cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A099116 Number of rhombus tilings of a hexagon with side lengths 2n+2,2n,2n+2,2n+2,2n,2n+2 which contain the rhombus above and next to the center of the hexagon.

Original entry on oeis.org

812, 579667803, 235948437342837440, 52366358060537007928863206000, 6262970727027052056580468670430288168750000, 401820562589647140572840734882930708995214500792163500000000
Offset: 1

Views

Author

Ralf Stephan, Oct 01 2004

Keywords

Crossrefs

Programs

  • Mathematica
    G = BarnesG; a[n_] := (G[2n+1]^(-2n-1) (G[2n+2] G[2n+4])^(2(n+1)) G[6n+5]( (4 Binomial[2n, n]^3)/Binomial[6n+4, 3n+2] + 1/3))/(G[2n+3]^(2n) (Gamma[ 2n+1] Gamma[2n+3])^(2(n+1))(G[4n+3]^2 G[4n+5])); Array[a, 6] (* Jean-François Alcover, Feb 20 2019 *)
  • PARI
    a(n)=(1/3+4*binomial(2*n,n)^3/binomial(6*n+4,3*n+2))*prod(i=1,2*n+2,prod(j=1,2*n,prod(k=1,2*n+2,(i+j+k-1)/(i+j+k-2))))

Formula

a(n) ~ exp(1/12) * 3^(83/12 + 24*n + 18*n^2) / (A * n^(1/12) * 2^(71/6 + 32*n + 24*n^2)), where A = A074962 is the Glaisher-Kinkelin constant. - Vaclav Kotesovec, Aug 29 2023

A099117 Number of rhombus tilings of a hexagon with side lengths 2n+3,2n-1,2n+3,2n+3,2n-1,2n+3 which contain the rhombus above and next to the center of the hexagon.

Original entry on oeis.org

152, 436381660, 574954797841668608, 388062759166540341977143692000, 137515819873369461005150742745259538637500000, 25797761881848486655895899589856317740988916476499759600000000
Offset: 1

Views

Author

Ralf Stephan, Oct 01 2004

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := (1/3+2(6n^2+9n+2)/(n+1)^2 Binomial[2n, n]^3/Binomial[6n+4, 3n+2]) Product[(i+j+k-1)/(i+j+k-2), {i, 1, 2n+3}, {j, 1, 2n-1}, {k, 1, 2n+3}];
    Array[a, 6] (* Jean-François Alcover, Nov 18 2018, from PARI *)
  • PARI
    a(n)=(1/3+2*(6*n*n+9*n+2)/(n+1)^2*binomial(2*n,n)^3/binomial(6*n+4,3*n+2))*prod(i=1,2*n+3,prod(j=1,2*n-1,prod(k=1,2*n+3,(i+j+k-1)/(i+j+k-2))))

Formula

a(n) ~ exp(1/12) * 3^(137/12 + 30*n + 18*n^2) / (A * n^(1/12) * 2^(131/6 + 40*n + 24*n^2)), where A = A074962 is the Glaisher-Kinkelin constant. - Vaclav Kotesovec, Aug 29 2023

A120258 Triangle of central coefficients of generalized Pascal-Narayana triangles.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 6, 3, 1, 1, 20, 20, 4, 1, 1, 70, 175, 50, 5, 1, 1, 252, 1764, 980, 105, 6, 1, 1, 924, 19404, 24696, 4116, 196, 7, 1, 1, 3432, 226512, 731808, 232848, 14112, 336, 8, 1, 1, 12870, 2760615, 24293412, 16818516, 1646568, 41580, 540, 9, 1
Offset: 0

Views

Author

Paul Barry, Jun 13 2006

Keywords

Comments

Columns are the central coefficients of the triangles T(n, k;r) with T(n, k;r)=Product{j=0..r, C(n+j, k+j)/C(n-k+j, j)}*[k<=n]; (r=0,A007318), (r=1;A001263),(r=2,A056939),(r=3,A056940),(r=4,A056941). Essentially A103905 as a number triangle with an extra diagonal of 1's. Central coefficients T(2n, n) are A008793. Row sums are A120259. Diagonal sums are A120260.

Examples

			Triangle begins:
  1;
  1,   1;
  1,   2,     1;
  1,   6,     3,     1;
  1,  20,    20,     4,    1;
  1,  70,   175,    50,    5,   1;
  1, 252,  1764,   980,  105,   6, 1;
  1, 924, 19404, 24696, 4116, 196, 7, 1;
  ...
		

Crossrefs

Row sums give A120259.

Programs

  • PARI
    T(n, k) = prod(j=0, k-1, binomial(2*n-2*k+j, n-k)/binomial(n-k+j, j)); \\ Seiichi Manyama, Apr 02 2021

Formula

Number triangle T(n, k)=[k<=n]*Product{j=0..k-1, C(2n-2k+j, n-k)/C(n-k+j, j)}
As a square array, this is T(n,m)=product{k=1..m, product{j=1..n, product{i=1..n, (i+j+k-1)/(i+j+k-2)}}}; - Paul Barry, May 13 2008

A071095 Number of ways to tile hexagon of edges n, n+1, n+1, n, n+1, n+1 with diamonds of side 1.

Original entry on oeis.org

1, 6, 175, 24696, 16818516, 55197331332, 872299918503728, 66345156372852988800, 24277282058281388285162560, 42730166102274086598901662210000, 361690697335823816369045433734882109375, 14721491647169381835282394824891766183125000000, 2880942480871157389699990094736740229925045312500000000
Offset: 0

Views

Author

N. J. A. Sloane, May 28 2002

Keywords

References

  • 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).

Crossrefs

Programs

  • Mathematica
    Table[Product[(i+j+k+2)/(i+j+k+1),{i,0,n-1},{j,0,n},{k,0,n}],{n,0,15}] (* Vaclav Kotesovec, Apr 26 2015 *)
  • PARI
    a(n) = prod(i=0, n-1, prod(j=0, n, prod(k=0, n, (i+j+k+2)/(i+j+k+1)))) \\ Michel Marcus, May 20 2013

Formula

a(n) = 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=c=n+1.
a(n) ~ exp(1/12) * 3^(9*n^2/2 + 6*n + 23/12) / (A * n^(1/12) * 2^(6*n^2 + 8*n + 11/4)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant. - Vaclav Kotesovec, Apr 26 2015
a(n) = (-1)^floor(n/2)*det(M(n)) where M(n) is the n X n matrix with m(i,j) = binomial(2*n+i+j,i+j). - Benoit Cloitre, Oct 22 2022

A181119 Number of transpose-complementary plane partitions of n.

Original entry on oeis.org

1, 2, 84, 81796, 1844536720, 962310111888300, 11608208114358751650000, 3236574482779383546336417240000, 20853456581643133066208521560263633137920, 3104385823530881109001458753652585998600603921849920, 10676554307318599842868990948461304923921623250562199975300214736
Offset: 0

Views

Author

Arvind Ayyer, Jan 21 2011

Keywords

Comments

The complement of a plane partition inside an m X m X m cube consists of the boxes which are within the cube, but not in the plane partition, rotated in an appropriate way.
a(n) is the number of plane partitions inside an 2n X 2n X 2n cube whose (matrix) transpose when written as an 2n X 2n array is the same as its complement.

Examples

			When n=2, there are two transpose-complementary plane partitions,
[1 1] and [2 1], both of whose transpose and complement is equal to themselves.
[1 1]     [1 0]
		

Crossrefs

Programs

  • Mathematica
    Table[Binomial[3n-1,n]Product[(2n+i+j+1)/(i+j+1),{i,1,2n-2}, {j,i,2n-2}], {n,0,10}] (* Harvey P. Dale, Jan 27 2012 *)
  • PARI
    a(n) = binomial(3*n-1,n)*prod(i=1,2*n-2,prod(j=i,2*n-2,(2*n+i+j+1)/(i+j+1))); \\ Michel Marcus, Jun 18 2015

Formula

a(n) = binomial(3n-1,n)*Product(i=1..2n-2,Product(j=i..2n-2,(2n+i+j+1)/(i+j+1))).
a(n) ~ exp(1/24) * 3^(9*n^2 - 3*n/2 - 1/24) / (sqrt(A) * n^(1/24) * 2^(12*n^2 - n - 1/3)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant. - Vaclav Kotesovec, Feb 28 2015

A380346 Number of corona for a hexagon of edge n with diamonds of side 1.

Original entry on oeis.org

18, 198, 1298, 5778, 19602, 54758, 132498, 287298, 571538, 1060902
Offset: 0

Views

Author

Craig Knecht, Jan 22 2025

Keywords

Comments

The number of diamonds that can surround a hexagon(n) fall into four categories: A016945(n), A016945(n) + 1, A016945(n) + 2, and A016945(n) + 3.
The number of coronal tilings for A016945(n) is 2.
The number of coronal tilings for A016945(n) + 1 is 9,36,81,144,225, see A016766.
The number of coronal tilings for A016945(n) + 2 is 6,96,486,1536,3750,7776,14406 = 6*A000583.
The number of coronal tilings for A016945(n) + 3 is 1,64,729,4096,15625, see A001014.
A008793 looks at the enumeration of diamonds inside the hexagon. In contrast this looks at the enumeration of diamond corona of the hexagon.

Crossrefs

Formula

a(n) = n^6 + 6*n^5 + 21*n^4 + 44*n^3 + 60*n^2 + 48*n + 18 (conjectured).

A066931 Number of ways to tile hexagon of edge n with diamonds of side 1, not counting rotations and reflections as different.

Original entry on oeis.org

1, 1, 6, 113, 20174, 22306955, 123222909271, 3283834214485890, 421263391026827547540, 260028731850596651411721718, 772086476515163830856527013278243, 11025620741283840573496993339545350520150, 757129347300072898736973484532998417574513923224
Offset: 0

Views

Author

R. K. Guy, Feb 05 2002

Keywords

Crossrefs

Cf. A008793.

Formula

From Peter J. Taylor, Jun 17 2015: (Start)
For odd n, a(n) = A008793(n)/12 + A049505(n)/4 + A006366(n)/6.
For even n, a(n) = A008793(n)/12 + A049505(n)/4 + A006366(n)/6 + A181119(n/2)/4 + A259049(n/2)/12 + A049503(n/2)/6.
See Taylor link.
(End)

Extensions

One more term from Don Reble, Feb 07 2002
More terms from Peter J. Taylor, Jun 17 2015

A071094 Number of ways to tile hexagon of edges n, n, n+1, n, n, n+1 with diamonds of side 1.

Original entry on oeis.org

1, 3, 50, 4116, 1646568, 3184461423, 29706808370096, 1335119245893326400, 288882990167192721013376, 300792059519113653077154558000, 1506680146887473588202049621593937500, 36298820709557430183399305000196605531250000, 4205446372314569673006362329181090368935937500000000, 2342761095072644391194625697884219372917666852341417500000000
Offset: 0

Views

Author

N. J. A. Sloane, May 28 2002

Keywords

References

  • 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).

Crossrefs

Programs

  • Mathematica
    Table[Product[(i+j+k+2)/(i+j+k+1),{i,0,n-1},{j,0,n-1},{k,0,n}],{n,0,15}] (* Vaclav Kotesovec, Apr 26 2015 *)
  • PARI
    a(n) = prod(k=0, n, binomial(2*n+k,n+k)/binomial(n+k,k)) \\ Michel Marcus, May 20 2013

Formula

a(n) = 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=b=n, c=n+1.
a(n) = Product_{k=0..n} C(2n+k,n+k)/C(n+k,k). - Paul Barry, May 13 2008
a(n) ~ exp(1/12) * 3^(9*n^2/2 + 3*n + 5/12) / (A * n^(1/12) * 2^(6*n^2 + 4*n + 3/4)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant. - Vaclav Kotesovec, Apr 26 2015

A291871 Number of standard Young tableaux of skew shape (3*n^(2*n), 2*n^n)/(n^n).

Original entry on oeis.org

1, 42, 14617044842400, 5458228515594914179387748450655273588000, 864891828322912925373153355728411014930091519471102108791040960580578545212124160000
Offset: 0

Views

Author

Alejandro H. Morales, Sep 04 2017

Keywords

Comments

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 a combinatorial model for the Selberg integral (by Kim and Oh) and using the Naruse hook-length formula for skew shapes (by Morales, Pak and Panova).

Examples

			a(1)=42 since there are 42 standard Young tableaux of skew shape 332/1 since this is the same as the number of standard Young tableaux of straight shape 332 given by the hook-length formula: 42 = 8!/(2^2*3*4^2*5).
		

Crossrefs

Programs

  • Maple
    b:=n->mul(factorial(i),i=1..n-1):
    a:=n->factorial(7*n^2)*b(n)^5*b(5*n)/(b(2*n)^2*b(6*n)):
    seq(a(i),i=0..9);
  • Mathematica
    b[n_] := Product[i!, {i, n - 1}]; Table[(7 n^2)!*b[n]^5*b[5 n]/(b[2 n]^2*b[6 n]), {n, 0, 4}] (* Michael De Vlieger, Sep 10 2017 *)
  • Sage
    def b(n): return mul([factorial(i) for i in range(1,n)])
    def a(n): return factorial(7*n^2)*b(n)^5*b(5*n)/(b(2*n)^2*b(6*n))
    [a(n) for n in range(10)]

Formula

a(n) = (7*n^2)!*b(n)^5*b(5*n)/(b(2*n)^2*b(6*n)) where b(n) = 1!*2!*...*(n-1)! is a superfactorial A000178(n-1).
a(n) = (7*n^2)!*c(n)*b(n)^2*b(2*n)*b(5*n)/(b(6*n)*b(3*n)) where b(n) = 1!*2!*...*(n-1)! is a superfactorial A000178(n-1) and c(n) = A008793.
log a(n) = 7*n^2*log(n) + (75/2 - 15*log(2) - 18*log(3) + 25/2*log(5) + 7*log(7))*n^2 + O(n*log(n)). (See Example 6.2 in Morales et al.)
a(n) ~ sqrt(Pi) * 5^(25*n^2/2 - 1/12) * 7^(7*n^2 + 1/2) * exp(7*n^2/2 + 1/4) * n^(7*n^2 + 3/4) / (A^3 * 2^(22*n^2 - 3/4) * 3^(18*n^2 - 1/12)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Apr 08 2021

A322759 Numerator of Product_{i=1..n, j=1..n, k=1..n, m=1..n} (i+j+k+m-2)/(i+j+k+m-3).

Original entry on oeis.org

1, 2, 500, 406594944, 17946021496671970416, 266763232240477060473636852903291418, 1455102310580730414248044171189633697513044431339233543733641216
Offset: 0

Views

Author

N. J. A. Sloane, Dec 28 2018

Keywords

Comments

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

Examples

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

References

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

Crossrefs

Programs

  • Maple
    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)];
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