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A005157
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Number of totally symmetric plane partitions that fit in an n X n X n box.
(Formerly M1499)
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8
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1, 2, 5, 16, 66, 352, 2431, 21760, 252586, 3803648, 74327145, 1885102080, 62062015500, 2652584509440, 147198472495020, 10606175914819584, 992340657705109416, 120567366227960791040, 19023173201224270401428, 3897937005297330777227264
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OFFSET
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0,2
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COMMENTS
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Also, number of 2-dimensional shifted complexes on n+1 nodes. [Klivans]
Also the number of totally symmetric partitions which fit in an (n-1)-dimensional box with side length 4 (for n>0). - Graham H. Hawkes, Jan 11 2014
Suppose we index this sequence slightly differently. Let the elements of a partition be represented by points rather than boxes, as in a Ferrers diagram. In this case, a 1 X 1 X 1 (closed) box would fit 8 points -— one at each vertex of the box, and we use the convention that a 0 X 0 X 0 (closed) box contains exactly one point. Using this indexing, the sequence begins (offset is still 0) 2,5,16,... rather than 1,2,5,... If we use the same method of indexing for all other dimensions, then we have the following remarkable result: The number of totally symmetric partitions which fit inside a d-dimensional box with side length n is equal to the number of totally symmetric partitions which fit inside an n-dimensional box of side length d. - Graham H. Hawkes, Jan 11 2014
For two other contexts where this sequence arises, see the Knuth (2019) link (noncrossing paths among the 2(2^n-1) paths defined in that note; independent sets of paths among the first 2^n-1 of those). - N. J. A. Sloane, Feb 09 2019, based on email from Don Knuth.
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REFERENCES
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D. M. Bressoud, Proofs and Confirmations, Camb. Univ. Press, 1999; Eq. (6.8), p. 198 (corrected).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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FORMULA
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a(n) = Product_{i=1..n} Product_{j=i..n} Product_{k=j..n} (i+j+k-1)/(i+j+k-2). - Paul Barry, May 13 2008
a(n) ~ exp(1/72) * GAMMA(1/3)^(2/3) * n^(7/72) * 3^(3*n*(n+1)/4 + 11/72) / (A^(1/6) * Pi^(1/3) * 2^(n*(2*n+1)/2 + 13/24)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant. - Vaclav Kotesovec, Mar 01 2015
a(n) = sqrt(A323848(n+1,n)) for n >= 1. [proof by Nikolai Beluhov; see Knuth (2019) link] - Alois P. Heinz, Feb 10 2019
Conjectures: if p == 1 (mod 6) is prime then a(p) == 2^((p+5)/6) (mod p^2); if p == 5 (mod 6) is prime then a(p) == 2^((p+1)/6) (mod p^2) (checked up to p = 1009). - Peter Bala, Feb 17 2023
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EXAMPLE
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a(2) = 5 because we have: void, 1, 21/1, 22/21, and 22/22.
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MAPLE
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A005157 := proc(n) local i, j; mul(mul((i+j+n-1)/(i+2*j-2), j=i..n), i=1..n); end;
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MATHEMATICA
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Table[Product[(i+j+k-1)/(i+j+k-2), {i, n}, {j, i, n}, {k, j, n}], {n, 0, 20}] (* Harvey P. Dale, Jul 17 2011 *)
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PROG
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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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STATUS
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approved
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