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A000186
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Number of 3 X n Latin rectangles in which the first row is in order.
(Formerly M2140 N0851)
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27
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1, 0, 0, 2, 24, 552, 21280, 1073760, 70299264, 5792853248, 587159944704, 71822743499520, 10435273503677440, 1776780700509416448, 350461958856515690496, 79284041282622163140608, 20392765404792755583221760, 5917934230798104348783083520, 1924427226324694427836833857536
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OFFSET
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0,4
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COMMENTS
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Or number of n X n matrices with exactly one 1 and one 2 in each row and column which are not in the main diagonal, other entries 0. - Vladimir Shevelev, Mar 22 2010
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REFERENCES
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L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 183.
Dulmage, A. L.; McMaster, G. E. A formula for counting three-line Latin rectangles. Proceedings of the Sixth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1975), pp. 279-289. Congressus Numerantium, No. XIV, Utilitas Math., Winnipeg, Man., 1975. MR0392611 (52 #13428). - From N. J. A. Sloane, Apr 06 2012
I. Gessel, Counting three-line Latin rectangles, Lect. Notes Math, 1234(1986), 106-111. [From Vladimir Shevelev, Mar 25 2010]
Goulden and Jackson, Combin. Enum., Wiley, 1983 p. 284.
S. M. Jacob, The enumeration of the Latin rectangle of depth three..., Proc. London Math. Soc., 31 (1928), 329-336.
S. M. Kerawala, The enumeration of the Latin rectangle of depth three by means of a difference equation, Bull. Calcutta Math. Soc., 33 (1941), 119-127.
S. M. Kerawala, The asymptotic number of three-deep Latin rectangles, Bull. Calcutta Math. Soc., 39 (1947), 71-72.
Koichi, Yamamoto, An asymptotic series for the number of three-line Latin rectangles, J. Math. Soc. Japan 1, (1950). 226-241.
W. Moser. A generalization of Riordan's formula for 3Xn Latin rectangles, Discrete Math., 40, 311-313 [From Vladimir Shevelev, Mar 25 2010]
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 210.
V. S. Shevelev, Reduced Latin rectangles and square matrices with identical sums in the rows and columns [Russian], Diskret. Mat., 4 (1992), no. 1, 91-110.
V. S. Shevelev, A generalized Riordan formula for three-rowed Latin rectangles and its applications, DAN of the Ukraine, 2 (1991), 8-12 (in Russian) [From Vladimir Shevelev, Mar 25 2010]
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
D. S. Stones, The many formulas for the number of Latin rectangles, Electron. J. Combin 17 (2010), A1.
D. S. Stones and I. M. Wanless, Divisors of the number of Latin rectangles, J. Combin. Theory Ser. A 117 (2010), 204-215.
RJ Stones, S Lin, X Liu, G Wang, On Computing the Number of Latin Rectangles, Graphs and Combinatorics, Graphs and Combinatorics (2016) 32:1187-1202; DOI 10.1007/s00373-015-1643-1
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LINKS
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FORMULA
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a(n) = n!*Sum_{k+j<=n} (2^j/j!)*k!*binomial(-3*(k+1), n-k-j).
Note that the formula Sum_{k=0..n, k <= n/2} binomial(n, k)*D(n-k)*D(k)*U(n-2*k), where D() = A000166 and U() represents the menage numbers given by Riordan, p. 209 gives the wrong answers unless we set U(1) = -1 (or in other words we must take U() = A000179). With U(1) = 0 (see A335700) it produces A170904. See the Maple code here. - N. J. A. Sloane, Jan 21 2010, Apr 04 2010. Thanks to Vladimir Shevelev for clarifying this comment. Additional changes from William P. Orrick, Aug 12 2020
E.g.f.: exp(2*x) Sum(n>=0; n! x^n /(1+x)^(3*n+3)) from Gessel reference. - Wouter Meeussen, Nov 02 2013
a(n+p)-2*a(n) is divisible by p for any prime p. - Mark van Hoeij, Jun 13 2019
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MAPLE
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for n from 1 to 250 do t0:=0; for j from 0 to n do for k from 0 to n-j do t0:=t0 + (2^j/j!)*k!*binomial(-3*(k+1), n-k-j); od: od: t0:=n!*t0; lprint(n, t0); od:
Maple code for A000186 based on Eq. (30) of Riordan, p. 205. Eq. (30a) on p. 206 is wrong. - N. J. A. Sloane, Jan 21 2010. Thanks to Neven Juric for correcting an error in the definition of fU, Mar 01 2010. Additional comment and modifications of code due to changes in underlying sequences from William P. Orrick, Aug 12 2020: Eq. (30) and Eq. (30a) are, in fact, related to each other by a trivial transformation and are both valid. Current code is based on Eq. (30a).
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)];
U := proc(n) if n=0 then 1 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)];
K:=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(K(n), n=0..30)];
# another Maple program:
a:= proc(n) option remember; `if`(n<5, [1, 0, 0, 2, 24][n+1],
((n-1)*(n^2-2*n+2)*a(n-1) +(n-1)*(n-2)*(n^2-2*n+2)*a(n-2)
+(n-1)*(n-2)*(n^2-2*n-2) *a(n-3)
+2*(n-1)*(n-2)*(n-3)*(n^2-5*n+3) *a(n-4)
-4*(n-2)*(n-3)*(n-4)*(n-1)^2 *a(n-5)) / (n-2))
end:
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MATHEMATICA
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a[n_] := (t0 = 0; Do[t0 = t0 + (2^j/j!)*k!*Binomial[-3*(k+1), n-k-j], {j, 0, n}, {k, 0, n-j}]; n!*t0); Table[a[n], {n, 0, 18}] (* Jean-François Alcover, Oct 13 2011, after Maple *)
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PROG
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(SageMath)
# after Maple code based on Riordan's Eq. (30a)
d = [1, 0]
for j in range(2, 31):
d.append((j - 1) * (d[-1] + d[-2]))
def u(n):
if n == 0:
return 1
else:
return sum((-1)^k * (2 * n) * binomial(2 * n - k, k) * factorial(n - k) / (2 * n - k) for k in range(0, n + 1))
def k(n):
return sum(binomial(n, k) * d[n - k] * d[k] * u(n - 2 * k) for k in range(0, floor(n / 2) + 1))
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CROSSREFS
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KEYWORD
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nonn,nice,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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