It is conjectured that for all n > 0 the following holds:
a(18*n) = 3/40*(243*n^5+405*n^4+35*n^3+395*n^2-318*n+40)
a(18*n+1) = 1/40*n*(729*n^4-405*n^3-615*n^2+225*n+106)
a(18*n+2) = 1/40*(729*n^5+1620*n^4+735*n^3+1320*n^2-684*n+40)
a(18*n+3) = 1/40*n*(729*n^4-705*n^2+136)
a(18*n+4) = 3/40*n*(243*n^4+675*n^3+515*n^2+565*n-118)
a(18*n+5) = 1/40*n*(729*n^4+405*n^3-615*n^2-225*n+106)
a(18*n+6) = 3/40*n*(243*n^4+810*n^3+845*n^2+790*n+32)
a(18*n+7) = 3/40*n*(n+1)*(243*n^3+27*n^2-142*n+12)
a(18*n+8) = 1/40*(729*n^5+2835*n^4+3705*n^3+3405*n^2+726*n+40)
a(18*n+9) = 3/40*n*(n+1)*(243*n^3+162*n^2-127*n-18)
a(18*n+10) = 1/40*(729*n^5+3240*n^4+5055*n^3+4860*n^2+1636*n+160)
a(18*n+11) = 3/40*n*(n+1)*(243*n^3+297*n^2-52*n-48)
a(18*n+12) = 1/40*(729*n^5+3645*n^4+6585*n^3+6795*n^2+2926*n+400)
a(18*n+13) = 3/40*n*(n+1)*(243*n^3+432*n^2+83*n-58)
a(18*n+14) = 1/40*(729*n^5+4050*n^4+8295*n^3+9270*n^2+4696*n+800)
a(18*n+15) = 3/40*n*(n+1)*(243*n^3+567*n^2+278*n-28)
a(18*n+16) = 3/40*(n+3)*(243*n^4+756*n^3+1127*n^2+734*n+160)
a(18*n+17) = 3/40*n*(n+1)*(243*n^3+702*n^2+533*n+62)