OFFSET
0,1
COMMENTS
Since 0 = Sin(Pi) = Sum_{n>=0}(-1)^n*Pi^(2n+1)/(2n+1)!, we can move the negative terms to the other side of the equation to get: Sum_{n>=0} Pi^(4n+1)/(4n+1)! = Sum_{n>=0}Pi^(4n+3)/(4n+3)!.
Now, if we let f(n) = Pi^(4n+1)/(4n+1)!, then the previous equation can be written as Sum_{n>=0}f(n) = Sum_{n>=0}(Pi^2/((4*n+3)*(4*n+2)))*f(n); a(n) is the n-th denominator on the right hand side.
LINKS
FORMULA
a(n) = 16*n^2 + 20*n + 6.
a(n) = 2*A033567(n+1).
G.f.: (6+24*x+2*x^2)/(1-x)^3. - Vincenzo Librandi, Apr 12 2015
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>2. - Vincenzo Librandi, Apr 12 2015
E.g.f.: 2*exp(x)*(3+18*x+8*x^2). - Wesley Ivan Hurt, Apr 29 2020
From Amiram Eldar, Jan 03 2022: (Start)
Sum_{n>=0} 1/a(n) = Pi/8 - log(2)/4.
Sum_{n>=0} (-1)^n/a(n) = sqrt(2)*log(sqrt(2)+1)/4 - (sqrt(2)-1)*Pi/8. (End)
MATHEMATICA
CoefficientList[Series[(6 + 24 x + 2 x^2) / (1 - x)^3, {x, 0, 40}], x] (* Vincenzo Librandi, Apr 12 2015 *)
PROG
(Magma) [16*n^2 + 20*n + 6: n in [0..40]]; // Vincenzo Librandi, Apr 12 2015
(PARI) vector(50, n, (4*n-1)*(4*n-2)) \\ Derek Orr, Apr 13 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bruce Zimov, Apr 10 2015
EXTENSIONS
More terms from Vincenzo Librandi, Apr 12 2015
STATUS
approved