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.
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,easy
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