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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A110103 a(n) is the number of 2-regular 4-hypergraphs on 2n labeled vertices. (In a r-hypergraph, each hyper-edge is a proper r-set; k-regular implies that each vertex is in exactly k hyperedges.)

Original entry on oeis.org

1, 0, 0, 15, 1855, 469980, 214402650, 160081596675, 182667234224475, 302414315250247200, 697372026302486234700, 2167773244010692751057625, 8842276105055583472501844625, 46275602006744820263447546152500
Offset: 0

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Author

Marni Mishna, Jul 11 2005

Keywords

Comments

P-recursive.

Examples

			One of the 15 2-regular 4-hypergraphs on 6 vertices: {{1234},{4561}, {2356}}.
		

Crossrefs

Formula

Differential equation satisfied by exponential generating function sum a(n) t^(2n)/(2n)! {F(0) = 1, -144*t^3*(-2 + t^2)^2*(d^2/dt^2)F(t) - 12*(-2 + t^2)*(2*t^8-t^6 + 72 + 6*t^4-108*t^2)*(d/dt)F(t) - t^5*(-2 + t^2)*(t^2-3)*(t^4 + 4*t^2 + 36)*F(t)}.
Linear recurrence for a(n): {(15067980*n + 10550232*n^6 + 2859384*n^7 + 522720*n^8 + 128*n^11 + 2494800 + 61600*n^9 + 4224*n^10 + 52629038*n^3 + 45995730*n^4 + 26679070*n^5 + 37729494*n^2)*a(n) + (3791790*n + 109368*n^6 + 13872*n^7 + 1008*n^8 + 1247400 + 32*n^9 + 3747208*n^3 + 1767087*n^4 + 543858*n^5 + 4994577*n^2)*a(n + 1) + (28354500*n + 154560*n^6 + 11712*n^7 + 384*n^8 + 15478428*n^3 + 5309976*n^4 + 1152480*n^5 + 27874680*n^2 + 12474000)*a(n + 2) + (-623700-794025*n-48*n^6-115380*n^3-17760*n^4-1440*n^5-416757*n^2)*a(n + 3) + (599130*n + 534600 + 267282*n^2 + 59328*n^3 + 6552*n^4 + 288*n^5)*a(n + 4) + (-14166*n-26730-2484*n^2-144*n^3)*a(n + 5) + 54*a(n + 6), a(3) = 15, a(4) = 1855, a(5) = 469980, a(0) = 1, a(1) = 0, a(2) = 0}
Recurrence (of order 5): 54*(3*n - 4)*a(n) = 18*(n-1)*(2*n - 1)*(12*n^2 - 16*n + 9)*a(n-1) + 18*(n-2)*(n-1)*(2*n - 3)*(2*n - 1)*(3*n + 1)*a(n-2) + 3*(n-2)*(n-1)*(2*n - 5)*(2*n - 3)*(2*n - 1)*(24*n^2 - 65*n + 24)*a(n-3) + 3*(n-3)*(n-2)*(n-1)*(2*n - 7)*(2*n - 5)*(2*n - 3)*(2*n - 1)*(3*n + 1)*a(n-4) + 2*(n-4)*(n-3)*(n-2)*(n-1)*(2*n - 9)*(2*n - 7)*(2*n - 5)*(2*n - 3)*(2*n - 1)*(3*n - 1)*a(n-5). - Vaclav Kotesovec, Mar 11 2014
a(n) ~ 2^(3*n+1) * n^(3*n) / (3^n * exp(3*n+3/2)). - Vaclav Kotesovec, Mar 11 2014

Extensions

Replaced broken link, Vaclav Kotesovec, Mar 11 2014

A318148 Coefficients of the Omega polynomials of order 4, triangle T(n,k) read by rows with 0<=k<=n.

Original entry on oeis.org

1, 0, 1, 0, -34, 35, 0, 11056, -16830, 5775, 0, -14873104, 27560780, -15315300, 2627625, 0, 56814228736, -119412815760, 84786627900, -24734209500, 2546168625, 0, -495812444583424, 1140896479608800, -948030209181000, 364143337057500, -65706427536750, 4509264634875
Offset: 0

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Author

Peter Luschny, Aug 22 2018

Keywords

Comments

The name 'Omega polynomial' is not a standard name.

Examples

			[0] [1]
[1] [0,           1]
[2] [0,         -34,            35]
[3] [0,       11056,        -16830,        5775]
[4] [0,   -14873104,      27560780,   -15315300,      2627625]
[5] [0, 56814228736, -119412815760, 84786627900, -24734209500, 2546168625]
		

Crossrefs

All row sums are 1, alternating row sums (taken absolute) are A211212.
T(n,1) ~ A273352(n), T(n,n) = A025036(n).
A023531 (m=1), A318146 (m=2), A318147 (m=3), this seq (m=4).

Programs

  • Maple
    # See A318146 for the missing functions.
    FL([seq(CL(OmegaPolynomial(4, n)), n=0..8)]);
  • Mathematica
    (* OmegaPolynomials are defined in A318146 *)
    Table[CoefficientList[OmegaPolynomial[4, n], x], {n, 0, 6}] // Flatten
  • Sage
    # See A318146 for the function OmegaPolynomial.
    [list(OmegaPolynomial(4, n)) for n in (0..6)]

Formula

Omega(m, n, z) = (m*n)!*[z^(n*m)] H(m, z)^x where H(m, z) = hypergeom([], [seq(i/m, i=1..m-1)], (z/m)^m). We consider here the case m = 4 (for other cases see the cross-references).

A361948 Array read by ascending antidiagonals. A(n, k) = Product_{j=0..k-1} binomial((j + 1)*n - 1, n - 1) if n >= 1, and A(0, k) = 1 for all k.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 10, 15, 1, 1, 1, 1, 35, 280, 105, 1, 1, 1, 1, 126, 5775, 15400, 945, 1, 1, 1, 1, 462, 126126, 2627625, 1401400, 10395, 1, 1, 1, 1, 1716, 2858856, 488864376, 2546168625, 190590400, 135135, 1, 1
Offset: 0

Views

Author

Peter Luschny, Apr 13 2023

Keywords

Comments

Row n gives the leading coefficients of the set partition polynomials of type n. The sequence of these polynomial sequences starts: A097805, A048993, A156289, A291451, A291452, ...

Examples

			Array A(n, k) starts:
  [0] 1, 1,   1,       1,           1,                 1, ...
  [1] 1, 1,   1,       1,           1,                 1, ...
  [2] 1, 1,   3,      15,         105,               945, ...  A001147
  [3] 1, 1,  10,     280,       15400,           1401400, ...  A025035
  [4] 1, 1,  35,    5775,     2627625,        2546168625, ...  A025036
  [5] 1, 1, 126,  126126,   488864376,     5194672859376, ...  A025037
  [6] 1, 1, 462, 2858856, 96197645544, 11423951396577720, ...  A025038
.
Triangle A(n-k, k) starts:
  [0] 1;
  [1] 1, 1;
  [2] 1, 1,  1;
  [3] 1, 1,  1,   1;
  [4] 1, 1,  3,   1,   1;
  [5] 1, 1, 10,  15,   1, 1;
  [6] 1, 1, 35, 280, 105, 1, 1;
		

Crossrefs

Cf. A060540 (subarray), A370407 (antidiagonal sums, row sums).
Cf. A001147 (row 2), A025035 (row 3), A025036 (row 4), A025037 (row 5), A025038 (row 6), A025039 (row 7), A025040 (row 8), A025041 (row 9).
Cf. A088218 (column 2), A060542 (column 3), A082368 (column 4), A322252 (column 5), A057599 (main diagonal).

Programs

  • Maple
    A := (n, k) -> mul(binomial((j + 1)*n - 1, n - 1), j = 0..k-1):
    seq(seq(A(n-k, k), k = 0..n), n = 0..9);
    # Alternative, using recursion:
    A := proc(n, k) local P; P := proc(n, k) option remember;
    if n = 0 then return x^k*k! fi; if k = 0 then 1 else add(binomial(n*k, n*j)*
    P(n,k-j)*x, j=1..k) fi end: coeff(P(n, k), x, k) / k! end:
    seq(print(seq(A(n, k), k = 0..5)), n = 0..6);
    # Alternative, using exponential generating function:
    egf := n -> ifelse(n=0, 1, exp(x^n/n!)): ser := n -> series(egf(n), x, 8*n):
    row := n -> local k; seq((n*k)!*coeff(ser(n), x, n*k), k = 0..6):
    for n from 0 to 6 do [n], row(n) od;  # Peter Luschny, Aug 15 2024
  • Mathematica
    A[n_, k_] := Product[Binomial[n (j + 1) - 1, n - 1], {j, 0, k - 1}]; Table[A[n - k, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Apr 13 2023 *)
  • SageMath
    def Arow(n, size):
        if n == 0: return [1] * size
        return [prod(binomial((j + 1)*n - 1, n - 1) for j in range(k)) for k in range(size)]
    for n in range(7): print(Arow(n, 7))
    # Alternative, using exponential generating function:
    def SetPolyLeadCoeff(m, n):
        x, z = var("x, z")
        if m == 0: return 1
        w = exp(2 * pi * I / m)
        o = sum(exp(z * w ** k) for k in range(m)) / m
        t = exp(x * (o - 1)).taylor(z, 0, m*n)
        p = factorial(m*n) * t.coefficient(z, m*n)
        return p.leading_coefficient(x)
    for m in range(7):
        print([SetPolyLeadCoeff(m, k) for k in range(6)])

Formula

A(n, k) = (1/k!) * [x^k] P(n, k), where P(n, k) = k!*x^k if n = 0 and otherwise 1 if k = 0 and otherwise Sum_{j=1..k} binomial(n*k, n*j)*P(n, k-j)*x.
A(n, k) = (n*k)!*[x^(n*k)] exp(x^n/n!) for n >= 1. - Peter Luschny, Aug 15 2024

A211311 a(n) = number |fdw(P,(n))| of entangled P-words with s=4.

Original entry on oeis.org

1, 68, 34236, 62758896, 304863598320, 3242854167461280, 66429116436728636640, 2389384600126093124110080
Offset: 1

Views

Author

N. J. A. Sloane, Apr 08 2012

Keywords

Comments

See Jenca and Sarkoci for the precise definition.

Crossrefs

Formula

From Peter Bala, Sep 05 2012: (Start)
Conjectural e.g.f.: 2 - 1/A(x), where A(x) = sum {n = 0..inf} (4*n)!/24^n*x^n/n! is the e.g.f. for A014608 (also the o.g.f. for A025036).
If true, this leads to the recurrence equation: a(n) = (4*n)!/24^n - sum {k = 1..n-1} (4*k)!/24^k*binomial(n,k)*a(n-k) with a(1) = 1.
(End)

A334061 Triangle read by rows: T(n,k) is the number of set partitions of {1..4n} into n sets of 4 with k disjoint strings of adjacent sets, each being a contiguous set of elements.

Original entry on oeis.org

1, 0, 1, 31, 4, 0, 5474, 292, 9, 0, 2554091, 72318, 1206, 10, 0, 2502018819, 43707943, 438987, 2871, 5, 0, 4456194509950, 52717010017, 351487598, 1622954, 4355, 1, 0, 13077453070386914, 111615599664989, 528618296314, 1764575884, 4080889, 4385, 0, 0
Offset: 0

Views

Author

Donovan Young, May 26 2020

Keywords

Comments

Number of configurations with k connected components (consisting of polyomino matchings) in the generalized game of memory played on the path of length 4n, see [Young].

Examples

			Triangle begins:
        1;
        0,     1;
       31,     4,   0;
     5474,   292,   9,  0;
  2554091, 72318,1206, 10, 0;
  ...
For n=2 and k=1 the configurations are (1,6,7,8),(2,3,4,5), as well as (1,2,7,8),(3,4,5,6) and also (1,2,3,8),(4,5,6,7) (i.e. configurations with a single contiguous set) and (1,2,3,4),(5,6,7,8) (i.e. two adjacent contiguous sets); hence T(2,1) = 4.
		

Crossrefs

Row sums are A025036.
Column k=0 is column 0 of A334057.

Programs

  • Mathematica
    CoefficientList[Normal[Series[Sum[y^j*(4*j)!/24^j/j!*((1-y*(1-z))/(1-y^2*(1-z)))^(4*j+1), {j, 0, 20}], {y, 0, 20}]], {y, z}]
  • PARI
    T(n)={my(v=Vec(sum(j=0, n, (4*j)! * x^j * (1-(1-y)*x + O(x*x^n))^(4*j+1) / (j! * 24^j * (1-(1-y)*x^2 + O(x*x^n))^(4*j+1))))); vector(#v, i, Vecrev(v[i], i))}
    { my(A=T(8)); for(n=1, #A, print(A[n])) }

Formula

G.f.: Sum_{j>=0} (4*j)! * y^j * (1-(1-z)*y)^(4*j+1) / (j! * 24^j * (1-(1-z)*y^2)^(4*j+1)).
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