%I
%S 1,0,0,2,24,572,21280,1074390,70299264,5792903144,587159944704,
%T 71822748886440,10435273503677440,1776780701352504408,
%U 350461958856515690496,79284041282799128098778,20392765404792755583221760,5917934230798152486136427600,1924427226324694427836833857536
%N Sequence obtained by a formal reading of Riordan's Eq. (30a), p. 206.
%C See the comments in A000186 for further discussion.
%C Neven Juric alerted me to the fact that Riordan's formula is misleading.
%C It is not error of Riordan, since, according to the rook theory, he considered U(1) to be 1. [_Vladimir Shevelev_, Apr 02 2010]
%C A combinatorial argument, valid for n >= 2, leads to Touchard's formula for the nth menage number, U(n), a formula which involves the coefficients of Chebyshev polynomials of the first kind. It is combinatorially reasonable to take U(0) = 1 and U(1) = 0, leading to A335700, but taking the connection with Chebyshev polynomials seriously instead gives U(0) = 2 and U(1) = 1, leading to A102761. Riordan derives equation (30) on page 205 for the number of reduced threeline Latin rectangles (A000186) by making use of product identities on Chebyshev polynomials, and therefore requires the second definition; it also requires extending the definition of menage numbers to negative index. Riordan then obtains equation (30a) on page 206 by eliminating the negative indices and redefining U(0) to be 1 (which leads to A000179). A170904 (this sequence) is what is obtained by mistakenly using A335700 instead of A000179 in Riordan's equation (30a).  _William P. Orrick_, Aug 11 2020
%D J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 206, 209.
%F One can enumerate 3 X n Latin rectangles by the formula A000186(2n)=a(2n) and A000186(2n+1)=a(2n+1)A001700(n)*A000166(n)*A000166(n+1).  Vladimir Shevelev, Apr 04 2010
%F a(2n)=A000186(2n), a(2n+1)=A000186(2n+1)+A001700(n)*A000166(n)*A000166(n+1). [From _Vladimir Shevelev_, Apr 02 2010]
%p # A000166
%p unprotect(D);
%p D := proc(n) option remember; if n<=1 then 1n else (n1)*(D(n1)+D(n2)); fi; end;
%p [seq(D(n),n=0..30)];
%p # A335700 (equals A000179 except that A335700(1) = 0)
%p U := proc(n) if n<=1 then 1n else add ((1)^k*(2*n)*binomial(2*nk, k)*(nk)!/(2*nk), k=0..n); fi; end;
%p [seq(U(n),n=0..30)];
%p # bad A000186 (A170904)
%p Kbad:=proc(n) local k; global D, U; add( binomial(n,k)*D(nk)*D(k)*U(n2*k), k=0..floor(n/2) ); end;
%p [seq(Kbad(n),n=0..30)];
%K nonn
%O 0,4
%A _N. J. A. Sloane_, Jan 21 2010
%E Edited by _N. J. A. Sloane_, Apr 04 2010 following a suggestion from _Vladimir Shevelev_
