2*x^3 + 2*x^2 - 3*x + 1) - Alexander R. Povolotsky (pevnev(AT)juno.com), Apr 26 2008
{-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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CROSSREFS
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Adjacent sequences: A011931 A011932 A011933 this_sequence A011935 A011936 A011937
Sequence in context: A038349 A100752 A143058 this_sequence A100206 A007044 A047862
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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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More terms from Henry Bottomley (se16(AT)btinternet.com), Nov 13 2000
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