A Quick proof of the Main Result of a 17-page Article that Appeard in the 
May 2017 issue of the journal Advances in Applied Mathematics



                              By Shalosh B. Ekhad



In this note we give  one-line, automatic proofs of the main result (Theorem 1.3) of the paper:

      "Identities involving weighted Catalan, Schroder, and Motkzin paths"

by Zhi Chen and Hao Pan, arXiv:1608.02448v2 [math.CO], that appeared recently in Advances in Applied Mathematics
v. 86 (May 2017), pp.81-98 .

   The left side of Eq. (1.14), let's call it L1(n), satisfies the recurrence



  2                2
(a  n - 2 b a n + b  n) L1(n) + (-2 n a - 2 b n - 3 b - 3 a) L1(n + 1) + (n + 3) L1(n + 2) = 0



  The right side of Eq. (1.14), let's call it R1(n), satisfies the recurrence

  2                2
(a  n - 2 b a n + b  n) R1(n) + (-2 n a - 2 b n - 3 b - 3 a) R1(n + 1) + (n + 3) R1(n + 2) = 0



Since L1(0)=R1(0), and L1(1)=R1(1) (check!), and the two sequences satisfy the same second-order recurrence this proves (1.14) of the above paper



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   The left side of Eq. (1.15), let's call it L2(n), satisfies the recurrence



   2
  a  n L2(n) + (-6 b - 3 a - 4 b n - 2 n a) L2(n + 1) + (n + 3) L2(n + 2) = 0



  The right side of Eq. (1.15), let's call it R2(n), satisfies the recurrence

   2
  a  n R2(n) + (-6 b - 3 a - 4 b n - 2 n a) R2(n + 1) + (n + 3) R2(n + 2) = 0


Since L2(0)=R2(0), and L2(1)=R2(1) (check!), and the two sequences satisfy 
the same second-order recurrence this proves (1.14) of the above paper

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   The left side of Eq. (1.16), let's call it L3(n), satisfies the recurrence




   2
  a  n L3(n) - (2 a n + 4 b n +3 a + 6 b) L3(n + 1) + (n + 3) L3(n + 2)  = 0



  The right side of Eq. (1.16), let's call it R3(n), satisfies the recurrence


   2
  a  n R3(n) - (2 a n + 4 b n +3 a + 6 b) R3(n + 1) + (n + 3) R3(n + 2)  = 0




Since L3(0)=R3(0), and L3(1)=R3(1) (check!), and the two sequences satisfy 
the same second-order recurrence this proves (1.16) of the above paper



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                     The whole think took, 0.972, seconds.