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A101089
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Second partial sums of fourth powers (A000583).
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15
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1, 18, 116, 470, 1449, 3724, 8400, 17172, 32505, 57838, 97812, 158522, 247793, 375480, 553792, 797640, 1125009, 1557354, 2120020, 2842686, 3759833, 4911236, 6342480, 8105500, 10259145, 12869766, 16011828, 19768546, 24232545, 29506544
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OFFSET
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1,2
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COMMENTS
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a(n-1)/n^5 is the "retention" of water on a 3 X 3 random surface of n levels - see Knecht et al., 2012, Schrenk et al., 2014. - Robert M. Ziff, Mar 08 2014
The general formula for the second partial sums of m-th powers is: b(n,m) = (n+1)*F(m) - F(m+1), where F(m) is the m-th Faulhaber’s polynomial. - Luciano Ancora, Jan 26 2015
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LINKS
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FORMULA
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a(n) = (1/60)*n*(n+1)^2*(n+2)*(2*n*(n+2)-1).
G.f.: x*(1+x)*(1+10*x+x^2)/(1-x)^7. - Colin Barker, Apr 04 2012
a(n) = Sum_{i=1..n} i*(n+1-i)^4, by the definition. - Bruno Berselli, Jan 31 2014
Sum_{n>=1} 1/a(n) = 85/3 + 10*Pi^2/3 - 20*sqrt(2/3)*Pi*cot(sqrt(3/2)*Pi). - Amiram Eldar, Jan 26 2022
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EXAMPLE
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a(7) = 8400 = 1*(8-1)^4 + 2*(8-2)^4 + 3*(8-3)^4 + 4*(8-4)^4 + 5*(8-5)^4 + 6*(8-6)^4 + 7*(8-7)^4. - Bruno Berselli, Jan 31 2014
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MAPLE
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f:=n->(2*n^6-5*n^4+3*n^2)/60;
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MATHEMATICA
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a[n_] := n(n+1)^2(n+2)(2n(n+2) -1)/60; Table[a[n], {n, 40}]
CoefficientList[Series[(1+x)*(1+10*x+x^2)/(1-x)^7, {x, 0, 40}], x] (* Vincenzo Librandi, Mar 24 2014 *)
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PROG
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(Magma) [(1/60)*n*(n+1)^2*(n+2)*(2*n*(n+2)-1): n in [1..40]]; // Vincenzo Librandi, Mar 24 2014
(Sage) [n*(n+1)^2*(n+2)*(2*n*(n+2)-1)/60 for n in range(1, 40)] # Danny Rorabaugh, Apr 20 2015
(GAP) List([1..40], n-> (n+1)^2*(2*(n+1)^4-5*(n+1)^2+3)/60); # G. C. Greubel, Jul 31 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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Cecilia Rossiter, Dec 14 2004
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EXTENSIONS
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STATUS
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approved
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