[seqfan] Should this be a rejection?

[Brad Klee]

NAME:
a(n) = Integral_{z=0..2*Pi}(1+2*cos(z)+4*Sin(z)^2+8*cos(z)*sin(z)^2)^n*dz.

DATA:
1, 3, 21, 135, 941, 6723, 49089, 363541, 2720797, 20525463, 155812101 . . .

COMMENTS:
The a(n) can at least be evaluated in terms of a binomial quadruple sum.
Zeilberger's algorithm does not readily apply; however, we may yet derive
a linear recurrence with polynomial coefficients from a Picard-Fuchs
type differential equation (Cf. formula).

FORMULA:
a(n)=Sum_{j=0..floor(n/2);k=0..2*j,l=0..n-2*j,m=0..j}
    binomial(n,2*j)*binomial(2*j,k)*binomial(n-2*j,l)
     *binomial(j,m)*binomial(2*(l+k+m),(l+k+m))*4^(j-m)*(-1)^m

Picard-Fuchs coefficients for generating function G(X).
c_{i,j}*x^j*d^i/dx^i; i=0..3, j=0..9 :
{{234, 14994, -136206, 2350578, 10556112, 16345728, 57922560, 0, 0, 0},
 {-78, 5688, -168168, -255912, 12144390, 24960432, 106614144, 173767680, 0,
0},
 {0, -702, 29439, -472101, -934233, 2111541, 9527784, 82095552, 86883840,
0},
 {0, 0, -351, 18974, -119350, 89550, -1663255, 775208, 12320448, 9653760}}

Coefficients of a recurrence zero sum for a(n).
c_{i,j}*n^j*a(n-i); i=0..4, j=0..6 :
{{0, -277014, 1738203, -3740133, 3529215, -1497015, 231660},
 {-1102200, 9606644, -30797857, 47082120, -36509105, 13739050, -1973400},
 {200760, -542338, -3026838, 8809121, -8157510, 3137225, -420420},
 {14394240, -108248064, 291844576, -374777152, 246556880, -79791920,
10021440},
 {14480640, -106899840, 283472640, -355266240, 225676800, -69700800,
8236800}}

Effectively this proposal is being rejected by submission limit sanctioning.

Cheers,

Brad