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A243456
a(n+6) = 6*a(n+4) - 12*a(n+2) + 8*a(n), a(0)..a(5) = 8,0,9,0,8,0.
3
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
OFFSET
0,1
COMMENTS
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.
REFERENCES
L. B. W. Jolley, Summation of Series, Dover (1961).
LINKS
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].
G. Myerson and A. J. van der Poorten, Some problems concerning recurrence sequences, Amer. Math. Monthly 102 (1995), no. 8, 698-705.
FORMULA
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.
MAPLE
A243456:=n->2^(n/2 - 4)*(1 + (-1)^n)*(n - 8)^2; seq(A243456(n), n=0..50); # Wesley Ivan Hurt, Jun 08 2014
MATHEMATICA
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 *)
CROSSREFS
Cf. A076788.
Sequence in context: A296182 A019863 A350747 * A246772 A198820 A138285
KEYWORD
nonn,easy,changed
AUTHOR
EXTENSIONS
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
STATUS
approved