A008793 -id:A008793 - cp's OEIS frontend

A322760 Denominator of Product_{i=1..n, j=1..n, k=1..n, m=1..n} (i+j+k+m-2)/(i+j+k+m-3).

1, 1, 3, 25, 343, 1701, 117406179, 1377144939510210299, 7319582701014978251, 68957488738377254912, 31655878065328599351648522128233122128960851, 509835253962108047546712884688217781173085736103608399459708858847501357, 254487582354743232569912323057838854214490688232159609663484703619

Offset: 0

N. J. A. Sloane, Dec 28 2018

The sequence of rationals A322759/A322760 is the natural successor to A000984 and A008793.

			1, 2, 500/3, 406594944/25, 17946021496671970416/343, 266763232240477060473636852903291418/1701, ...

Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 576, line 3, with a=b=c=d=n.

Cf. A000984, A008793, A322759.

f4:=proc(n) local a,i,j,k,m;
a:=1;
for i from 1 to n do
for j from 1 to n do
for k from 1 to n do
for m from 1 to n do
a:=a*(i+j+k+m-2)/(i+j+k+m-3); od: od: od: od: a; end;
t1 := [seq(f4(n),n=0..12)];

a(n) = denominator(prod(i=1,n,prod(j=1,n,prod(k=1,n,prod(m=1,n,(i+j+k+m-2)/(i+j+k+m-3)))))) \\ Georg Fischer, Mar 16 2025

1 prepended by Georg Fischer, Mar 16 2025

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

Original entry on oeis.org

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

Offset: 0

Peter Bala, Apr 22 2022

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 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 + ....

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

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

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);

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

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)

A352657 The number of lozenge tilings of a semiregular hexagon of side lengths n, n, 3n, n, n and 3n; equivalently, the number of plane partitions whose solid Young diagram fits inside an n X n X 3*n box.

Original entry on oeis.org

1, 4, 336, 572572, 19571505408, 13365232267026024, 182001937855822420050000, 49372092168218024268166702560000, 266640931683989945767062736068603511111680, 28657545169614835585678719963104037818950931553412096, 61277278161726929232430881966673334396569563602790616552072890176

Offset: 0

Peter Bala, Apr 22 2022

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).

			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 + ....

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, Plane Partition
Wikipedia, Superfactorial

Cf. A000178, A005810, A008793, A074962, A352656.

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);

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 *)

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
From Peter Bala, Feb 15 2023: (Start)
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)

A259049 Number of self-complementary plane partitions in a (2n)-cube.

Original entry on oeis.org

1, 4, 400, 960400, 54218191104, 71410553858811024, 2186315392560559723530496, 1552832545847343203950118294425600, 25554649541466337940020968722797075170918400, 9736551559782513812975251884508283964266367033264640000

Offset: 0

Peter J. Taylor, Jun 17 2015

Odd cubes have no self-complementary plane partitions.

R. P. Stanley, Symmetries of Plane Partitions, J. Comb. Theory Ser. A 43 (1986), 103-113.
P. J. Taylor, Counting distinct dimer hex tilings, Preprint, 2015.

Cf. A008793.

a(n) = prod(i=0, n-1, i!^2*(i+2*n)!^2 / (i+n)!^4) \\ Michel Marcus, Jun 18 2015

a(n) = Product_{i=0..n-1} i!^2 (i+2n)!^2 / (i+n)!^4.

a(n) = A008793(n)^2.

A291879 Number of monomials of the Schubert polynomial of the permutation 351624 tensor 1^n.

Original entry on oeis.org

1, 8, 6720, 561120560, 4557185891241984, 3571558033324129373292768, 269111599998006391761541640176800000, 1945556482213500279178010210766074095827609600000, 1347912754604769492992184400055703948513202427323999206349209600

Offset: 0

Alejandro H. Morales, Sep 05 2017

nonn

The permutation 351624 tensor 1^n is the permutation whose permutation matrix is obtained from that of 351624 by replacing each 1 with an n X n identity matrix.

			For n=1 we have that a(1)=8 since the Schubert polynomial of 351624 equals the following sum of eight monomials: x0^3*x1^3*x2 + x0^3*x1^2*x2^2 + x0^2*x1^3*x2^2 + x0^3*x1^3*x3 + x0^3*x1^2*x2*x3 + x0^2*x1^3*x2*x3 + x0^3*x1^2*x3^2 + x0^2*x1^3*x3^2.

A. H. Morales, I. Pak, G. Panova, Hook formulas for skew shapes III. Multivariate and product formulas, arXiv:1707.00931 [math.CO], 2017.
R. P. Stanley, Some Schubert shenanigans, arXiv:1704.00851 [math.CO], 2017.

Cf. A000178, A008793, A291871.

Table[BarnesG[n + 1]^5 * BarnesG[3*n + 1]^2 * BarnesG[5*n + 1] / (BarnesG[2*n + 1]^4 * BarnesG[4*n + 1]^2), {n, 0, 10}] (* Vaclav Kotesovec, Apr 08 2021 *)

b(n) = prod(k=1, n-1, k!);
a(n) = b(n)^5*b(3*n)^2*b(5*n)/(b(2*n)^4*b(4*n)^2); \\ Michel Marcus, Sep 07 2017

def b(n): return mul([factorial(i) for i in range(1,n)])
def a(n): return b(n)^5*b(3*n)^2*b(5*n)/(b(2*n)^4*b(4*n)^2)
[a(n) for n in range(10)]

a(n) = b(n)^5*b(3*n)^2*b(5*n)/(b(2*n)^4*b(4*n)^2) where b(n) = 1!*2!*...*(n-1)! is a superfactorial A000178(n-1). [corrected by Vaclav Kotesovec, Apr 08 2021]
a(n) = c(n)*b(3*n)^2*b(6*n)/((7*n^2)!*b(2*n)^2*b(4*n)^2) where b(n) = 1!*2!*...*(n-1)! is a superfactorial A000178(n-1) and c(n) = A291871.
a(n) ~ exp(1/6) * 3^(9*n^2 - 1/6) * 5^(25*n^2/2 - 1/12) / (A^2 * n^(1/6) * 2^(40*n^2 - 2/3)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Apr 08 2021

A291908 Number of standard Young tableaux of skew shape lambda/mu where lambda is the staircase (4n-1,4n-2,...,2,1) and mu is the square n^n.

Original entry on oeis.org

1, 16, 4362327818240, 19265181532031090042534736325278852710400, 830325323503973129435791248069702287019820905338483131168940909920954227594481411031040

Offset: 0

Alejandro H. Morales, Sep 05 2017

nonn

The number of standard Young tableaux of a fixed skew shape has a determinantal formula, the Jacobi-Trudi formula. It is rare when a family of skew shapes has a product formula for the number of standard Young tableaux. This product formula has independently been proved using P-Schur functions (by DeWitt) and using the Naruse hook-length formula for skew shapes (by Morales, Pak and Panova).

			a(1)=16 since there are 16 standard Young tableaux of skew shape 321/1 since this is the same as the number of standard Young tableaux of straight shape 321 given by the hook-length formula: 16 = 6!/(3^2*5).

E. A. DeWitt, Identities Relating Schur s-Functions and Q-Functions, Ph.D. thesis, University of Michigan, 2012, 73 pp.
A. H. Morales, I. Pak, G. Panova, Hook formulas for skew shapes III. Multivariate and product formulas, arXiv:1707.00931 [math.CO], 2017.

Cf. A000178, A057863, A008793, A291871, A000085, A061343, A005118.

b:=n->mul(factorial(i),i=1..n-1):
c:=n->mul(doublefactorial(2*i-1),i=1..n-1):
a:=n->factorial(binomial(4*n,2)-n^2)*b(n)^3*b(3*n)*c(n)*c(3*n)/(b(2*n)^3*c(2*n)^2*c(4*n)):
seq(a(n),n=0..9);

def b(n): return mul([factorial(i) for i in range(1,n)])
def d(n): return factorial(n+1)/(2^((n+1)/2)*factorial((n+1)/2))
def c(n): return mul([d(2*i-1) for i in range(1,n)])
def a(n):
    return factorial(binomial(4*n,2)-n^2)*b(n)^3*b(3*n)*c(n)*c(3*n)/(b(2*n)^3*c(2*n)^2*c(4*n))
[a(n) for n in range(10)]

a(n) = (binomial(4*n,2)-n^2)!*b(n)^3*b(3*n)*c(n)*c(3*n)/(b(2*n)^3*c(2*n)^2*c(4*n)) where b(n) = 1!*2!*...*(n-1)! is the superfactorial A000178(n-1), and c(n) = 1!!*3!!*...*(2*n-3)!! is super doublefactorial A057863(n-1).

a(n) ~ sqrt(Pi) * 3^(9*n^2 - 3*n/2 - 1/24) * 7^(7*n^2 - 2*n + 1/2) * exp(7*n^2/2 - 2*n + 23/56) * n^(7*n^2 - 2*n + 7/8) / (A^(3/2) * 2^(33*n^2 - 6*n - 7/8)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Apr 08 2021

A071096 Number of ways to tile hexagon of edges n, n+1, n+2, n, n+1, n+2 with diamonds of side 1.

Original entry on oeis.org

1, 10, 490, 116424, 133613766, 739309710568, 19702998159210080, 2527580342020127455360, 1560172391098377453031770400, 4632518859090968506120863642225000, 66153724447703043353053979949899667187500, 4542776083800437392420665771479758969781250000000, 1499928882906010042230116408158354282455601808812500000000

Offset: 0

N. J. A. Sloane, May 28 2002

nonn

J. Propp, Enumeration of matchings: problems and progress, pp. 255-291 in L. J. Billera et al., eds, New Perspectives in Algebraic Combinatorics, Cambridge, 1999 (see page 261).

J. Propp, Updated article
J. Propp, Enumeration of matchings: problems and progress, in L. J. Billera et al. (eds.), New Perspectives in Algebraic Combinatorics

Cf. A008793, A071094, A071095.

Table[Product[Product[Product[(i+j+k-1)/(i+j+k-2),{i,1,n+1}],{j,1,n+1}],{k,1,n+1}],{n,0,10}]/2 (* Alexander Adamchuk, Jul 10 2006 *)

{a(n) = abs(matdet(matrix(n+1, n+1, i, j, binomial(2*(n+1), i+j))))}; \\ Shifted by Georg Fischer, Jun 19 2022

Product_{i=0..a-1} Product_{j=0..b-1} Product_{k=0..c-1} (i+j+k+2)/(i+j+k+1) with a=n, b=n+1, c=n+2.
a(n) = (-1)^floor((n+1)/2)*det(M(n+1)) where M(n) is the n X n matrix m(i, j)=C(2n, i+j), with i and j ranging from 1 to n. - Benoit Cloitre, Jan 31 2003
a(n) = (1/2)*Product[Product[Product[(i+j+k-1)/(i+j+k-2),{i,1,n+1}],{j,1,n+1}],{k,1,n+1}]. a(n) = A008793(n+1)/2. - Alexander Adamchuk, Jul 10 2006
a(n) ~ exp(1/12) * 3^(9*n^2/2 + 9*n + 53/12) / (A * n^(1/12) * 2^(6*n^2 + 12*n + 27/4)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant. - Vaclav Kotesovec, Apr 26 2015

A115965 Number of planar subpartitions of size n pyramidal planar partition.

Original entry on oeis.org

1, 2, 9, 96, 2498, 161422, 26217833, 10794429504

Offset: 0

Franklin T. Adams-Watters, Mar 14 2006

This is a 2-dimensional analog of the Catalan numbers C_n (A000108). The number of subpartitions of the triangular partition [n,n-1,...,1] is C_{n+1}. The planar partition having its subpartitions counted is:
n n-1 ... 2 1
n-1 n-2 ... 1
... ...
2 1
1

			The 9 planar subpartitions of [2,1|1] are [], [1], [2], [1,1], [1|1], [2,1], [2|1], [1,1|1] and [2,1|1] itself, so a(2)=9. (Here "," separates values on the same line and "|" separates lines.)

Oleg Lazarev, Matt Mizuhara, Ben Reid, Some Results in Partitions, Plane Partitions, and Multipartitions

Cf. A115728, A115729, A000219, A000108, A008793.

A128477 Structure counts for certain radical benzenoidal structures (see Cash and Dias paper for precise definition).

Original entry on oeis.org

306, 59886, 39045432, 100645429215

Offset: 0

N. J. A. Sloane, May 08 2007

Gordon G. Cash and Jerry Ray Dias, Computation, Properties and Resonance Topology of Benzenoid Monoradicals and Polyradicals and the Eigenvectors Belonging to Their Zero Eigenvalues, J. Math. Chem., 30 (2001), 429-444.

Cf. A008793, A128476-A128484.

A128479 Structure counts for certain radical benzenoidal structures (see Cash and Dias paper for precise definition).

Original entry on oeis.org

273956, 604073400, 3859415491248

Offset: 0

N. J. A. Sloane, May 08 2007

Gordon G. Cash and Jerry Ray Dias, Computation, Properties and Resonance Topology of Benzenoid Monoradicals and Polyradicals and the Eigenvectors Belonging to Their Zero Eigenvalues, J. Math. Chem., 30 (2001), 429-444.

Cf. A008793, A128476-A128484.

cp's OEIS Frontend

A322760 Denominator of Product_{i=1..n, j=1..n, k=1..n, m=1..n} (i+j+k+m-2)/(i+j+k+m-3).

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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.

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A352657 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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A259049 Number of self-complementary plane partitions in a (2n)-cube.

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A291879 Number of monomials of the Schubert polynomial of the permutation 351624 tensor 1^n.

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A291908 Number of standard Young tableaux of skew shape lambda/mu where lambda is the staircase (4*n-1,4*n-2,...,2,1) and mu is the square n^n.

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A071096 Number of ways to tile hexagon of edges n, n+1, n+2, n, n+1, n+2 with diamonds of side 1.

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

A352657 The number of lozenge tilings of a semiregular hexagon of side lengths n, n, 3n, n, n and 3n; equivalently, the number of plane partitions whose solid Young diagram fits inside an n X n X 3*n box.

A291908 Number of standard Young tableaux of skew shape lambda/mu where lambda is the staircase (4n-1,4n-2,...,2,1) and mu is the square n^n.