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A170904
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Sequence obtained by a formal reading of Riordan's Eq. (30a), p. 206.
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2
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1, 0, 0, 2, 24, 572, 21280, 1074390, 70299264, 5792903144, 587159944704, 71822748886440, 10435273503677440, 1776780701352504408, 350461958856515690496, 79284041282799128098778, 20392765404792755583221760, 5917934230798152486136427600, 1924427226324694427836833857536
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OFFSET
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0,4
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COMMENTS
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See the comments in A000186 for further discussion.
Neven Juric alerted me to the fact that Riordan's formula is misleading.
It is not error of Riordan, since, according to the rook theory, he considered A000179(1) as -1. [From Vladimir Shevelev, Apr 02 2010]
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REFERENCES
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J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 206, 209.
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LINKS
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Table of n, a(n) for n=0..18.
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FORMULA
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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
a(2n)=A000186(2n), a(2n+1)=A000186(2n+1)+A001700(n)*A000166(n)*A000166(n+1). [From Vladimir Shevelev, Apr 02 2010]
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MAPLE
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# A000166
unprotect(D);
D := proc(n) option remember; if n<=1 then 1-n else (n-1)*(D(n-1)+D(n-2)); fi; end;
[seq(D(n), n=0..30)];
# A000179
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;
[seq(U(n), n=0..30)];
# bad A000186 (A170904)
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;
[seq(Kbad(n), n=0..30)];
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CROSSREFS
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Sequence in context: A210905 A012113 A156525 * A090732 A014298 A280794
Adjacent sequences: A170901 A170902 A170903 * A170905 A170906 A170907
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KEYWORD
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nonn
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AUTHOR
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N. J. A. Sloane, Jan 21 2010
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EXTENSIONS
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Edited by N. J. A. Sloane, Apr 04 2010 following a suggestion from Vladimir Shevelev.
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STATUS
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approved
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