A102094 a(n) = (2*n-1)*(2*n+1)^2.
9, 75, 245, 567, 1089, 1859, 2925, 4335, 6137, 8379, 11109, 14375, 18225, 22707, 27869, 33759, 40425, 47915, 56277, 65559, 75809, 87075, 99405, 112847, 127449, 143259, 160325, 178695, 198417, 219539, 242109, 266175, 291785, 318987, 347829, 378359, 410625
Offset: 1
References
- G. Boros and V. Moll, Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals, Cambridge University Press, Cambridge, 2004, p. 123.
- J. Ewell, An Eulerian Method for Representing Pi^2 by Series, The Rocky Mountain Journal of Mathematics 1992 v.22, pp. 165-168.
Links
- G. C. Greubel, Table of n, a(n) for n = 1..1000
- J. Ewell, An Eulerian Method for Representing Pi^2 by Series, The Rocky Mountain Journal of Mathematics 1992 v.22, pp. 165-168.
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Crossrefs
Cf. A002388.
Programs
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GAP
List([1..40], n-> (2*n-1)*(2*n+1)^2); # G. C. Greubel, Oct 27 2019
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Magma
[(2*n-1)*(2*n+1)^2: n in [1..40]]; // G. C. Greubel, Oct 27 2019
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Maple
seq((2*n-1)*(2*n+1)^2, n=1..40); # G. C. Greubel, Oct 27 2019
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Mathematica
Table[(2n-1)(2n+1)^2,{n,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{9,75,245,567},40] (* Harvey P. Dale, Jul 24 2012 *)
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PARI
vector(40, n, (2*n-1)*(2*n+1)^2) \\ G. C. Greubel, Oct 27 2019
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Sage
[(2*n-1)*(2*n+1)^2 for n in (1..40)] # G. C. Greubel, Oct 27 2019
Formula
Sum_{n>=1} 1/a(n) = (12 - Pi^2)/16.
Sum_{n>=1} n/a(n) = (Pi^2 - 4)/32. - Sign flipped by Bernard Schott, May 06 2020
From Harvey P. Dale, Jul 24 2012: (Start)
a(1)=9, a(2)=75, a(3)=245, a(4)=567, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
G.f.: (9 + 39*x - x^2 + x^3)/(1-x)^4. (End)
E.g.f.: 1 + (-1 + 10*x + 28*x^2 + 8*x^3)*exp(x). - G. C. Greubel, Oct 27 2019
Extensions
More terms from Harvey P. Dale, Jul 24 2012
Comments