Here’s a solution.
Series: 15, 15, 13, 13, 13, 14, ...
Let T(n)=a₀+a₁n+a₂n²+a₃n³+a₄n⁴+a₅n⁵.
T(0)=T(1)=15, T(2)=T(3)=T(4)=13, T(5)=14.
When n=0, T(0)=15, so a₀=15
Let t(n)=T(n)-a₀ then the series for t(n) is 0, 0, -2, -2, -2, -1.
t(n)=a₁n+a₂n²+a₃n³+a₄n⁴+a₅n⁵ for n>0.
LEVEL 1
(1) t(1)=a₁+a₂+a₃+a₄+a₅=0
(2) t(2)=2a₁+4a₂+8a₃+16a₄+32a₅=-2
(3) t(3)=3a₁+9a₂+27a₃+81a₄+243a₅=-2
(4) t(4)=4a₁+16a₂+64a₃+256a₄+1024a₅=-2
(5) t(5)=5a₁+25a₂+125a₃+625a₄+3125a₅=-1
LEVEL 2
(6) t(2)-2t(1)=2a₂+6a₃+14a₄+30a₅=-2 becomes
(6) a₂+3a₃+7a₄+15a₅=-1
(7) t(3)-3t(1)=6a₂+24a₃+78a₄+240a₅=-2 becomes
(7) 3a₂+12a₃+39a₄+120a₅=-1
(8) t(4)-4t(1)=12a₂+60a₃+252a₄+1020a₅=-2 becomes
(8) 6a₂+30a₃+126a₄+510a₅=-1
(9) t(5)-5t(1)=20a₂+120a₃+620a₄+3120a₅=-1
LEVEL 3
(10) (7)-3(6)=3a₃+18a₄+75a₅=2
(11) (8)-2(7)=6a₃+48a₄+270a₅=1
(12) (9)-20(6)=60a₃+480a₄+2820a₅=19
LEVEL 4
(13) (11)-2(10)=12a₄+120a₅=-3 becomes
(13) 4a₄+40a₅=-1
(14) (12)-10(11)=120a₅=9
SOLUTION
a₅=9/120=3/40 from (14);
4a₄=-1-40a₅=-4, a₄=-1 from (13);
3a₃=2-18a₄-75a₅=115/8, a₃=115/24 from (10);
a₂=-(1+3a₃+7a₄+15a₅)=-19/2 from (6);
a₁=-(a₂+a₃+a₄+a₅)=169/30 from (1).
T(n)=15+169n/30-19n²/2+115n³/24-n⁴+3n⁵/40 for n≥0.
The series generated is 15, 15, 13, 13, 13, 14, 29, 92, 267, 657, 1413, ...
This is probably only one of many solutions.