|
%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 n-th 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 three-line 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 1-n else (n-1)*(D(n-1)+D(n-2)); 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 1-n else add ((-1)^k*(2*n)*binomial(2*n-k, k)*(n-k)!/(2*n-k), 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(n-k)*D(k)*U(n-2*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_
|
