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A352657
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The number of lozenge tilings of a semiregular hexagon of side lengths n, n, 3*n, n, n and 3*n; equivalently, the number of plane partitions whose solid Young diagram fits inside an n X n X 3*n box.
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2
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1, 4, 336, 572572, 19571505408, 13365232267026024, 182001937855822420050000, 49372092168218024268166702560000, 266640931683989945767062736068603511111680, 28657545169614835585678719963104037818950931553412096, 61277278161726929232430881966673334396569563602790616552072890176
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OFFSET
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0,2
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COMMENTS
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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 (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 = n, c = 3*n gives the entries for the present sequence, a superfactorial analog of A005810(n) = binomial(4*n,n).
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LINKS
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FORMULA
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a(n) = S(n)^2*S(3*n)*S(5*n)/(S(2*n)*S(4*n)^2), where S(n) = Product_{k = 0..n-1} k! with S(0) = 1.
a(n) = Product_{i = 1..3*n} (2*n+i-1)!*(i-1)!/(n+i-1)!^2.
a(n) = Product_{i = 1..n} (4*n+i-1)!*(i-1)!/((3*n+i-1)!*(n+i-1)!).
a(n) = Product_{i = 1..3*n} Product_{1 <= j, k <= n) (i + j + k - 1)/(i + j + k - 2).
a(n) = Product_{i = 1..n} Product_{j = 1..n} (3*n + i + j - 1)/(i + j - 1).
a(n) = Product_{i = 1..3*n} Product_{j = 1..n} (n + i + j - 1)/(i + j - 1).
For n >= 1, a(n) = det( (binomial(4*n,n+i-j)) ) for 1 <= i, j <= n. Apply Krattenhaller, Theorem 4 with a = n, b = 3*n and c = n.
a(n) ~ 1/A*(32/(15*n))^(1/12)*exp(B*n^2 + 1/12), where A = 1.2824271291... is the Glaisher-Kinkelin constant A074962 and B = (25/2)*log(5) + (9/2)*log(3) - 34*log(2).
Conjecture 1): 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 2): 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.
a(n) ~ exp(1/12) * 3^(9*n^2/2 - 1/12) * 5^(25*n^2/2 - 1/12) / (A * n^(1/12) * 2^(34*n^2 - 5/12)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, May 16 2022
a(n) = Product_{i = 1..n} Product_{j = 3*n..4*n-1} (i+j) / Product_{j = 0..n-1} (i+j).
a(n) = Product_{i = 1..3*n} Product_{j = n..2*n-1} (i+j) / Product_{j = 0..n-1} (i+j). (End)
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EXAMPLE
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Examples of supercongruences:
p = 5, n = 1, r = 1:
a(5) - a(1)^5 = 13365232267026024 - 4^5 = (2^3)*(5^5)*534609290681 == 0 (mod 5^5).
p = 7, n = 1, r = 1:
a(7) - a(1)^7 = 49372092168218024268166702560000 - 4^7 = (2^8)*(7^4)*42153329 *1905537621534581059 == 0 (mod 7^4).
p = 3, n = 1, r = 2:
a(3^2) - a(3)^3 = 28657545169614835585678719963104037818950931553412096 - 572572^3 = (2^6)*(3^9)*7*13*36206433373771931*6904632711001213215426713099 == 0 (mod 3^9).
exp(Sum_{n >= 1} a(n)*x^n/n) = 1 + 4*x + 176 x^2 + 191540*x^3 + 4893655248*x^4 + 2673066058559752*x^5 + 30333667002369040991520*x^6 + 7053156145366242954671905412736*x^7 + 33330116488711372656254906993570075436704*x^8 + 3184171685646079976603214029980784880572652377971904*x^9 + 6127727816185429609991005336553574169498938182021433716145181760*x^10 + ....
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MAPLE
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S := proc(n) local i; mul(i!, i = 0..n-1) end proc:
a := n -> S(n)^2*S(3*n)*S(5*n)/(S(2*n)*S(4*n)^2):
seq(a(n), n = 0..10);
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MATHEMATICA
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Table[BarnesG[n + 1]^2 * BarnesG[3*n + 1] * BarnesG[5*n + 1] / (BarnesG[2*n + 1] * BarnesG[4*n + 1]^2), {n, 0, 10}] (* Vaclav Kotesovec, May 16 2022 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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