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
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
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.
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-12,0,8).
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
KEYWORD
nonn
AUTHOR
Alexander R. Povolotsky, Jun 05 2014
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