[seqfan] Riordan's classical formula and Juric's calculations

Vladimir Shevelev shevelev at bgu.ac.il
Fri Apr 2 09:53:53 CEST 2010

Dear seqfans,
I was very confused when in comment to Riordan's reference A000186   I read that, due to N.Juric, the famous Riordan's explicit formula for the number of 3Xn Latin rectangles is wrong...although in my paper "Reduced Latin rectangles and square indentical sums in the rows and columns" (Journal of Academy of Russia " Discrete Math, 4 (1992), 91-110) I gave another proof of it (with some generalizations). If to read Riordan's formula (30a)  in his "An Introduction to Combinatorial Analysis" , Wiley, 1967 (copyright from 1958)  with u_n=A000179(n), then it, indeed, seems incorrect. But instead of A000179(1)=0, Riordan accepted u_1=-1 and such value is justified with the rook point of view (see note on p.197 in his book: "However, by equation (4),
 M_1(x)=1+2x, which implies u_1(t)=-1+2t (here there is a trivial misprint: it should be, of course, u_1(t) (not u_2(t)) and thusd u_1=-1. Since in his further arguments Riordan uses the rook theory,
u_1=-1 appears again and again.  For example, if to put n=3 in the Cayley recursion (19) on p. 201 :
(n-2)*u_n=n*(n-2)*u_(n-1)+n*u_(n-2)+4*(-1)^(n+1), which is proved also by the rook tecnique, then we have 1*u_3=3*1*u_2+3*u_1+4 and on view of u_2=0, u_3=1, we again find u_1=-1.
Therefore, in my mentioned above paper I inderlined in formula (3) of Introduction that  u_0=1,u_1=-1,u_2=0. Thus Juric made a mistake ascribing sequence A170904 to Riordan.
Actually,  it is easy to see that A170904(2n)=A000186(2n), while A170904(2n+1)=
Best regards,
Vladimir Shevelev

 Shevelev Vladimir‎

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