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%I #51 Jan 25 2025 12:15:49
%S 1,78,2366,41405,496860,4504864,32821152,200443464,1057896060,
%T 4936848280,20734762776,79483257308,281248448936,927192688800,
%U 2869882132000,8394405236100,23331508670925,61912369414350,157496378334750,385451662766625,910400117772600
%N Number of 2 X 11 matrices with elements in 0..n with each row and each column in nondecreasing order. 2,11,n can be permuted, see formula.
%C In the definition, 2,11,n can be permuted, see formula.
%C Conjecture: 12th column (and diagonal) of the triangle A001263. - _Bruno Berselli_, May 07 2012
%D S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; Prop. 8.4, case n=13. - _N. J. A. Sloane_, Aug 28 2010.
%H Feihu Liu, Guoce Xin, and Chen Zhang, <a href="https://arxiv.org/abs/2412.18744">Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS</a>, arXiv:2412.18744 [math.CO], 2024. See pp. 1, 3, 25, 27.
%F Empirical: Set p,q,r to n,11,2 (in any order) in s=p+q+r-1; a(n) = product {i in 0..r-1} (binomial(s,p+i)*i!/(s-i)^(r-i-1))
%F G.f. conjectured: (1 + 55*x + 825*x^2 + 4950*x^3 + 13860*x^4 + 19404*x^5 + 13860*x^6 + 4950*x^7 + 825*x^8 + 55*x^9 + x^10)/(1 - x)^23. - _Bruno Berselli_, May 07 2012
%F Conjecture: a(n) = ((n+12)/(12*n+12))*binomial(n+11,11)^2. - _Bruno Berselli_, May 07 2012
%F Conjecture: a(n) = Product_{i=1..11} A002378(n+i)/A002378(i). - _Bruno Berselli_, Sep 01 2016
%F From _Amiram Eldar_, Oct 19 2020: (Start)
%F Conjecture: Sum_{n>=0} 1/a(n) = 3538258540001/8820 - 40646320*Pi^2.
%F Conjecture: Sum_{n>=0} (-1)^n/a(n) = 1678950598/2205 - 23068672*log(2)/21. (End)
%Y Cf. A001263, A134291, A140925.
%K nonn,changed
%O 0,2
%A _R. H. Hardin_, Jul 05 2008