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A243456
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a(n+6) = 6*a(n+4) - 12*a(n+2) + 8*a(n), a(0)..a(5) = 8,0,9,0,8,0.
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3
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8, 0, 9, 0, 8, 0, 4, 0, 0, 0, 16, 0, 128, 0, 576, 0, 2048, 0, 6400, 0, 18432, 0, 50176, 0, 131072, 0, 331776, 0, 819200, 0, 1982464, 0, 4718592, 0, 11075584, 0, 25690112, 0, 58982400, 0, 134217728, 0, 303038464, 0, 679477248, 0
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OFFSET
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0,1
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COMMENTS
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The infinite sum of the reciprocals of the even terms a(2*k) for k>=5 yields (1/96)*(Pi^2 - 6*log^2(2)). That is, sum(k>=1, 1/(k^2*2^k) = dilog(1/2) = (Pi^2 - 6*(log(2))^2)/12. See the Jolley reference, pp. 66-69, (360) (c), and Abramowitz-Stegun, p. 1004, 27.7.3 for x=1/2. For the decimal expansion of dilog(1/2) see A076788.
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REFERENCES
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L. B. W. Jolley, Summation of Series, Dover (1961).
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LINKS
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
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FORMULA
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a(n) = 0, if n is odd, a(n) = (n - 8)^2*2^((n-6)/2), if n is even.
a(n) = 2^(n/2 - 4)*(1 + (-1)^n)*(n - 8)^2.
a(n+2) = (2*(n-6)^2*a(n))/(n-8)^2, n>=0, with a(0) = 8, a(1) = 0 and a(10) = 0/0 := 16.
G.f.: (8 - 39*x^2 + 50*x^4)/(1-2*x^2)^3.
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MAPLE
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MATHEMATICA
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Table[2^(n/2 - 4)*(1 + (-1)^n)*(n - 8)^2, {n, 0, 50}] (* Wesley Ivan Hurt, Jun 08 2014 *)
CoefficientList[Series[(8 - 39 x^2 + 50 x^4)/(1 - 2 x^2)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Jun 16 2014 *)
LinearRecurrence[{0, 6, 0, -12, 0, 8}, {8, 0, 9, 0, 8, 0}, 50] (* Harvey P. Dale, Mar 02 2016 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Comment on the sum reformulated and Jolley and Abramowitz-Stegun reference added. G.f. corrected for offset 0. In the recurrence a(10) defined. - Wolfdieter Lang, Jun 16 2014
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STATUS
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approved
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