A005157 Number of totally symmetric plane partitions that fit in an n X n X n box.
1, 2, 5, 16, 66, 352, 2431, 21760, 252586, 3803648, 74327145, 1885102080, 62062015500, 2652584509440, 147198472495020, 10606175914819584, 992340657705109416, 120567366227960791040, 19023173201224270401428, 3897937005297330777227264
Offset: 0
Examples
a(2) = 5 because we have: void, 1, 21/1, 22/21, and 22/22.
References
- 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).
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..130 (first 51 terms from T. D. Noe)
- Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
- R. K. Guy, Letter to N. J. A. Sloane, Dec 5 1988.
- Graham H. Hawkes, Totally symmetric partitions in boxes
- Seth Ireland, A bijection between strongly stable and totally symmetric partitions, arXiv:2302.02505 [math.CO], 2023.
- C. Klivans, Obstructions to shiftedness, preprint.
- C. Klivans, Obstructions to shiftedness, Discrete Comput. Geom., 33 (2005), 535-545.
- Don Knuth, A conjecture about noncrossing paths, Feb 06 2019.
- R. P. Stanley, A baker's dozen of conjectures concerning plane partitions, pp. 285-293 of "Combinatoire Enumerative (Montreal 1985)", Lect. Notes Math. 1234, 1986.
- R. P. Stanley, A baker's dozen of conjectures concerning plane partitions, pp. 285-293 of "Combinatoire Enumerative (Montreal 1985)", Lect. Notes Math. 1234, 1986. Preprint. [Annotated scanned copy]
Programs
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Maple
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
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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PARI
A005157(n)=prod(i=1,n,prod(j=i,n,(i+j+n-1)/(i+2*j-2))) \\ M. F. Hasler, Sep 26 2018
Formula
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
Apparently, a(n) = Sum_{k=0..n} A184173(n,k). - Alois P. Heinz, Feb 11 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
Comments