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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A008793 The problem of the calissons: number of ways to tile a hexagon of edge n with diamonds of side 1. Also number of plane partitions whose Young diagrams fit inside an n X n X n box.

Original entry on oeis.org

1, 2, 20, 980, 232848, 267227532, 1478619421136, 39405996318420160, 5055160684040254910720, 3120344782196754906063540800, 9265037718181937012241727284450000, 132307448895406086706107959899799334375000
Offset: 0

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Keywords

Comments

The 3-dimensional analog of A000984. - William Entriken, Aug 06 2013
The largest prime factor of a(n) is the largest prime p < 3*n. Its multiplicity is equal to 3*n-p. This can be proved with the formula of Michel Marcus, for example. - Walter Trump, Feb 11 2023
a(n) is also the number of resonance structures of circumcircum...coronene, where circum is repeated n-2 times where a(1) is the number of resonance structures of benzene (see Gutman et al.). - Yuan Yao, Oct 29 2023

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 545, also p. 575 line -1 with a=b=c=n.
  • D. M. Bressoud, Proofs and Confirmations, Camb. Univ. Press, 1999; Eq. (6.8), p. 198. The first printing of Eq. (6.8) is wrong (see A049505 and A005157), but if one changes the limits in the formula (before it is corrected) to {1 <= i <= r, 1 <= j <= r}, one obtains the present sequence. - N. J. A. Sloane, Jun 30 2013
  • 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. [See K, p. 442.]
  • Sebastien Desreux, Martin Matamala, Ivan Rapaport, Eric Remila, Domino tilings and related models: space of configurations of domains with holes, arXiv:math/0302344, 27 Feb 2003
  • Anne S. Meeussen, Erdal C. Oguz, Yair Shokef, Martin van Hecke1, Topological defects produce exotic mechanics in complex metamaterials, arXiv preprint 1903.07919, 2019 [See Section "Compatible metamaterials with fully antiferromagnetic interactions" - N. J. A. Sloane, Mar 23 2019]
  • 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 p. 261).

Crossrefs

Cf. A000984, A066931, A352656, A352657. Main diagonal of array A103905.

Programs

  • Maple
    A008793 := proc(n) local i; mul((i - 1)!*(i + 2*n - 1)!/((i + n - 1)!)^2, i = 1 .. n) end proc;
  • Mathematica
    Table[ Product[ (i+j+k-1)/(i+j+k-2), {i, n}, {j, n}, {k, n} ], {n, 10} ]
  • PARI
    a(n) = prod(i=1,n, prod(j=1, n, (n+i+j-1)/(i+j-1))); \\ Michel Marcus, Jul 13 2020

Formula

a(n) = Product_{i = 0..n-1} (i^(-i)*(n+i)^(2*i-n)*(2*n+i)^(n-i)).
a(n) = Product_{i = 1..n} Product_{j = 0..n-1} (3*n-i-j)/(2*n-i-j).
a(n) = Product_{i = 1..n} Gamma[i]*Gamma[i+2*n]/Gamma[i+n]^2.
a(n) = Product_{i = 0..n-1} i!*(i+2*n)!/(i+n)!^2.
a(n) = Product_{i = 1..n} Product_{j = n..2*n-1} i+j / Product_{j = 0..n-1} i+j. - Paul Barry, Jun 13 2006
For n >= 1, a(n) = det(binomial(2*n,n+i-j)) for 1<=i,j<=n [Krattenhaller, Theorem 4, with a = b = c = n].
Let H(n) = Product_{k = 1..n-1} k!. Then for a,b,c nonnegative integers (H(a)*H(b)*H(c)*H(a+b+c))/(H(a+b)*H(b+c)*H(c+a)) is an integer [MacMahon, Chapter II, Section 429, p. 182, with x -> 1]. Setting a = b = c = n gives the entries for this sequence. - Peter Bala, Dec 22 2011
a(n) ~ exp(1/12) * 3^(9*n^2/2 - 1/12) / (A * n^(1/12) * 2^(6*n^2 - 1/4)), where A = A074962 = 1.28242712910062263687534256886979... is the Glaisher-Kinkelin constant. - Vaclav Kotesovec, Feb 27 2015
a(n) = Product_{i = 1..n} Product_{j = 1..n} (n+i+j-1)/(i+j-1). - Michel Marcus, Jul 13 2020
Conjecture: 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. - Peter Bala, Apr 07 2022

Extensions

More terms from Eric W. Weisstein

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.

Original entry on oeis.org

1, 4, 336, 572572, 19571505408, 13365232267026024, 182001937855822420050000, 49372092168218024268166702560000, 266640931683989945767062736068603511111680, 28657545169614835585678719963104037818950931553412096, 61277278161726929232430881966673334396569563602790616552072890176
Offset: 0

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Author

Peter Bala, Apr 22 2022

Keywords

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 (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

			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 + ....
		

Crossrefs

Programs

  • Maple
    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);
  • Mathematica
    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 *)

Formula

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)
Showing 1-2 of 2 results.