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.

Previous Showing 41-48 of 48 results.

A269122 Number of sequences with n copies each of 1,2,3,4 avoiding the pattern 1234.

Original entry on oeis.org

1, 23, 1879, 173891, 16140983, 1474050783, 132279480115, 11690616534627, 1020149742404727, 88094027572412735, 7541872062618216079, 641058122723345275763, 54163773797158984485203, 4553249681522845632696611, 381122944171897823109474291
Offset: 0

Views

Author

Alois P. Heinz, Feb 19 2016

Keywords

Crossrefs

Row n=4 of A269129.

Formula

a(n) = A008977(n) - A268840(n).

A377218 Expansion of the o.g.f. A(x) defined by [x^n] A(x)^(24*n) = (4*n)!/n!^4 for n >= 0.

Original entry on oeis.org

1, 1, 29, 2246, 239500, 30318701, 4271201506, 647359627557, 103476937050223, 17223017775652625, 2959285397777331751, 521687007046376376544, 93932798602803741121051, 17215649571517858590782737, 3203146941738318544432065500, 603763082812549420389330837978, 115095760617137117019641563685386
Offset: 0

Views

Author

Peter Bala, Oct 20 2024

Keywords

Comments

Compare with A000984(n) = [x^n] (1 + x)^(2*n) = (2*n)!/n!^2.
The central binomial coefficients A000984(n) satisfy the supercongruences u(n*p^k) == u(n*p^(k-1)) (mod p^(3*k)) for all primes p >= 5 and positive integers n and k.
More generally, for positive integers r and s, the sequence {u(r,s; n) : n >= 0} defined by u(r,s; n) = [x^(s*n)] (1 + x)^(r*n) = binomial(r*n, s*n) satisfies the same supercongruences (Meštrović, Section 6, equation 39).
Conjecture: for positive integers r and s, the sequence {v(r,s; n) : n >= 0} defined by v(r,s; n) = [x^(s*n)] A(x)^(r*n) also satisfies the same supercongruences.

Crossrefs

Programs

  • Maple
    Order := 25:
    E(x) := exp(add((4*n)!/n!^4 * x^n/n, n = 1..25)):
    solve(series(x*E(x),x) = y, x):
    convert(%, polynom):
    g := taylor(y/%, y = 0, 25):
    seq(coeftayl(g^(1/24), y = 0,  n), n = 0..20);

Formula

O.g.f.: A(x) = ( x/(x * series_reversion(E(x)))^(1/24), where E(x) = exp(Sum_{n >= 1} (4*n)!/n!^4 *x^n/n) is the o.g.f. of A333042.

A059068 Card-matching numbers (Dinner-Diner matching numbers).

Original entry on oeis.org

1, 9, 8, 6, 0, 1, 297, 672, 736, 480, 246, 64, 24, 0, 1, 13833, 49464, 84510, 90944, 69039, 38448, 16476, 5184, 1431, 216, 54, 0, 1, 748521, 3662976, 8607744, 12880512, 13731616, 11042688, 6928704, 3458432, 1395126
Offset: 0

Views

Author

Barbara Haas Margolius (margolius(AT)math.csuohio.edu)

Keywords

Comments

This is a triangle of card matching numbers. A deck has 4 kinds of cards, n of each kind. The deck is shuffled and dealt in to 4 hands with each with n cards. A match occurs for every card in the j-th hand of kind j. Triangle T(n,k) is the number of ways of achieving exactly k matches (k=0..4n). The probability of exactly k matches is T(n,k)/((4n)!/n!^4).
Rows have lengths 1,5,9,13,...
Analogous to A008290 - Zerinvary Lajos, Jun 22 2005

Examples

			There are 736 ways of matching exactly 2 cards when there are 2 cards of each kind and 4 kinds of card so T(2,2)=736.
Triangle begins:
      1;
      9,     8,     6,     0,     1;
    297,   672,   736,   480,   246,    64,    24,    0,    1;
  13833, 49464, 84510, 90944, 69039, 38448, 16476, 5184, 1431, 216, 54, 0, 1;
  ...
		

References

  • F. N. David and D. E. Barton, Combinatorial Chance, Hafner, NY, 1962, Ch. 7 and Ch. 12.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 174-178.
  • R. P. Stanley, Enumerative Combinatorics Volume I, Cambridge University Press, 1997, p. 71.

Crossrefs

Cf. A008290.
Row sums give A008977.

Programs

  • Maple
    p := (x,k)->k!^2*sum(x^j/((k-j)!^2*j!),j=0..k); R := (x,n,k)->p(x,k)^n; f := (t,n,k)->sum(coeff(R(x,n,k),x,j)*(t-1)^j*(n*k-j)!,j=0..n*k);
    for n from 0 to 5 do seq(coeff(f(t,4,n),t,m)/n!^4,m=0..4*n); od;
  • Mathematica
    p[x_, k_] := k!^2*Sum[x^j/((k-j)!^2*j!), {j, 0, k}]; r[x_, n_, k_] := p[x, k]^n; f[t_, n_, k_] := Sum[ Coefficient[r[x, n, k], x, j]*(t-1)^j*(n*k-j)!, {j, 0, n*k}]; Table[ Coefficient[f[t, 4, n], t, m]/n!^4, {n, 0, 4}, {m, 0, 4*n}] // Flatten (* Jean-François Alcover, Dec 17 2012, translated from Maple *)

Formula

G.f.: sum(coeff(R(x, n, k), x, j)*(t-1)^j*(n*k-j)!, j=0..n*k) where n is the number of kinds of cards (4 in this case), k is the number of cards of each kind and R(x, n, k) is the rook polynomial given by R(x, n, k)=(k!^2*sum(x^j/((k-j)!^2*j!))^n (see Stanley or Riordan). coeff(R(x, n, k), x, j) indicates the coefficient for x^j of the rook polynomial.

A248600 G.f.: Sum_{n>=0} R_n(x+x*y) * x^(2*n)*y^n / (1-x-x*y)^(4*n+1) = Sum_{n>=0} Sum_{k=0..n} C(n,k)^4 * x^n*y^k, where R_n(x+x*y) equals the n-th row polynomial R_n(z) = Sum_{k=0..2*n} T(n,k)*z^k at z = x+x*y.

Original entry on oeis.org

1, 14, 8, 2, 786, 1056, 576, 96, 6, 61340, 131760, 117900, 48320, 9540, 720, 20, 5562130, 16481920, 20917120, 13847680, 5118400, 1025920, 105280, 4480, 70, 549676764, 2079579600, 3444581700, 3165926400, 1755532800, 598123008, 123656400, 14716800, 926100, 25200, 252, 57440496036
Offset: 0

Views

Author

Paul D. Hanna, Oct 11 2014

Keywords

Examples

			Triangle begins:
[1],
[14, 8, 2],
[786, 1056, 576, 96, 6],
[61340, 131760, 117900, 48320, 9540, 720, 20],
[5562130, 16481920, 20917120, 13847680, 5118400, 1025920, 105280, 4480, 70],
[549676764, 2079579600, 3444581700, 3165926400, 1755532800, 598123008, 123656400, 14716800, 926100, 25200, 252],
[57440496036, 264565490112, 542687590368, 640299696960, 477284304420, 233110386432, 75243589344, 15835792896, 2103157980, 165802560, 7051968, 133056, 924],
[6242164112184, 33895475918304, 83073660613944, 119912994225024, 112698387745944, 72172565713248, 32111980788888, 9951304416768, 2124873478728, 305035899168, 28270554312, 1584815232, 48600552, 672672, 3432],
[698300344311570, 4368053451041280, 12465205610457600, 21305587665922560, 24216302627637120, 19255941998092800, 10989839486545920, 4550117424652800, 1366687981264320, 295074717949440, 44954858108160, 4691645038080, 320878958400, 13445752320, 311351040, 3294720, 12870], ...
where this triangle forms the coefficients in the series
B(x,y) = 1/(1-x-x*y) +
(14 + 8*(x+x*y) + 2*(x+x*y)^2) * x^2*y/(1-x-x*y)^5 +
(786 + 1056*(x+x*y) + 576*(x+x*y)^2 + 96*(x+x*y)^3 + 6*(x+x*y)^4) * x^4*y^2/(1-x-x*y)^9 +
(61340 + 131760*(x+x*y) + 117900*(x+x*y)^2 + 48320*(x+x*y)^3 + 9540*(x+x*y)^4 + 720*(x+x*y)^5 + 20*(x+x*y)^6) * x^6*y^3/(1-x-x*y)^13 +...
such that the sum may be expressed using binomial coefficients C(n,k)^4 like so:
B(x,y) =  1 +
x*(1 + y) +
x^2*(1 + 2^4*y + y^2) +
x^3*(1 + 3^4*y + 3^4*y^2 + y^3) +
x^4*(1 + 4^4*y + 6^4*y^2 + 4^4*y^3 + y^4) +
x^5*(1 + 5^4*y + 10^4*y^2 + 10^4*y^3 + 5^4*y^4 + y^5) +
x^6*(1 + 6^4*y + 15^4*y^2 + 20^4*y^3 + 15^4*y^4 + 6^4*y^5 + y^6) +...
The central terms of the rows begin:
[1, 8, 576, 48320, 5118400, 598123008, 75243589344, 9951304416768, 1366687981264320, ...].
		

Crossrefs

Formula

Leftmost border equals A050983, de Bruijn's S(4,n):
T(n,0) = Sum_{k=0..2*n} (-1)^(n+k) * C(2*n,k)^4.
Rightmost border equals A000984, the central binomial coefficients:
T(n,2*n) = Sum_{k=0..2*n} (-1)^(n+k)* C(2*n,k)^2 = (2*n)!/(n!)^2.
Row sums equal A008977(n) = (4*n)!/(n!)^4.
Sum_{k=0..n} (-1)^k * T(n,k) = A002897(n) = C(2*n,n)^3.

A307618 A Calabi-Yau period integral: a(n) = C(4*n,2*n)*C(2*n,n)^3.

Original entry on oeis.org

1, 48, 15120, 7392000, 4414410000, 2956651746048, 2133278987583744, 1621682968820428800, 1281351259836532170000, 1043032815185819858400000, 869343653096068540955685120, 738637974389826550020188712960, 637665137404661719206664998969600
Offset: 0

Views

Author

Bradley Klee, Jun 04 2019

Keywords

Comments

Entry number six in the "Big Table" of Almkvist et al. (see links). The period T(x) = Sum_{n>=0} a(n)*x^(2*n) is also the first x-derivative of the 6-volume associated to the algebraic variety V6 = P1 & P2 & P3, with P1 : X1^2 + Y1^2 = X2^2 + Y2^2, P2 : X2^2 + Y2^2 = X3^2 + Y3^2, P3 : x=(X1^2 + X2^2 + X3^2 + Y1^2 + Y2^2 + Y3^2)^3*(1 - X1*X2*X3*Y1*Y2*Y3). The small x limit reduces V6 to a 6-ball with 6-volume proportional to x. Similar constructions are known to exist for a few other geometries on Almkvist's list, most notably #3: A186420, and #16: A039699.

Crossrefs

Hadamard Factors: A000984, A002894, A002897, A001448, A000897, A008977.
Calabi-Yau Periods: A008978, A186420, A268553, A039699.

Programs

  • Mathematica
    Binomial[4*#,2*#]*Binomial[2*#,#]^3&/@Range[0,10]

Formula

G.f.: 4F3({1/4, 3/4, 1/2, 1/2}, {1, 1, 1}, 1024*x).
Define the period integral:
dt(x) = dz1*dz2*dz3/sqrt(1-32*x*cos(z1)*cos(z2)*cos(z3)).
T(x)=1/(2*Pi)^3*Integral_{0..2*Pi,0..2*Pi,0..2*Pi} dt(x),
the Picard-Fuchs coefficients:(c0,c1,c2,c3,c4)=
(768*x, 14592*x^2-1, x*(25344*x^2-7), 2*x^2*(5120*x^2-3), x^3*(32*x-1)*(32*x+1)),
and the certificate function:
G(z1,z2,z3)=(16*sin(z1)*(
48*x*cos(z1)
+ cos(z2)*cos(z3)
+ 48*x*cos(z1)*(cos(z3)^2 + cos(z2)^2)
+ 2304*x^2*cos(z1)^2*cos(z2)*cos(z3)
+ 80*x*cos(z1)*cos(z2)^2*cos(z3)^2
+ 384*x^2*cos(z1)^2*(cos(z2)*cos(z3)^3 + cos(z2)^3*cos(z3))
+ 256*x^2*cos(z1)^2*cos(z2)^3*cos(z3)^3)
)/(3*(1 - 32*x*cos(z1)*cos(z2)*cos(z3))^(7/2)),
Then: 0 = Sum_{n=0..4}cn*d^n/dx^n dt(x) + d/dz1 G(z1,z2,z3) + d/dz2 G(z2,z3,z1) + d/dz3 G(z3,z1,z2), thus: 0 = Sum_{n=0..4} cn*d^n/dx^n T(x).
Furthermore, let (a1,a2,a3)=(c1,c2,c3)/c0, then also: 0 = (1/2)*a2*a3 - (1/8)*a3^3 + d/dx(a2) - (3/4)*a3*d/dx(a3) - (1/2)*d^2/dx^2(a3) - a1.
D-finite with recurrence: n^4*a(n) -16*(4*n-1)*(4*n-3)*(-1+2*n)^2*a(n-1)=0. - R. J. Mathar, Jan 27 2020

A318105 Triangle read by rows: T(n,k) = (4*n - 3*k)!/((n-k)!^4*k!).

Original entry on oeis.org

1, 24, 1, 2520, 120, 1, 369600, 22680, 360, 1, 63063000, 4804800, 113400, 840, 1, 11732745024, 1072071000, 33633600, 415800, 1680, 1, 2308743493056, 246387645504, 9648639000, 168168000, 1247400, 3024, 1, 472518347558400, 57718587326400, 2710264100544, 61108047000, 672672000, 3243240, 5040, 1
Offset: 0

Views

Author

Gheorghe Coserea, Oct 15 2018

Keywords

Comments

Diagonal of rational function R(x,y,z,w,t) = 1/(1 - (x+y+z+w + t*x*y*z*w)) with respect to x,y,z,w, i.e., T(n,k) = [(xyzw)^n*t^k] R(x,y,z,w,t).
Annihilating differential operator: x^2*(3*t*x + 1)^2*((t*x - 1)^4 - 256*x)*Dx^3 + 3*x*(3*t*x + 1)*((t*x - 1)^3*(6*t^2*x^2 + 3*t*x - 1) - 384*x*(t*x + 1))*Dx^2 + (t*x - 1)*((t*x - 1)*(63*t^4*x^4 + 66*t^3*x^3 - 18*t*x + 1) + 48*x*(15*t*x + 17))*Dx + (t*x - 1)*(t*(9*t^4*x^4 + 12*t^3*x^3 + 6*t^2*x^2 - 12*t*x + 1) - 24*(15*t*x - 1)).

Examples

			A(x;t) = 1 + (24 + t)*x + (2520 + 120*t + t^2)*x^2 + (369600 + 22680*t + 360*t^2 + t^3)*x^3 + ...
Triangle starts:
n\k [0]            [1]           [2]         [3]        [4]      [5]   [6]
[0] 1;
[1] 24,            1;
[2] 2520,          120,          1;
[3] 369600,        22680,        360,        1;
[4] 63063000,      4804800,      113400,     840,       1;
[5] 11732745024,   1072071000,   33633600,   415800,    1680,    1;
[6] 2308743493056, 246387645504, 9648639000, 168168000, 1247400, 3024, 1;
[7] ...
		

Crossrefs

Programs

  • Mathematica
    t[n_,k_] := (4*n - 3*k)!/((n-k)!^4*k!); Table[t[n, k], {n, 0, 10}, {k , 0,n}] // Flatten  (* Amiram Eldar, Nov 07 2018 *)
  • PARI
    T(n, k) = (4*n-3*k)!/(k!*(n-k)!^4);
    concat(vector(8, n, vector(n, k, T(n-1, k-1))))
    /*
    test:
    P(n, v='t) = subst(Polrev(vector(n+1, k, T(n, k-1)), 't), 't, v);
    diag(expr, N=22, var=variables(expr)) = {
      my(a = vector(N));
      for (k = 1, #var, expr = taylor(expr, var[#var - k + 1], N));
      for (n = 1, N, a[n] = expr;
        for (k = 1, #var, a[n] = polcoef(a[n], n-1)));
      return(a);
    };
    apply_diffop(p, s) = {
      s=intformal(s);
      sum(n=0, poldegree(p, 'Dx), s=s'; polcoef(p, n, 'Dx) * s);
    };
    \\ diagonal property:
    x='x; y='y; z='z; w='w; t='t;
    diag(1/(1 - (x+y+z+w + t*x*y*z*w)), 9, [x, y, z, w]) == vector(9, n, P(n-1))
    \\ annihilating diffop:
    y = Ser(vector(101, n, P(n-1)), 'x);
    p = x^2*(3*t*x + 1)^2*((t*x - 1)^4 - 256*x)*Dx^3 + 3*x*(3*t*x + 1)*((t*x - 1)^3*(6*t^2*x^2 + 3*t*x - 1) - 384*x*(t*x + 1))*Dx^2 + (t*x - 1)*((t*x - 1)*(63*t^4*x^4 + 66*t^3*x^3 - 18*t*x + 1) + 48*x*(15*t*x + 17))*Dx + (t*x - 1)*(t*(9*t^4*x^4 + 12*t^3*x^3 + 6*t^2*x^2 - 12*t*x + 1) - 24*(15*t*x - 1));
    0 == apply_diffop(p, y)
    */

Formula

Let P_n(t) = Sum_{k=0..n} T(n,k)*t^k. Then A125143(n) = P_n(-27), A008977(n) = P_n(0), A082488(n) = P_n(1).

A324469 Exponent of highest power of 3 that divides multinomial(4*n;n,n,n,n).

Original entry on oeis.org

0, 1, 2, 1, 2, 4, 2, 5, 6, 1, 2, 3, 2, 3, 6, 4, 6, 7, 2, 3, 4, 5, 6, 8, 6, 8, 9, 1, 2, 3, 2, 3, 5, 3, 6, 7, 2, 3, 4, 3, 4, 8, 6, 8, 9, 4, 5, 6, 6, 7, 9, 7, 9, 10, 2, 3, 4, 3, 4, 6, 4, 9, 10, 5, 6, 7, 6, 7, 10, 8, 10, 11, 6, 7, 8, 8, 9, 11, 9, 11, 12, 1, 2
Offset: 0

Views

Author

N. J. A. Sloane, Mar 03 2019

Keywords

Crossrefs

Analogs for binomial and trinomials: A000989, A053735. See also A324467.
Cf. A007949 (3-adic valuation of n), A008977.

Programs

  • Maple
    [seq(padic[ordp](combinat[multinomial](4*n, n$4), 3), n=0..128)];
  • Mathematica
    s[n_] := Plus @@ IntegerDigits[n, 3]; a[n_] := 2*s[n] - s[4*n]/2; Array[a, 100, 0] (* Amiram Eldar, Feb 21 2021 *)

Formula

a(n) = 2*A000989(n) + A000989(2*n). - Charlie Neder, Mar 09 2019
From Amiram Eldar, Feb 21 2021: (Start)
a(n) = A007949(A008977(n)).
a(n) = 2*A053735(n) - A053735(4*n)/2. (End)

A342107 a(n) = Sum_{k=0..n} (4*k)!/k!^4.

Original entry on oeis.org

1, 25, 2545, 372145, 63435145, 11796180169, 2320539673225, 474838887231625, 100035931337622625, 21552788197602942625, 4726913659271173170145, 1051798742538350304851425, 236861100204680963085573025
Offset: 0

Views

Author

Seiichi Manyama, Feb 28 2021

Keywords

Comments

Partial sums of A008977.
In general, for m > 1, Sum_{k=0..n} (m*k)!/k!^m ~ m^(m*n + m + 1/2) / ((m^m - 1) * (2*Pi*n)^((m-1)/2)). - Vaclav Kotesovec, Feb 28 2021

Crossrefs

Programs

  • Maple
    A342107 := proc(n)
        add((4*k)!/k!^4,k=0..n) ;
    end proc:
    seq(A342107(n),n=0..70) ; # R. J. Mathar, Dec 04 2023
  • Mathematica
    Table[Sum[(4*k)!/k!^4, {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Feb 28 2021 *)
  • PARI
    a(n) = sum(k=0, n, (4*k)!/k!^4);

Formula

a(n) ~ 2^(8*n + 15/2) / (255 * Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Feb 28 2021
D-finite with recurrence n^3*a(n) +(-257*n^3+384*n^2-176*n+24)*a(n-1) +8*(4*n-3)*(2*n-1)*(4*n-1)*a(n-2)=0. - R. J. Mathar, Dec 04 2023
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