A352656 The number of lozenge tilings of a semiregular hexagon of side lengths n, n, 2*n, n, n and 2*n; equivalently, the number of plane partitions whose solid Young diagram fits inside an n X n X 2*n box.

1, 3, 105, 41580, 184225041, 9095857138368, 4995284546047230864, 30483011847732623089267500, 2065715788914012182693991725390625, 1553908887541345830681718185939775035000000, 12971921694089364427957671958722080861704163596800000

Offset: 0

Author

Peter Bala, Apr 22 2022

Keywords

• nonn
• easy

Comments

A lozenge is a unit rhombus with internal angles of 60 and 120 degrees. A hexagon is semiregular if its internal angles are 120 degrees and opposite sides are of equal length. Let S(n) = Product_{k = 0..n-1} k! = A000178(n-1) for n >= 1. S(n) equals the superfactorial of n-1. Then for a, b and c nonnegative integers a semiregular hexagon with side-lengths a, b, c, a, b, c can be tiled by lozenges in exactly S(a+b+c)*S(a)*S(b)*S(c)/(S(a+b)*S(a+c)*S(b+c)) ways.

The superfactorial ratio F(a,b,c) := (S(a)*S(b)*S(c)*S(a+b+c))/ (S(a+b)*S(a+c)*S(b+c)) is an integer (see MacMahon, Chapter II, Section 429, p. 182, with x -> 1) and can be viewed as the superfactorial analog of the binomial coefficient (a + b)!/(a!*b!).

Setting a = b = c = n, gives S(3*n)*S(n)^3/S(2*n)^3 = A008793(n), a superfactorial analog of A000984(n) = binomial(2*n,n); setting a = b = n, c = 2*n gives the entries for the present sequence, a superfactorial analog of A005809(n) = binomial(3*n,n).

Conjecture 1: the supercongruences F(a*p^r,b*p^r,c*p^r) == F(a*p^(r-1),b*p^(r-1),c*p^(r-1))^p (mod p^(4*k)) hold for all primes p, where r is a positive integer and a, b and c are nonnegative integers.

Examples

			Examples of supercongruences:
p = 5, n = 1, r = 1:
a(5) - a(1)^5 = 9095857138368 - 3^5 = (3^2)*(5^4)*109*367*40423 == 0 (mod 5^4)
p = 7, n = 1, r = 1:
a(7) - a(1)^7 = 30483011847732623089267500 - 3^7 = (3^2)*(7^4)*1716943* 3007843*273156893 = 0 (mod 7^4)
p = 3, n = 1, r = 2:
a(3^2) - a(3)^3 = 1553908887541345830681718185939775035000000 - 41580^3 = (2^10)*(3^17)*(5^3)*7*43*78233*3992066532482127207049 == 0 (mod 3^17)
exp(Sum_{n >= 1} a(n)*x^n/n) = 1 + 3*x + 57*x^2 + 14022*x^3 + 46099458*x^4 + 1819310390847*x^5 + 832552884579020616*x^6 + 4354718475994129490705199*x^7 + 258214486678446939353495542546848*x^8 + 172656543834793205815736306409587678877597*x^9 + 1297192169926906086694501903974161495745648027761154*x^10 + ....
		

Links

• C. Krattenthaler, Advanced Determinant Calculus: A Complement, Linear Algebra Appl. 411 (2005), 68-166; arXiv:math/0503507v2 [math.CO], 2005.
• P. A. MacMahon, Combinatory Analysis, vol. 2, Cambridge University Press, 1916; reprinted by Chelsea, New York, 1960.
• Eric Weisstein's World of Mathematics, Barnes G-function
• Eric Weisstein's World of Mathematics, Plane Partition
• Wikipedia, Superfactorial

Crossrefs

Cf. A000178, A005809, A008793, A074962, A342972, A352657.

Programs

Maple

S := proc(n) local i; mul(i!, i = 0..n-1) end proc:
a := n -> S(4*n)*S(n)^2/S(3*n)^2;
seq(a(n), n = 0..10);

Mathematica

Table[BarnesG[4*n + 1]*BarnesG[n + 1]^2/BarnesG[3*n + 1]^2, {n, 0, 10}] (* Vaclav Kotesovec, May 16 2022 *)

Formula

a(n) = S(4*n)*S(n)^2/S(3*n)^2, where S(n) = Product_{k = 0..n-1} k! with S(0) = 1.

a(n) = G(4*n+1)*G(n+1)^2/G(3*n+1)^2, where G(n) is Barnes G-function.

a(n) = Product_{i = 1..2*n} (2*n+i-1)!*(i-1)!/(n+i-1)!^2.

a(n) = Product_{i = 1..n} (3*n+i-1)!*(i-1)!/((2*n+i-1)!*(n+i-1)!).

a(n) = Product_{i = 1..2*n} Product_{1 <= j, k <= n} (i + j + k - 1)/(i + j + k - 2).

a(n) = Product_{i = 1..n} Product_{j = 1..n} (2*n + i + j - 1)/(i + j - 1).

a(n) = Product_{i = 1..2*n} Product_{j = 1..n} (n + i + j - 1)/(i + j - 1).

a(n) = A342972(2*n,n).

For n >= 1, a(n) = det( (binomial(3*n,n+i-j)) ) for 1 <= i, j <= n. Apply Krattenhaller, Theorem 4 with a = n, b = 2*n and c = n.

a(n+1) = n!^2*(4*n)!*(4*n+1)!*(4*n+2)!*(4*n+3)!/((3*n)!*(3*n+1)!*(3*n+2)!)^2 * a(n) with a(0) = 1.

a(n) ~ 1/A*(9/(4*n))^(1/12)*exp(B*n^2 + 1/12), where A = 1.2824271291... is the Glaisher-Kinkelin constant A074962 and B = 16*log(2) - 9*log(3).

Conjecture 2: the Gauss congruences a(n*p^r) == a(n*p^(r-1)) (mod p^r) hold for all primes p and positive integers n and r. If true, then the expansion of exp(Sum_{n >= 1} a(n)*x^n/n) has integer coefficients.

Conjecture 3: the supercongruences a(n*p^r) == a(n*p^(r-1))^p (mod p^(4*r)) hold for all primes p and positive integers n and r.

From Peter Bala, Feb 14 2023: (Start)

a(n) = Product_{i = 1..2*n} Product_{j = n..2*n-1} (i+j) / Product_{j = 0..n-1} (i+j).

a(n) = Product_{i = 1..n} Product_{j = 2*n..3*n-1} (i+j) / Product_{j = 0..n-1} (i+j). (End)