30, Challenging Problems about The limit as n goes to infinity to Solutions of q-Recurrence equations with polynomial coefficients in q^n Problem Number, 1, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n n 2 -q (q ) (q q - 1) (q q + 1) (q (q ) - 1) a(n) 7 n 3 6 n 3 5 n 3 n 4 4 n + (q (q ) + q (q ) + q (q ) - q q - q - 1) q a(1 + n) 4 n 2 3 n 2 n 2 5 + (-q (q ) - q (q ) - (q ) q - 1) a(2 + n) + a(3 + n) = 0 subject to the initial conditions 4 3 2 a(1) = 1 + q, a(2) = 1 + 1/q + q + 2 q + 2 q + q, 4 3 2 9 8 7 6 5 a(3) = 2 + 5 q + 5 q + 3 q + 2 q + 1/q + q + 2 q + 3 q + 4 q + 4 q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (j + 2 j) ) q / ----- j = -infinity ------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 2, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n n 2 -q (q ) (q q - 1) (q q + 1) (q (q ) - 1) a(n) 8 n 3 7 n 3 6 n 3 6 n 5 n (q (q ) + q (q ) + q (q ) - q - q q - 1) q a(1 + n) + ------------------------------------------------------------- q 4 n 2 3 n 2 n 2 5 + (-q (q ) - q (q ) - (q ) q - 1) a(2 + n) + a(3 + n) = 0 subject to the initial conditions 1 4 3 2 3 2 a(1) = 1 + q, a(2) = 1 + ---- + 2 q + q + 2 q + q, a(3) = 2 + 4 q + 3 q 2 q 1 9 8 7 6 5 4 + 2 q + 1/q + ---- + 2 q + 2 q + 3 q + 4 q + 4 q + 4 q 2 q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (j + 3 j) ) q / ----- j = -infinity ------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 3, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n n 2 -q (q ) (q q - 1) (q q + 1) (q (q ) - 1) a(n) 10 n 3 9 n 3 8 n 3 10 7 n n (q (q ) + q (q ) + q (q ) - q - q q - 1) q a(1 + n) + --------------------------------------------------------------- 3 q 4 n 2 3 n 2 n 2 5 + (-q (q ) - q (q ) - (q ) q - 1) a(2 + n) + a(3 + n) = 0 subject to the initial conditions 1 4 3 2 6 a(1) = 1 + q, a(2) = 1 + ---- + q + q + 2 q + q + q , a(3) = 2 + 2 q + 1/q 4 q 1 1 1 9 8 7 6 5 4 3 + ---- + ---- + ---- + 2 q + 2 q + 3 q + 4 q + 3 q + 3 q + 3 q 2 3 4 q q q 2 11 10 + 2 q + q + q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (j + 5 j) ) q / ----- j = -infinity ------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 4, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n n 2 -q (q ) (q q - 1) (q q + 1) (q (q ) - 1) a(n) n 13 n 3 16 12 n 3 11 n 3 10 n q (q (q ) - q + q (q ) + q (q ) - q q - 1) a(1 + n) + ------------------------------------------------------------------ 6 q 4 n 2 3 n 2 n 2 5 + (-q (q ) - q (q ) - (q ) q - 1) a(2 + n) + a(3 + n) = 0 subject to the initial conditions 1 4 3 2 9 1 1 a(1) = 1 + q, a(2) = 1 + ---- + q + q + 2 q + q + q , a(3) = 1 + ---- + ---- 7 2 3 q q q 1 1 1 1 9 8 7 6 5 4 3 + ---- + ---- + ---- + ---- + 2 q + q + 2 q + 3 q + 3 q + 3 q + 3 q 4 5 6 7 q q q q 2 14 13 12 11 10 + 2 q + q + q + q + q + q + q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (j + 8 j) ) q / ----- j = -infinity ------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 5, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n n 2 -q (q ) (q q - 1) (q q + 1) (q (q ) - 1) a(n) n 22 16 n 3 15 n 3 14 n 3 n 13 q (-q + q (q ) + q (q ) + q (q ) - q q - 1) a(1 + n) + ------------------------------------------------------------------- 9 q 4 n 2 3 n 2 n 2 5 + (-q (q ) - q (q ) - (q ) q - 1) a(2 + n) + a(3 + n) = 0 subject to the initial conditions 1 4 3 2 12 1 1 a(1) = 1 + q, a(2) = 1 + --- + q + q + 2 q + q + q , a(3) = 1 + ---- + ---- 10 5 6 q q q 1 1 1 1 9 8 7 6 5 4 3 + ---- + ---- + ---- + --- + q + q + 2 q + 3 q + 3 q + 3 q + 3 q 7 8 9 10 q q q q 2 17 16 15 14 13 12 + 2 q + q + q + q + q + q + q + q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (j + 11 j) ) q / ----- j = -infinity -------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 6, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n n 2 -q (q ) (q q - 1) (q q + 1) (q (q ) - 1) a(n) n 32 21 n 3 20 n 3 19 n 3 18 n q (-q + q (q ) + q (q ) + q (q ) - q q - 1) a(1 + n) + ------------------------------------------------------------------- 14 q 4 n 2 3 n 2 n 2 5 + (-q (q ) - q (q ) - (q ) q - 1) a(2 + n) + a(3 + n) = 0 subject to the initial conditions 1 4 3 2 17 1 1 a(1) = 1 + q, a(2) = 1 + --- + q + q + 2 q + q + q , a(3) = 1 + --- + --- 15 10 11 q q q 1 1 1 1 9 8 7 6 5 4 3 + --- + --- + --- + --- + q + q + 2 q + 3 q + 3 q + 3 q + 3 q 12 13 14 15 q q q q 2 22 21 20 19 18 17 + 2 q + q + q + q + q + q + q + q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (j + 16 j) ) q / ----- j = -infinity -------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 7, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n n 2 -q (q ) (q q - 1) (q q + 1) (q (q ) - 1) a(n) n 34 22 n 3 21 n 3 20 n 3 19 n q (-q + q (q ) + q (q ) + q (q ) - q q - 1) a(1 + n) + ------------------------------------------------------------------- 15 q 4 n 2 3 n 2 n 2 5 + (-q (q ) - q (q ) - (q ) q - 1) a(2 + n) + a(3 + n) = 0 subject to the initial conditions 1 4 3 2 18 1 1 a(1) = 1 + q, a(2) = 1 + --- + q + q + 2 q + q + q , a(3) = 1 + --- + --- 16 11 12 q q q 1 1 1 1 9 8 7 6 5 4 3 + --- + --- + --- + --- + q + q + 2 q + 3 q + 3 q + 3 q + 3 q 13 14 15 16 q q q q 2 23 22 21 20 19 18 + 2 q + q + q + q + q + q + q + q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (j + 17 j) ) q / ----- j = -infinity -------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 8, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n n 2 -q (q ) (q q - 1) (q q + 1) (q (q ) - 1) a(n) n 36 23 n 3 22 n 3 21 n 3 n 20 q (-q + q (q ) + q (q ) + q (q ) - q q - 1) a(1 + n) + ------------------------------------------------------------------- 16 q 4 n 2 3 n 2 n 2 5 + (-q (q ) - q (q ) - (q ) q - 1) a(2 + n) + a(3 + n) = 0 subject to the initial conditions 1 4 3 2 19 1 1 a(1) = 1 + q, a(2) = 1 + --- + q + q + 2 q + q + q , a(3) = 1 + --- + --- 17 12 13 q q q 1 1 1 1 9 8 7 6 5 4 3 + --- + --- + --- + --- + q + q + 2 q + 3 q + 3 q + 3 q + 3 q 14 15 16 17 q q q q 2 24 23 22 21 20 19 + 2 q + q + q + q + q + q + q + q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (j + 18 j) ) q / ----- j = -infinity -------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 9, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n n 2 -q (q ) (q q - 1) (q q + 1) (q (q ) - 1) a(n) n 38 24 n 3 23 n 3 22 n 3 n 21 q (-q + q (q ) + q (q ) + q (q ) - q q - 1) a(1 + n) + ------------------------------------------------------------------- 17 q 4 n 2 3 n 2 n 2 5 + (-q (q ) - q (q ) - (q ) q - 1) a(2 + n) + a(3 + n) = 0 subject to the initial conditions 1 4 3 2 20 1 1 a(1) = 1 + q, a(2) = 1 + --- + q + q + 2 q + q + q , a(3) = 1 + --- + --- 18 13 14 q q q 1 1 1 1 9 8 7 6 5 4 3 + --- + --- + --- + --- + q + q + 2 q + 3 q + 3 q + 3 q + 3 q 15 16 17 18 q q q q 2 25 24 23 22 21 20 + 2 q + q + q + q + q + q + q + q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (j + 19 j) ) q / ----- j = -infinity -------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 10, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n 2 n 2 n 3 n 2 -q (q (q ) - 1) (q q - 1) (q q + 1) (q q - 1) (q q + 1) (q (q ) - 1) a(n) + 4 n 2 2 3 n 2 2 n 2 n q (q (q ) + q + q + 1) (q (q ) - 1) (q q - 1) (q q + 1) a(1 + n) 8 n 3 7 n 3 n 2 6 2 + (-q (q ) - q (q ) + (q ) q - q - q - 1) a(2 + n) + a(3 + n) = 0 subject to the initial conditions 5 4 3 2 12 11 a(1) = 1 + q, a(2) = q + 3 q + 3 q + 4 q + 2 q + 1, a(3) = q + q 10 9 8 7 6 5 4 3 2 + 3 q + 5 q + 6 q + 8 q + 9 q + 9 q + 7 q + 6 q + 4 q + 2 q + 1 Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (3/2 j + 1/2 j) ) q / ----- j = -infinity ------------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 11, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n 2 n 2 n 3 n 2 -q (q (q ) - 1) (q q - 1) (q q + 1) (q q - 1) (q q + 1) (q (q ) - 1) a(n) + 4 n 2 2 3 n 2 2 n 2 n q (q (q ) + q + q + 1) (q (q ) - 1) (q q - 1) (q q + 1) a(1 + n) 9 n 3 6 n 3 n 2 6 2 + (-q (q ) - q (q ) + (q ) q - q - q - 1) a(2 + n) + a(3 + n) = 0 subject to the initial conditions 6 5 4 3 2 14 13 a(1) = 1 + q, a(2) = q + q + 2 q + 3 q + 3 q + 2 q + 2, a(3) = q + q 12 11 10 9 8 7 6 5 4 3 + q + 2 q + 2 q + 4 q + 5 q + 7 q + 8 q + 8 q + 8 q + 7 q 2 + 4 q + 2 q + 2 Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (3/2 j + 3/2 j) ) q / ----- j = -infinity ------------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 12, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n 2 n 2 n 3 n 2 -q (q (q ) - 1) (q q - 1) (q q + 1) (q q - 1) (q q + 1) (q (q ) - 1) a(n) + 4 n 2 2 3 n 2 2 n 2 n q (q (q ) + q + q + 1) (q (q ) - 1) (q q - 1) (q q + 1) a(1 + n) 11 n 3 n 2 6 4 n 3 2 + (-q (q ) + (q ) q - q (q ) - q - q - 1) a(2 + n) + a(3 + n) = 0 subject to the initial conditions 1 4 3 2 8 7 6 5 a(1) = 1 + q, a(2) = 2 q + 2 + 1/q + ---- + q + q + 2 q + q + q + q + q , 2 q 16 8 1 10 18 17 15 14 a(3) = 4 + 2/q + q + 3 q + ---- + 2 q + 4 q + q + q + q + q 2 q 3 2 9 13 12 11 4 7 6 5 + 6 q + 6 q + 4 q + 2 q + q + 2 q + 6 q + 4 q + 5 q + 5 q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (3/2 j + 7/2 j) ) q / ----- j = -infinity ------------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 13, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n 2 n 2 n 3 n 2 -q (q (q ) - 1) (q q - 1) (q q + 1) (q q - 1) (q q + 1) (q (q ) - 1) a(n) + 4 n 2 2 3 n 2 2 n 2 n q (q (q ) + q + q + 1) (q (q ) - 1) (q q - 1) (q q + 1) a(1 + n) 12 n 3 n 2 6 3 n 3 2 + (-q (q ) + (q ) q - q (q ) - q - q - 1) a(2 + n) + a(3 + n) = 0 subject to the initial conditions a(1) = 1 + q, 1 1 4 3 2 9 8 7 6 a(2) = 2 + 1/q + ---- + ---- + q + q + 2 q + q + q + q + q + q , a(3) 2 3 q q 2 19 16 8 2 20 10 18 17 = 4 + 3/q + ---- + q + q + 3 q + ---- + q + 3 q + 5 q + q + q 3 2 q q 15 14 3 2 9 13 12 11 4 7 + q + q + 5 q + 5 q + 3 q + q + 2 q + 2 q + 4 q + 3 q 6 5 + 4 q + 4 q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (3/2 j + 9/2 j) ) q / ----- j = -infinity ------------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 14, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n 2 n 2 n 3 n 2 -q (q (q ) - 1) (q q - 1) (q q + 1) (q q - 1) (q q + 1) (q (q ) - 1) a(n) + 4 n 2 2 3 n 2 2 n 2 n q (q (q ) + q + q + 1) (q (q ) - 1) (q q - 1) (q q + 1) a(1 + n) 13 n 3 n 2 6 2 n 3 2 + (-q (q ) + (q ) q - q (q ) - q - q - 1) a(2 + n) + a(3 + n) = 0 subject to the initial conditions a(1) = 1 + q, a(2) = 1 1 1 4 3 2 10 9 8 7 1/q + ---- + ---- + ---- + q + q + 2 q + q + 1 + q + q + q + q , 2 3 4 q q q 2 19 8 3 2 1 20 10 22 a(3) = 5 + 3/q + ---- + q + 2 q + ---- + ---- + ---- + q + 2 q + q 3 2 4 5 q q q q 21 18 17 15 14 3 2 9 13 12 + 3 q + q + q + q + q + q + 4 q + 4 q + 3 q + 2 q + 2 q 11 4 7 6 5 + 3 q + 4 q + 3 q + 3 q + 3 q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (3/2 j + 11/2 j) ) q / ----- j = -infinity -------------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 15, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n 2 n 2 n 3 n 2 -q (q (q ) - 1) (q q - 1) (q q + 1) (q q - 1) (q q + 1) (q (q ) - 1) a(n) + 4 n 2 2 3 n 2 2 n 2 n q (q (q ) + q + q + 1) (q (q ) - 1) (q q - 1) (q q + 1) a(1 + n) 15 n 3 n 2 6 n 3 2 + (-q (q ) + (q ) q - (q ) - q - q - 1) a(2 + n) + a(3 + n) = 0 subject to the initial conditions a(1) = 1 + q, a(2) = 1 1 1 1 4 3 2 12 11 10 9 ---- + ---- + ---- + ---- + q + q + 2 q + q + 1 + q + q + q + q , 3 4 5 6 q q q q 2 1 26 25 16 8 3 3 2 a(3) = 3 + 2/q + ---- + ---- + q + q + q + q + ---- + ---- + ---- 3 7 2 4 5 q q q q q 2 1 1 10 24 23 22 21 17 15 + ---- + ---- + ---- + q + q + q + q + 2 q + q + q + 2 q 6 8 9 q q q 14 3 2 9 13 12 11 4 7 6 + 2 q + 3 q + 3 q + 2 q + 3 q + 2 q + 2 q + 3 q + 2 q + 3 q 5 + 3 q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (3/2 j + 15/2 j) ) q / ----- j = -infinity -------------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 16, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n 2 n 2 n 3 n 2 -q (q (q ) - 1) (q q - 1) (q q + 1) (q q - 1) (q q + 1) (q (q ) - 1) a(n) + 4 n 2 2 3 n 2 2 n 2 n q (q (q ) + q + q + 1) (q (q ) - 1) (q q - 1) (q q + 1) a(1 + n) 19 n 3 8 n 2 4 3 n 3 2 (q (q ) - q (q ) + q + q + (q ) + q ) a(2 + n) - ------------------------------------------------------ + a(3 + n) = 0 2 q subject to the initial conditions a(1) = 1 + q, a(2) = 1 1 1 1 4 3 2 14 13 12 11 ---- + ---- + ---- + ---- + q + q + 2 q + q + 1 + q + q + q + q , 5 6 7 8 q q q q 29 30 28 27 2 19 1 26 25 a(3) = 2 + 1/q + q + q + q + q + ---- + q + ---- + q + q 3 7 q q 16 8 2 3 2 2 2 1 1 1 1 + 2 q + q + ---- + ---- + ---- + ---- + ---- + ---- + --- + --- + --- 2 4 5 6 8 9 10 11 12 q q q q q q q q q 1 18 17 15 14 3 2 9 13 12 + --- + q + q + 2 q + 3 q + 2 q + 3 q + 2 q + q + 2 q + q 13 q 11 4 7 6 5 + q + 3 q + 2 q + 3 q + 3 q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (3/2 j + 19/2 j) ) q / ----- j = -infinity -------------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 17, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n 2 n 2 n 3 n 2 -q (q (q ) - 1) (q q - 1) (q q + 1) (q q - 1) (q q + 1) (q (q ) - 1) a(n) + 4 n 2 2 3 n 2 2 n 2 n q (q (q ) + q + q + 1) (q (q ) - 1) (q q - 1) (q q + 1) a(1 + n) 27 n 3 12 n 2 8 7 6 n 3 (q (q ) - q (q ) + q + q + q + (q ) ) a(2 + n) - ------------------------------------------------------- + a(3 + n) = 0 6 q subject to the initial conditions a(1) = 1 + q, a(2) = 1 1 1 1 4 3 2 18 17 16 15 ---- + --- + --- + --- + q + q + 2 q + q + 1 + q + q + q + q , 9 10 11 12 q q q q 1 1 1 1 1 36 37 1 19 2 a(3) = 1 + --- + --- + --- + --- + --- + q + q + --- + 3 q + ---- 17 16 18 20 21 19 7 q q q q q q q 35 16 34 8 33 38 1 1 2 3 2 + q + q + q + q + q + q + ---- + ---- + ---- + ---- + ---- 4 5 6 8 9 q q q q q 2 1 1 20 23 22 21 18 17 15 + --- + --- + --- + 2 q + q + q + q + 2 q + 2 q + 2 q + q 10 11 12 q q q 3 2 9 4 7 6 5 + 3 q + 2 q + q + 3 q + 2 q + 3 q + 3 q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (3/2 j + 27/2 j) ) q / ----- j = -infinity -------------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 18, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n 2 n 2 n 3 n 2 -q (q (q ) - 1) (q q - 1) (q q + 1) (q q - 1) (q q + 1) (q (q ) - 1) a(n) + 4 n 2 2 3 n 2 2 n 2 n q (q (q ) + q + q + 1) (q (q ) - 1) (q q - 1) (q q + 1) a(1 + n) 29 n 3 13 n 2 9 8 7 n 3 (q (q ) - q (q ) + q + q + q + (q ) ) a(2 + n) - ------------------------------------------------------- + a(3 + n) = 0 7 q subject to the initial conditions a(1) = 1 + q, a(2) = 1 1 1 1 4 3 2 19 18 17 16 --- + --- + --- + --- + q + q + 2 q + q + 1 + q + q + q + q , 10 11 12 13 q q q q 1 1 1 1 1 36 37 1 19 39 a(3) = 1 + --- + --- + --- + --- + --- + q + q + --- + 2 q + q 18 20 21 22 23 19 q q q q q q 2 35 16 8 40 38 1 1 2 3 2 + ---- + q + q + q + q + q + ---- + ---- + ---- + ---- + --- 7 5 6 8 9 10 q q q q q q 2 1 1 20 24 23 22 21 18 17 + --- + --- + --- + 3 q + q + q + 2 q + q + 2 q + 2 q + q 11 12 13 q q q 3 2 9 4 7 6 5 + 3 q + 2 q + q + 3 q + 2 q + 3 q + 3 q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (3/2 j + 29/2 j) ) q / ----- j = -infinity -------------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 19, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n 2 n 2 n 3 n 2 -q (q (q ) - 1) (q q - 1) (q q + 1) (q q - 1) (q q + 1) (q (q ) - 1) a(n) + 4 n 2 2 3 n 2 2 n 2 n q (q (q ) + q + q + 1) (q (q ) - 1) (q q - 1) (q q + 1) a(1 + n) 31 n 3 14 n 2 10 9 8 n 3 (q (q ) - q (q ) + q + q + q + (q ) ) a(2 + n) - -------------------------------------------------------- + a(3 + n) = 0 8 q subject to the initial conditions a(1) = 1 + q, a(2) = 1 1 1 1 4 3 2 20 19 18 17 --- + --- + --- + --- + q + q + 2 q + q + 1 + q + q + q + q , 11 12 13 14 q q q q 1 1 1 1 37 41 42 19 1 1 a(3) = 1 + --- + --- + --- + --- + q + q + q + 2 q + --- + --- 20 21 22 23 25 24 q q q q q q 39 1 25 8 40 38 1 1 2 2 3 2 + q + ---- + q + q + q + q + --- + ---- + ---- + ---- + --- + --- 7 14 6 8 9 10 11 q q q q q q q 2 1 20 24 23 22 21 18 17 3 + --- + --- + 2 q + q + 2 q + 2 q + q + 3 q + q + q + 3 q 12 13 q q 2 9 4 7 6 5 + 2 q + q + 3 q + 2 q + 3 q + 3 q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (3/2 j + 31/2 j) ) q / ----- j = -infinity -------------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 20, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n 2 n 2 n 3 n 2 -q (q (q ) - 1) (q q - 1) (q q + 1) (q q - 1) (q q + 1) (q (q ) - 1) a(n) + 4 n 2 2 3 n 2 2 n 2 n q (q (q ) + q + q + 1) (q (q ) - 1) (q q - 1) (q q + 1) a(1 + n) 33 n 3 15 n 2 11 10 9 n 3 (q (q ) - q (q ) + q + q + q + (q ) ) a(2 + n) - --------------------------------------------------------- + a(3 + n) = 0 9 q subject to the initial conditions a(1) = 1 + q, a(2) = 1 1 1 1 4 3 2 21 20 19 18 --- + --- + --- + --- + q + q + 2 q + q + 1 + q + q + q + q , 12 13 14 15 q q q q 1 1 1 41 42 19 1 1 39 1 26 a(3) = 1 + --- + --- + --- + q + q + q + --- + --- + q + ---- + q 15 22 23 25 24 7 q q q q q q 25 8 40 1 1 1 2 2 3 2 2 20 + q + q + q + --- + --- + ---- + ---- + --- + --- + --- + --- + 2 q 14 27 8 9 10 11 12 13 q q q q q q q q 24 23 22 21 18 3 2 9 44 43 + 2 q + 2 q + 3 q + q + 2 q + q + 3 q + 2 q + q + q + q 4 7 6 5 1 + 3 q + 2 q + 3 q + 3 q + --- 26 q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (3/2 j + 33/2 j) ) q / ----- j = -infinity -------------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 21, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence 3 n 2 n n 2 n 2 n 3 n 2 -q (q (q ) - 1) (q q - 1) (q q + 1) (q q - 1) (q q + 1) (q (q ) - 1) a(n) + 4 n 2 2 3 n 2 2 n 2 n q (q (q ) + q + q + 1) (q (q ) - 1) (q q - 1) (q q + 1) a(1 + n) 35 n 3 16 n 2 12 11 10 n 3 (q (q ) - q (q ) + q + q + q + (q ) ) a(2 + n) - ---------------------------------------------------------- + a(3 + n) = 10 q 0 subject to the initial conditions a(1) = 1 + q, a(2) = 1 1 1 1 4 3 2 22 21 20 19 --- + --- + --- + --- + q + q + 2 q + q + 1 + q + q + q + q , 13 14 15 16 q q q q 1 27 1 1 1 46 45 41 42 19 1 a(3) = 1 + --- + q + --- + --- + --- + q + q + q + q + q + --- 29 16 15 28 25 q q q q q 1 26 25 8 2 1 1 1 2 2 3 2 + --- + q + 2 q + q + --- + --- + ---- + ---- + --- + --- + --- + --- 24 14 27 8 9 10 11 12 13 q q q q q q q q q 20 24 23 22 21 3 2 9 44 43 + q + 2 q + 3 q + 2 q + q + 2 q + 3 q + 2 q + q + q + q 4 7 6 5 1 + 3 q + 2 q + 3 q + 3 q + --- 26 q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (3/2 j + 35/2 j) ) q / ----- j = -infinity -------------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 22, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence n n n 2 q (q q - 1) (q q + 1) (q (q ) - 1) a(n) n 2 6 n 2 + (-(q ) q - (q ) - q - 1) a(1 + n) + a(2 + n) = 0 subject to the initial conditions 1 4 a(1) = ---- + 1 + q + q , 2 q 2 4 3 2 7 6 5 10 a(2) = ---- + 2 q + 2 + 1/q + 2 q + q + 2 q + q + q + q + q 2 q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (j + 3 j) ) q / ----- j = -infinity ------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 23, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence n n n 2 q (q q - 1) (q q + 1) (q (q ) - 1) a(n) 8 n 2 2 n 2 (q (q ) + q + (q ) + q) a(1 + n) - ------------------------------------ + a(2 + n) = 0 q subject to the initial conditions 1 5 a(1) = ---- + 1 + q + q , a(2) = 3 q 1 1 1 4 3 2 8 7 6 5 12 ---- + 2 + 1/q + ---- + ---- + q + q + 2 q + q + q + q + q + q + q 4 2 3 q q q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (j + 4 j) ) q / ----- j = -infinity ------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 24, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence n n n 2 q (q q - 1) (q q + 1) (q (q ) - 1) a(n) 20 n 2 8 7 n 2 (q (q ) + q + q + (q ) ) a(1 + n) - -------------------------------------- + a(2 + n) = 0 7 q subject to the initial conditions 1 11 1 1 1 1 1 4 3 a(1) = ---- + 1 + q + q , a(2) = --- + ---- + ---- + ---- + ---- + q + q 9 16 6 7 8 9 q q q q q q 2 14 13 12 11 24 + 2 q + q + 1 + q + q + q + q + q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (j + 10 j) ) q / ----- j = -infinity -------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 25, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence n n n 2 q (q q - 1) (q q + 1) (q (q ) - 1) a(n) 22 n 2 9 8 n 2 (q (q ) + q + q + (q ) ) a(1 + n) - -------------------------------------- + a(2 + n) = 0 8 q subject to the initial conditions 1 12 1 1 1 1 1 4 3 a(1) = --- + 1 + q + q , a(2) = --- + ---- + ---- + ---- + --- + q + q 10 18 7 8 9 10 q q q q q q 2 15 14 13 12 26 + 2 q + q + 1 + q + q + q + q + q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (j + 11 j) ) q / ----- j = -infinity -------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 26, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence n n n 2 q (q q - 1) (q q + 1) (q (q ) - 1) a(n) 24 n 2 10 9 n 2 (q (q ) + q + q + (q ) ) a(1 + n) - --------------------------------------- + a(2 + n) = 0 9 q subject to the initial conditions 1 13 1 1 1 1 1 4 3 2 a(1) = --- + 1 + q + q , a(2) = --- + ---- + ---- + --- + --- + q + q + 2 q 11 20 8 9 10 11 q q q q q q 16 15 14 13 28 + q + 1 + q + q + q + q + q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (j + 12 j) ) q / ----- j = -infinity -------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 27, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence n n n 2 q (q q - 1) (q q + 1) (q (q ) - 1) a(n) 28 n 2 12 11 n 2 (q (q ) + q + q + (q ) ) a(1 + n) - ---------------------------------------- + a(2 + n) = 0 11 q subject to the initial conditions 1 15 1 1 1 1 1 4 3 2 a(1) = --- + 1 + q + q , a(2) = --- + --- + --- + --- + --- + q + q + 2 q 13 24 10 11 12 13 q q q q q q 18 17 16 15 32 + q + 1 + q + q + q + q + q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (j + 14 j) ) q / ----- j = -infinity -------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 28, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence n n n 2 q (q q - 1) (q q + 1) (q (q ) - 1) a(n) n 2 34 15 14 n 2 ((q ) q + q + q + (q ) ) a(1 + n) - ---------------------------------------- + a(2 + n) = 0 14 q subject to the initial conditions 1 18 1 1 1 1 1 4 3 2 a(1) = --- + 1 + q + q , a(2) = --- + --- + --- + --- + --- + q + q + 2 q 16 30 13 14 15 16 q q q q q q 21 20 19 18 38 + q + 1 + q + q + q + q + q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (j + 17 j) ) q / ----- j = -infinity -------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 29, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence n n n 2 q (q q - 1) (q q + 1) (q (q ) - 1) a(n) 36 n 2 16 15 n 2 (q (q ) + q + q + (q ) ) a(1 + n) - ---------------------------------------- + a(2 + n) = 0 15 q subject to the initial conditions 1 19 1 1 1 1 1 4 3 2 a(1) = --- + 1 + q + q , a(2) = --- + --- + --- + --- + --- + q + q + 2 q 17 32 14 15 16 17 q q q q q q 22 21 20 19 40 + q + 1 + q + q + q + q + q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (j + 18 j) ) q / ----- j = -infinity -------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Problem Number, 30, : Let , a(n), be the sequence of polynomials in q defined by the linear recurrence n n n 2 q (q q - 1) (q q + 1) (q (q ) - 1) a(n) 38 n 2 17 16 n 2 (q (q ) + q + q + (q ) ) a(1 + n) - ---------------------------------------- + a(2 + n) = 0 16 q subject to the initial conditions 1 20 1 1 1 1 1 4 3 2 a(1) = --- + 1 + q + q , a(2) = --- + --- + --- + --- + --- + q + q + 2 q 18 34 15 16 17 18 q q q q q q 23 22 21 20 42 + q + 1 + q + q + q + q + q Prove that , surprisingly, the limit of a(n) as n goes to infinity equals infinity ----- 2 \ (j + 19 j) ) q / ----- j = -infinity -------------------------- infinity --------' ' | | k | | (1 - q ) | | | | k = 1 ------------------------------------------------------------------------ Solutions to the Problems Proof of Problem, 1 It so happened that a(n) equals n ----- 2 \ (j + 2 j) ) qbinomial(2 n, n - 2 j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 2 It so happened that a(n) equals n ----- 2 \ (j + 3 j) ) qbinomial(2 n, n - 2 j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 3 It so happened that a(n) equals n ----- 2 \ (j + 5 j) ) qbinomial(2 n, n - 2 j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 4 It so happened that a(n) equals n ----- 2 \ (j + 8 j) ) qbinomial(2 n, n - 2 j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 5 It so happened that a(n) equals n ----- 2 \ (j + 11 j) ) qbinomial(2 n, n - 2 j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 6 It so happened that a(n) equals n ----- 2 \ (j + 16 j) ) qbinomial(2 n, n - 2 j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 7 It so happened that a(n) equals n ----- 2 \ (j + 17 j) ) qbinomial(2 n, n - 2 j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 8 It so happened that a(n) equals n ----- 2 \ (j + 18 j) ) qbinomial(2 n, n - 2 j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 9 It so happened that a(n) equals n ----- 2 \ (j + 19 j) ) qbinomial(2 n, n - 2 j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 10 It so happened that a(n) equals n ----- 2 \ (3/2 j + 1/2 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 11 It so happened that a(n) equals n ----- 2 \ (3/2 j + 3/2 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 12 It so happened that a(n) equals n ----- 2 \ (3/2 j + 7/2 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 13 It so happened that a(n) equals n ----- 2 \ (3/2 j + 9/2 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 14 It so happened that a(n) equals n ----- 2 \ (3/2 j + 11/2 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 15 It so happened that a(n) equals n ----- 2 \ (3/2 j + 15/2 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 16 It so happened that a(n) equals n ----- 2 \ (3/2 j + 19/2 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 17 It so happened that a(n) equals n ----- 2 \ (3/2 j + 27/2 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 18 It so happened that a(n) equals n ----- 2 \ (3/2 j + 29/2 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 19 It so happened that a(n) equals n ----- 2 \ (3/2 j + 31/2 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 20 It so happened that a(n) equals n ----- 2 \ (3/2 j + 33/2 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 21 It so happened that a(n) equals n ----- 2 \ (3/2 j + 35/2 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 22 It so happened that a(n) equals n ----- 2 \ (j + 3 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 23 It so happened that a(n) equals n ----- 2 \ (j + 4 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 24 It so happened that a(n) equals n ----- 2 \ (j + 10 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 25 It so happened that a(n) equals n ----- 2 \ (j + 11 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 26 It so happened that a(n) equals n ----- 2 \ (j + 12 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 27 It so happened that a(n) equals n ----- 2 \ (j + 14 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 28 It so happened that a(n) equals n ----- 2 \ (j + 17 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 29 It so happened that a(n) equals n ----- 2 \ (j + 18 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ Proof of Problem, 30 It so happened that a(n) equals n ----- 2 \ (j + 19 j) ) qbinomial(2 n, n - j) q / ----- j = -n By the q-Zeilberger algorithm (pre-processed with the amazing Peter Paule sy\ mmetrization) this sum satisfies the given recurrence and since the initial conditions mat\ ch, the solution of the recurrence indeed coincides with the above sum Now take the limit as n goes to infinity, and you are done! ------------------------------------------------------------------------ ----------------------------------