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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A274783 Diagonal of the rational function 1/(1 - (w*x*y*z + w*x*y + w*x*z + w*y*z + x*y*z)).

Original entry on oeis.org

1, 1, 1, 25, 121, 361, 3361, 24361, 116425, 790441, 6060121, 36888721, 238815721, 1760983225, 11968188961, 79763351305, 570661612585, 4040282139625, 27901708614985, 198090585115105, 1420583920034161, 10056659775872161, 71730482491962361, 517012699162717825, 3713833648541268121
Offset: 0

Views

Author

Gheorghe Coserea, Jul 13 2016

Keywords

Comments

Diagonal of the rational function 1/(1 - (x^3 + y^3 + z^3 + w^3 + x*y*z*w)). - Seiichi Manyama, Jul 04 2025

Crossrefs

Programs

  • Maple
    with(combinat):
    seq(add((n+k)!/(k!^4*(n-3*k)!), k = 0..floor(n/3)), n = 0..20); # Peter Bala, Jan 27 2018
  • PARI
    my(x='x, y='y, z='z, w='w);
    R = 1/(1-(w*x*y*z+w*x*y+w*x*z+w*y*z+x*y*z));
    diag(n, expr, var) = {
      my(a = vector(n));
      for (i = 1, #var, expr = taylor(expr, var[#var - i + 1], n));
      for (k = 1, n, a[k] = expr;
           for (i = 1, #var, a[k] = polcoeff(a[k], k-1)));
      return(a);
    };
    diag(20, R, [x,y,z,w])

Formula

0 = x^2*(x+3)^2*(x^4 - 260*x^3 + 6*x^2 - 4*x + 1)*y''' + 3*x*(x+3)*(2*x^5 - 381*x^4 - 1944*x^3 + 34*x^2 - 18*x + 3)*y'' + (7*x^6 - 764*x^5 - 9101*x^4 - 27264*x^3 + 381*x^2 - 132*x + 9)*y' + (x^5 - 13*x^4 - 246*x^3 - 5946*x^2 + 69*x - 9)*y, where y is the g.f.
a(n) = Sum_{k = 0..floor(n/3)} (n+k)!/(k!^4*(n-3*k)!) = Sum_{k = 0..floor(n/3)} binomial(n,3*k)*binomial(n+k,k)*(3*k)!/k!^3 (apply Eger, Theorem 3 to the set of column vectors S = {[1,1,1,1], [1,1,0,1], [1,0,1,1], [0,1,1,1], [1,1,1,0]}). - Peter Bala, Jan 27 2018
G.f.: Sum_{k>=0} (4*k)!/k!^4 * x^(3*k)/(1-x)^(4*k+1). - Seiichi Manyama, Mar 19 2023
From Vaclav Kotesovec, Mar 19 2023: (Start)
Recurrence: n^3*(2*n - 5)*(4*n - 11)*(4*n - 7)*a(n) = (4*n - 11)*(32*n^5 - 184*n^4 + 368*n^3 - 327*n^2 + 147*n - 27)*a(n-1) - (192*n^6 - 1920*n^5 + 7628*n^4 - 15366*n^3 + 16567*n^2 - 9117*n + 2025)*a(n-2) + (4*n - 9)*(4*n - 3)*(520*n^4 - 4420*n^3 + 13809*n^2 - 18769*n + 9367)*a(n-3) - (n-3)^3*(2*n - 3)*(4*n - 7)*(4*n - 3)*a(n-4).
a(n) ~ sqrt(9/8 + 3/(32*sqrt(2)) + sqrt(1085/32 + 161/(2*sqrt(2)))/8) * (1 + 2*sqrt(2) + 2*sqrt(2*(2*sqrt(2) - 1)))^n / (Pi^(3/2) * n^(3/2)). (End)

A001460 a(n) = (5*n)!/((2*n)!*(n!)^3).

Original entry on oeis.org

1, 60, 18900, 8408400, 4364860500, 2473653742560, 1483630051503600, 925833064837824000, 594927307937311420500, 391004487919622186610000, 261614105944603801295306400, 177601637048592673099585584000, 122027661025630720013771117910000
Offset: 0

Views

Author

Keywords

Crossrefs

Programs

  • Magma
    [Factorial(5*n)/(Factorial(2*n)*Factorial(n)^3):n in [0..15]]; // Marius A. Burtea, Feb 17 2020
    
  • Maple
    f := n->(5*n)!/((2*n)!*(n!)^3);
    seq((5*n)!/(n!)^5/binomial(2*n,n), n=0..15); # Zerinvary Lajos, Jun 28 2007
  • Mathematica
    Table[(5 n)!/((2 n)! (n!)^3), {n, 0, 15}] (* or *)
    Table[(5 n)!/(n!)^5/Binomial[2 n, n], {n, 0, 15}] (* Michael De Vlieger, Jul 18 2016 *)
  • PARI
    a(n) = (5*n)!/((2*n)!*n!^3);  \\ Gheorghe Coserea, Jul 18 2016
    
  • SageMath
    f=factorial; [f(5*n)/(f(2*n)*f(n)^3) for n in range(16)] # G. C. Greubel, Sep 03 2023

Formula

a(n) = A008978(n)/A000984(n). - Zerinvary Lajos, Jun 28 2007
From Gheorghe Coserea, Jul 18 2016: (Start)
a(n) = [(xyzw)^(3n)] 1/(1-(w*x*y+w*z+x*z+y*z)).
a(n) ~ sqrt(5)/(4*Pi^(3/2)) * n^(-3/2) * (3125/4)^n.
0 = (-4*x^3+3125*x^6)*y'''' + (-18*x^2+37500*x^5)*y''' + (-10*x+117500*x^4)*y'' + (2+95000*x^3)*y' + (9720*x^2)*y, where y(x) = A(x^3). (End)
From Peter Bala, Dec 30 2019: (Start)
a(n) = binomial(3*n,n)*binomial(4*n,n)*binomial(5*n,n).
a(n) = ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) * ( [x^n](1 + x)^(5*n) ).
a(n) = [x^n]( F(x)^(60*n) ), where [x^n] is the coefficient extraction operator and where F(x) = 1 + x + 98*x^2 + 23861*x^3 + 7987534*x^4 + 3169655645*x^5 + 1398711076599*x^6 + ... appears to have integer coefficients. Cf. A008978. (End)
From Peter Bala, Feb 16 2020: (Start)
Congruences: a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k - apply Mestrovic, equation 39, p. 12.
a(n) = [(x*y*z)^n] (1 + x + y + z)^(5*n). (End)
a(n) = a(n-1)*5*(5*n - 1)*(5*n - 2)*(5*n - 3)*(5*n - 4)/(2*n^3*(2*n - 1)). - Neven Sajko, Jul 22 2023

A181418 a(n) = A000984(n)*A000172(n), which is the term-wise product of the Central binomial coefficients and Franel numbers, respectively.

Original entry on oeis.org

1, 4, 60, 1120, 24220, 567504, 14030016, 360222720, 9513014940, 256758913840, 7051260776560, 196403499277440, 5535202897806400, 157551884911456000, 4522682234563776000, 130783762623673221120, 3806221127760278029980
Offset: 0

Views

Author

Paul D. Hanna, Jan 28 2011

Keywords

Comments

This sequence is s_6 in Cooper's paper. - Jason Kimberley, Nov 25 2012
Diagonal of the rational function R(x,y,z,w)=1/(1-(w*x*y+w*z+x*y+x*z+y+z)). - Gheorghe Coserea, Jul 13 2016

Examples

			E.g.f.: A(x) = 1 + 4*x/2! + 60*x^2/(2!*4!) + 1120*x^3/(3!*6!) + 24220*x^4/(4!*8!) + 567504*x^5/(5!*10!) +....
where A(x)^(1/2) = 1 + x + x^2/2!^3 + x^3/3!^3 + x^4/4!^3 +x^5/5!^3 +...
		

Crossrefs

Related to diagonal of rational functions: A268545-A268555.

Programs

  • Mathematica
    Table[Binomial[2n,n]*Sum[Binomial[n,k]^3,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Mar 06 2014 *)
  • PARI
    {a(n)=binomial(2*n,n)*sum(k=0,n,binomial(n,k)^3)}
    
  • PARI
    {a(n)=(2*n)!*n!*polcoeff(sum(m=0, n, x^m/m!^3+x*O(x^n))^2, n)}

Formula

a(n) = C(2n,n) * Sum_{k=0..n} C(n,k)^3.
E.g.f.: Sum_{n>=0} a(n)*x^n/(n!*(2*n)!) = ( Sum_{n>=0} x^n/n!^3 )^2.
From Jason Kimberley, Nov 26 2012: (Start)
1/Pi
= (2/25)*Sum_{n>=0} a(n)*(9n+2)/50^n. [Cooper, equation (5)]
= (2/25)*Sum_{n>=0} a(n)*A017185(n)/A165800(n). (End)
G.f.: 4*hypergeom([1/6, 1/3],[1],(27/2)*(1+(1-32*x)^(1/2))*(1-(1-32*x)^(1/2))^2/(3+(1-32*x)^(1/2))^3)^2/(3+(1-32*x)^(1/2)). - Mark van Hoeij, May 07 2013
Recurrence: n^3*a(n) = 2*(2*n-1)*(7*n^2 - 7*n + 2)*a(n-1) + 32*(n-1)*(2*n-3)*(2*n-1)*a(n-2). - Vaclav Kotesovec, Mar 06 2014
a(n) ~ 2^(5*n+1) / (sqrt(3) * (Pi*n)^(3/2)). - Vaclav Kotesovec, Mar 06 2014
0 = (-x^2+28*x^3+128*x^4)*y''' + (-3*x+126*x^2+768*x^3)*y'' + (-1+92*x+864*x^2)*y' + (4+96*x)*y, where y is g.f. - Gheorghe Coserea, Jul 13 2016

A275047 Diagonal of the rational function 1/(1-(1+w)(xy + xz + yz)) [even-indexed terms only].

Original entry on oeis.org

1, 18, 1350, 141120, 17151750, 2272538268, 318430816704, 46404203788800, 6961609406993670, 1068002895589987500, 166779781860762170100, 26422986893371642828800, 4236593267629481817240000, 686167053247777413372681600, 112093956900827388909570240000
Offset: 0

Views

Author

Gheorghe Coserea, Jul 18 2016

Keywords

Comments

Odd-order terms are zero since R(x,y,z,w) = R(-x,-y,-z,w), where R(x,y,z,w) = 1/(1-(1+w)*(x*y + x*z + y*z)).
From Peter Bala, Jun 22 2023: (Start)
a(n) = A(n,n,2*n,2*n) (= A(2*n,2*n,n,n)) in the notation of Straub, equation 8, where it is shown that the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r. This also follows from Meštrović equation 39, since a(n) = binomial(3*n,n)^2 * binomial(2*n,n).
Inductively define a family of sequences {a(i,n) : n >= 0}, i >= 1, by setting a(1,n) = a(n) and, for i >= 2, a(i,n) = [x^n] ( exp(Sum_{k >= 1} a(i-1,k)*x^k/k) )^n.
We conjecture that the sequences {a(i,n) : n >= 0}, i >= 2, also satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5, and positive integers n and r. Cf. A362725 and A362732. (End)

Examples

			1 + 18*x^2 + 1350*x^4 + 141120*x^6 + ...
		

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; `if`(n=0, 1,
          9*(3*n-1)^2*(3*n-2)^2*a(n-1)/((4*n-2)*n^3))
        end:
    seq(a(n), n=0..20);  # Alois P. Heinz, Jul 25 2016
  • Mathematica
    Table[(3*n)!^2 / (n!^4*(2*n)!), {n, 0, 20}] (* Vaclav Kotesovec, Aug 03 2016 *)
    CoefficientList[Series[HypergeometricPFQ[{1/3, 1/3, 2/3, 2/3}, {1/2, 1, 1}, 729x/4], {x, 0, 10}], x] (* Benedict W. J. Irwin, Aug 05 2016 *)
  • PARI
    my(x='x, y='y, z='z, w='w);
    R = 1/(1-(1+w)*(x*y+x*z+y*z));
    diag(n, expr, var) = {
      my(a = vector(n));
      for (i = 1, #var, expr = taylor(expr, var[#var - i + 1], n));
      for (k = 1, n, a[k] = expr;
           for (i = 1, #var, a[k] = polcoeff(a[k], k-1)));
      return(a);
    };
    diag(23, R, [x,y,z,w])

Formula

0 = (-4*x^2+729*x^4)*y'''' + (-20*x+7290*x^3)*y''' + (-16+18063*x^2)*y'' + 10449*x*y' + 576*y, where y = 1 + 18*x^2 + 1350*x^4 + ...
From Vaclav Kotesovec, Aug 03 2016: (Start)
a(n) = (3*n)!^2 / (n!^4 * (2*n)!).
a(n) ~ 3^(6*n+1) / (Pi^(3/2) * n^(3/2) * 2^(2*n+2)).
(End)
G.f.: 4F3(1/3,1/3,2/3,2/3;1/2,1,1;729x/4). - Benedict W. J. Irwin, Aug 05 2016
From Peter Bala, Sep 20 2021: (Start)
a(n) = 9*(3*n - 1)^2*(3*n - 2)^2/(2*n^3*(2*n - 1))*a(n-1).
a(n) = Sum_{k = n..3*n} (-1)^k*binomial(3*n,k)^2*binomial(k,n)^2. (End)
From Peter Bala, Jun 22 2023: (Start)
a(n) = binomial(3*n,n)^2 * binomial(2*n,n) = A188662(n) * A000984(n).
a(n) = Sum_{k = 0..n} binomial(n,k)*binomial(2*n,k)*binomial(2*n-k,n)* binomial(4*n-k,2*n).
a(n) = [(x*y)^n * (z*t)^(2*n)] 1/((1 - x - y)*(1 - z - t) - x*y*z*t). (End)

A268554 Diagonal of the rational function 1/((1 - w - u v) * (1 - x y - x z - y z)).

Original entry on oeis.org

1, 36, 6300, 1552320, 445945500, 139815211536, 46384755633216, 16009450307136000, 5689533506261190300, 2067982222137781950000, 765185639177176836418800, 287266309673587605560908800, 109149488451384203661831720000
Offset: 0

Views

Author

N. J. A. Sloane, Feb 29 2016

Keywords

Comments

Each second element (which is zero) is skipped. - R. J. Mathar, Mar 10 2016
Annihilating differential operator: (-x^2+432*x^4)*Dx^4 + (-5*x+4320*x^3)*Dx^3 + (-4+10644*x^2)*Dx^2 + 6012*x*Dx + 288.

Crossrefs

Programs

  • Maple
    A268554 := proc(n)
        1/(1-w-u*v)/(1-x*y-x*z-y*z) ;
        coeftayl(%,x=0,n) ;
        coeftayl(%,y=0,n) ;
        coeftayl(%,z=0,n) ;
        coeftayl(%,u=0,n) ;
        coeftayl(%,v=0,n) ;
        coeftayl(%,w=0,n) ;
    end proc:
    seq(A268554(2*n),n=0..40) ; # R. J. Mathar, Mar 10 2016
  • Mathematica
    Table[(4*n)!*(3*n)!/((n!)^3*(2*n)!^2), {n, 0, 15}] (* Vaclav Kotesovec, Jul 01 2016 *)
  • PARI
    my(x1='x1, x2='x2, x3='x3, y1='y1, y2='y2, y3='y3);
    R = 1/((1 - y1 - y2*y3) * (1 - x1*x2 - x1*x3 - x2*x3));
    diag(n, expr, var) = {
      my(a = vector(n));
      for (i = 1, #var, expr = taylor(expr, var[#var - i + 1], n));
      for (k = 1, n, a[k] = expr;
           for (i = 1, #var, a[k] = polcoeff(a[k], k-1)));
      return(a);
    };
    diag(11, R, [x1,x2,x3,y1,y2,y3]) \\ Gheorghe Coserea, Jun 30 2016

Formula

Conjecture: n^3*(2*n-1)*a(n) -6*(4*n-1)*(3*n-1)*(3*n-2)*(4*n-3)*a(n-1)=0. - R. J. Mathar, Mar 10 2016
From Vaclav Kotesovec, Jul 01 2016: (Start)
a(n) = (4*n)! * (3*n)! / ((n!)^3 * (2*n)!^2).
a(n) ~ 2^(4*n - 3/2) * 3^(3*n + 1/2) / (Pi^(3/2) * n^(3/2)).
(End)
0 = (-x^2+432*x^4)*y'''' + (-5*x+4320*x^3)*y''' + (-4+10644*x^2)*y'' + 6012*x*y' + 288*y, where y = 1 + 36*x^2 + 6300*x^4 + ... is the g.f. - Gheorghe Coserea, Jul 03 2016
From Peter Bala, Oct 16 2024: (Start)
a(n) = 4 * Sum_{k = 0..2*n-1} (-1)^(n+k) * binomial(2*n-1, k) * binomial(4*n+k-1, k) * A108625(2*n, 2*n-k) for n >= 1 (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). Cf. A002897.
a(n) = binomial(4*n, 2*n)*binomial(3*n, n)*binomial(2*n, n).
The supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k (apply Meštrović, Section 6, equation 39).
a(n) = [x^(2*n)] (1 + x)^(4*n) * [x^n] (1 + x)^(3*n) * [x^n] (1 + x)^(2*n) = [x^n] F(x)^(36*n), where F(x) = 1 + x + 52*x^2 + 6919*x^3 + 1266837*x^4 + 275133604*x^5 + 66468858333*x^6 + 17272069128056*x^7 + 4732687104502730*x^8 + 1350192483617697301*x^9 + 397617338885817524186*x^10 + ... appears to have integer coefficients (checked up to O(x^500)).
Let E(x) = exp(Sum_{n >= 1} (1/36) *a(n)*x^n/n). Then E(x) = 1 + x + 88*x^2 + 14461*x^3 + 3115089*x^4 + 781116715*x^5 + 215898182457*x^6 + 63857605571783*x^7 + 19853845202113934*x^8 + 6413541401057933731*x^9 + 2135530251738770328084*x^10 + ... appears to have integer coefficients (checked up to O(x^500)).
a(n) = 36 * [x^n] ( x/series_reversion(E(x)) )^n.
For integer r and positive integer s, define sequences {u(n) : n >= 0} and {v(n) : n >= 0} by setting u(n) = [x^(s*n)] F(x)^(r*n) and v(n) = [x^(s*n)] E(x)^(r*n). We conjecture that both u(n) and v(n) satisfy the above supercongruences. (End)

A274668 Diagonal of the rational function 1/(1 - x - y - z - x y + x z - y z + x y z).

Original entry on oeis.org

1, 7, 109, 2095, 44401, 995647, 23161909, 552919423, 13454515585, 332268466327, 8302478659069, 209447296631503, 5325782947464721, 136330694520639535, 3509805380065157989, 90806156097601965055, 2359490223343888886785, 61541525049445532797735, 1610570872210945422212365
Offset: 0

Views

Author

Gheorghe Coserea, Jul 05 2016

Keywords

Comments

Annihilating differential operator: x*(5*x+4)*(x^3+19*x^2+27*x-1)*Dx^2 + (15*x^4+206*x^3+363*x^2+216*x-4)*Dx + 5*x^3+33*x^2+36*x+28.

Crossrefs

Programs

  • Mathematica
    gf = Hypergeometric2F1[1/12, 5/12, 1, 13824*x^4*(1 - 27*x - 19*x^2 - x^3) / (1 - 28*x + 54*x^2 + 20*x^3 + x^4)^3]/(1 - 28*x + 54*x^2 + 20*x^3 + x^4)^(1/4);
    CoefficientList[gf + O[x]^20, x] (* Jean-François Alcover, Dec 01 2017 *)
  • PARI
    my(x='x, y='y, z='z);
    R = 1/(1 - x - y - z - x*y + x*z - y*z + x*y*z);
    diag(n, expr, var) = {
      my(a = vector(n));
      for (i = 1, #var, expr = taylor(expr, var[#var - i + 1], n));
      for (k = 1, n, a[k] = expr;
           for (i = 1, #var, a[k] = polcoeff(a[k], k-1)));
      return(a);
    };
    diag(10, R, [x, y, z])
    
  • PARI
    \\ system("wget http://www.jjj.de/pari/hypergeom.gpi");
    read("hypergeom.gpi");
    N = 21; x = 'x + O('x^N);
    Vec(hypergeom([1/12, 5/12],[1],13824*x^4*(1-27*x-19*x^2-x^3)/(1-28*x+54*x^2+20*x^3+x^4)^3, N)/(1-28*x+54*x^2+20*x^3+x^4)^(1/4))

Formula

G.f.: hypergeom([1/12, 5/12],[1],13824*x^4*(1-27*x-19*x^2-x^3)/(1-28*x+54*x^2+20*x^3+x^4)^3)/(1-28*x+54*x^2+20*x^3+x^4)^(1/4).
0 = x*(5*x+4)*(x^3+19*x^2+27*x-1)*y'' + (15*x^4+206*x^3+363*x^2+216*x-4)*y' + (5*x^3+33*x^2+36*x+28)*y, where y is the g.f.
From Peter Bala, Jan 15 2020: (Start)
a(n) = Sum_{0 <= j, k <= n} (-1)^(n+k)*C(n,k)*C(n,j)*C(n+k,k)*C(n+k+j,k+j). Cf. A001850 and A126086.
n^2*(37*n - 49)*a(n) = (999*n^3 - 2322*n^2 + 1567*n - 328)*a(n-1) + (703*n^3 - 2337*n^2 + 2295*n - 536)*a(n-2) + (n - 2)^2*(37*n - 12)*a(n-3). (End)

A274785 Diagonal of the rational function 1/(1-(w*x*y*z + w*x*z + w*y + x*y + z)).

Original entry on oeis.org

1, 1, 25, 121, 2881, 23521, 484681, 5223625, 97949041, 1243490161, 22061635465, 309799010665, 5331441539425, 79799232449665, 1352284119871465, 21095036702450281, 355125946871044561, 5694209222592780625, 95705961654403180201, 1563714140278617173641, 26311422169994777663761
Offset: 0

Views

Author

Gheorghe Coserea, Jul 13 2016

Keywords

Comments

Diagonal of the rational function 1/(1 - (x^2 + y^2 + z^2 + w^2 + x*y*z*w)). - Seiichi Manyama, Jul 04 2025

Crossrefs

Programs

  • Maple
    seq(add(binomial(n+2*k, 2*k)*binomial(n, 2*k)*binomial(2*k, k)^2, k = 0..floor(n/2)), n = 0..20); # Peter Bala, Jan 27 2018
  • Mathematica
    Table[Sum[Binomial[n + 2*k, 2*k]*Binomial[n, 2*k]*Binomial[2*k, k]^2, {k, 0, n/2}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 17 2023 *)
  • PARI
    my(x='x, y='y, z='z, w='w);
    R = 1/(1-(w*x*y*z+w*x*z+w*y+x*y+z));
    diag(n, expr, var) = {
      my(a = vector(n));
      for (i = 1, #var, expr = taylor(expr, var[#var - i + 1], n));
      for (k = 1, n, a[k] = expr;
           for (i = 1, #var, a[k] = polcoeff(a[k], k-1)));
      return(a);
    };
    diag(12, R, [x,y,z,w])
    
  • PARI
    a(n) = sum(k=0, n\2, binomial(n + 2*k,2*k) * binomial(n,2*k) * binomial(2*k,k)^2) \\ Andrew Howroyd, Mar 18 2023

Formula

0 = (-x^2+2*x^3+257*x^4+508*x^5+257*x^6+2*x^7-x^8)*y''' + (-3*x+15*x^2+1524*x^3+2286*x^4+789*x^5+3*x^6-6*x^7)*y'' + (-1+16*x+1687*x^2+1168*x^3+217*x^4-8*x^5-7*x^6)*y' + (1+183*x-178*x^2-2*x^3-3*x^4-x^5)*y, where y is the g.f.
a(n) = Sum_{k = 0..floor(n/2)} C(n + 2*k,2*k)*C(n,2*k)*C(2*k,k)^2 (apply Eger, Theorem 3 to the set of column vectors S = {[0,0,1,0], [1,1,0,0], [0,1,0,1], [1,0,1,1],[1,1,1,1]}). - Peter Bala, Jan 27 2018
n^3*(n - 2)*(2*n - 5)*a(n) = (2*n - 5)*(2*n - 1)*(2*n^3 - 6*n^2 + 4*n - 1)*a(n-1) + (2*n - 3)*(250*n^4 - 1500*n^3 + 3066*n^2 - 2448*n + 629)*a(n-2) + (2*n - 5)*(2*n - 1)*(2*n^3 - 12*n^2 + 22*n - 11)*a(n-3) - (2*n - 1)*(n - 1)*(n - 3)^3*a(n-4). - Peter Bala, Mar 17 2023
a(n) ~ 5^(1/4) * phi^(6*n + 3) / (2^(5/2) * Pi^(3/2) * n^(3/2)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Mar 17 2023

A268550 Diagonal of 1/(1 - x - y - z - x y - x z - y z).

Original entry on oeis.org

1, 12, 366, 13800, 574650, 25335072, 1159174464, 54435558672, 2606102756730, 126634810078920, 6226427808402516, 309095505195676992, 15466884541698962736, 779158533743408851200, 39476348002042199114400, 2010009672816216740255520
Offset: 0

Views

Author

N. J. A. Sloane, Feb 29 2016

Keywords

Comments

Annihilating differential operator: x*(x+2)*(x+1)*(27*x^2+54*x-1)*Dx^2 + (81*x^4+324*x^3+431*x^2+214*x-2)*Dx + 24*x^3+72*x^2+72*x+24. - Gheorghe Coserea, Jul 03 2016

Crossrefs

Programs

  • Maple
    A268550 := proc(n)
        1/(1-x-y-z-x*y-x*z-y*z) ;
        coeftayl(%,x=0,n) ;
        coeftayl(%,y=0,n) ;
        coeftayl(%,z=0,n) ;
    end proc:
    seq(A268550(n),n=0..20) ; # R. J. Mathar, Mar 10 2016
  • Mathematica
    gf = Hypergeometric2F1[1/12, 5/12, 1, 1728*x^3*(x + 2)^3*(27*x^2 + 54*x - 1)/(-1 + 48*x + 24*x^2)^3]/(1 - 48*x - 24*x^2)^(1/4);
    CoefficientList[gf + O[x]^20, x] (* Jean-François Alcover, Dec 03 2017, after Gheorghe Coserea *)
  • PARI
    my(x='x, y='y, z='z);
    R =  1/(1 - x - y - z - x*y - x*z - y*z);
    diag(n, expr, var) = {
      my(a = vector(n));
      for (i = 1, #var, expr = taylor(expr, var[#var - i + 1], n));
      for (k = 1, n, a[k] = expr;
           for (i = 1, #var, a[k] = polcoeff(a[k], k-1)));
      return(a);
    };
    diag(10, R, [x,y,z])
    
  • PARI
    \\ system("wget http://www.jjj.de/pari/hypergeom.gpi");
    read("hypergeom.gpi");
    N = 16; x = 'x + O('x^N);
    Vec(hypergeom([1/12, 5/12], [1], 1728*x^3*(x+2)^3*(27*x^2+54*x-1)/(-1+48*x+24*x^2)^3, N)/(1-48*x-24*x^2)^(1/4)) \\ Gheorghe Coserea, Jul 03 2016

Formula

Conjecture: 2*n^2*(n-1)*a(n) -(n-1)*(107*n^2-107*n+24)*a(n-1) +12*(-9*n^3+27*n^2-22*n+2)*a(n-2) -3*n*(3*n-5)*(3*n-7)*a(n-3)=0. - R. J. Mathar, Mar 10 2016
a(n) ~ sqrt(3) * (27 + 6*sqrt(21))^n / (2*Pi*n). - Vaclav Kotesovec, Jul 01 2016
G.f.: hypergeom([1/12, 5/12], [1], 1728*x^3*(x+2)^3*(27*x^2+54*x-1)/(-1+48*x+24*x^2)^3)/(1-48*x-24*x^2)^(1/4). - Gheorghe Coserea, Jul 01 2016
0 = x*(x+2)*(x+1)*(27*x^2+54*x-1)*y'' + (81*x^4+324*x^3+431*x^2+214*x-2)*y' + (24*x^3+72*x^2+72*x+24)*y, where y is g.f. - Gheorghe Coserea, Jul 03 2016
a(n) = Sum_{i = 0..n} Sum_{k = 0..n-i} Sum_{j = 0..n-k} (n+i+j+k)!/(i!*j!*k!*(n-k-i)!*(n-k-j)!*(2*k+i+j-n)!). Apply [Eger, Theorem 3] to the set of column vectors S = {[1,0,0], [0,1,0], [0,0,1], [1,1,0], [1,0,1], [0,1,1]}. - Peter Bala, Apr 28 2022

A268552 Diagonal of the rational function 1/((1 - u v - u w - v w - u v w) * (1 - x y - x z - y z)).

Original entry on oeis.org

1, 42, 13590, 7410480, 5074665750, 3931541906292, 3290234596753104, 2903719469665734720, 2664470557292509315350, 2519171054960424071707500, 2438942560726150616825027340, 2407205497290665533295978551680, 2414129024898590207617757303701200
Offset: 0

Views

Author

N. J. A. Sloane, Feb 29 2016

Keywords

Comments

Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r. - Peter Bala, Jun 22 2023

Crossrefs

Programs

  • Maple
    A268552 := proc(n)
        1/(1-u*v-u*w-v*w-u*v*w)/(1-x*y-x*z-y*z) ;
        coeftayl(%,x=0,n) ;
        coeftayl(%,y=0,n) ;
        coeftayl(%,z=0,n) ;
        coeftayl(%,u=0,n) ;
        coeftayl(%,v=0,n) ;
        coeftayl(%,w=0,n) ;
    end proc:
    seq(A268552(2*n),n=0..40) ; # R. J. Mathar, Mar 10 2016
  • Mathematica
    f = 1/((1 - u v - u w - v w - u v w)*(1 - x y - x z - y z));
    a[n_] := Fold[SeriesCoefficient[#1, {#2, 0, n}]&, f, {x, y, z, u, v, w}];
    Table[a[2n], {n, 0, 20}] (* Jean-François Alcover, Dec 03 2017 *)

Formula

Recurrence: (n-1)^2*n^4*(2*n - 1)^2*(3504*n^5 - 30952*n^4 + 107572*n^3 - 183680*n^2 + 154013*n - 50763)*a(n) = 3*(n-1)^2*(3*n - 2)*(3*n - 1)*(798912*n^9 - 8654880*n^8 + 39805024*n^7 - 101548488*n^6 + 157599040*n^5 - 153473292*n^4 + 93212371*n^3 - 33837996*n^2 + 6644187*n - 547020)*a(n-1) - 9*(3*n - 5)*(3*n - 4)*(3*n - 2)*(3*n - 1)*(7989120*n^9 - 94537920*n^8 + 478555808*n^7 - 1351912112*n^6 + 2330100416*n^5 - 2513561312*n^4 + 1671938288*n^3 - 648448700*n^2 + 130707255*n - 10800900)*a(n-2) + 27*(2*n - 5)^2*(3*n - 8)*(3*n - 7)*(3*n - 5)*(3*n - 4)*(3*n - 2)*(3*n - 1)*(3504*n^5 - 13432*n^4 + 18804*n^3 - 11636*n^2 + 3081*n - 306)*a(n-3). - Vaclav Kotesovec, Jul 01 2016
a(n) ~ c * d^n / (Pi^2 * n^2), where d = 1189.7580084904576415418942340231454997... is the root of the equation -19683 + 415530*d - 1539*d^2 + d^3 = 0 and c = 0.44151111077974450880059816263885416848395811427... is the root of the equation -1 + 72*c - 384*c^2 + 512*c^3 = 0. - Vaclav Kotesovec, Jul 01 2016
From Peter Bala, Oct 31 2024: (Start)
a(n) = [x^n] F(x)^n, where F(x)^(1/6) = 1 + 7*x + 863*x^2 + 293447*x^3 + 143872942*x^4 + 85385169968*x^5 + 57059886430771*x^6 + 41372830809612345*x^7 + 31846446970675137307*x^8 + 25661130820830021920913*x^9 + 21438532416924289799815363*x^10 + ... appears to have integer coefficients.
Conjecture 1. Let m be an integer. The sequence defined by u(n) = [x^n] F(x)^(m*n/6) satisfies the supercongruences u(n*p^r) === u(n*p^(r-1)) (mod p^r) for all primes p >= 5 and all positive integers n and r.
Let E(x) = exp(Sum_{n >= 1} a(n)*x^n/n). Then E(x)^(1/6) = 1 + 7*x + 1157*x^2 + 419678*x^3 + 214995384*x^4 + 133012390735*x^5 + 92645697263188*x^6 + 70017181144513849*x^7 + 56177705597073431689*x^8 + 47186677673368534079592*x^9 + 41097380151878683734374690*x^10 + ... appears to have integer coefficients.
Conjecture 2. Let m be an integer. The sequence defined by v(n) = [x^n] E(x)^(m*n/6) satisfies the supercongruences v(n*p^r) === v(n*p^(r-1)) (mod p^r) for all primes p >= 5 and all positive integers n and r. (End)

A274671 Diagonal of the rational function 1/(1 - x - y - z - x y + x z + y z - x y z).

Original entry on oeis.org

1, 5, 49, 593, 7921, 111965, 1641865, 24705665, 378931585, 5898329045, 92893693729, 1477015762865, 23671209428881, 381902943661517, 6196712221450009, 101043805539177473, 1654726991239056385, 27201210101330189477, 448652354194417534609, 7422254208909904273553
Offset: 0

Views

Author

Gheorghe Coserea, Jul 05 2016

Keywords

Comments

Annihilating differential operator: x*(4*x-5)*(x^3+27*x^2-19*x+1)*Dx^2 + (12*x^4+196*x^3-481*x^2+190*x-5)*Dx + 4*x^3+18*x^2-95*x+25.
Also diagonal of rational functions 1/(1 + z + 2*x*y - y*z + x*z - x*y*z), 1/(1 + 2*z + x*y - 2*y*z + x*z - x*y*z), 1/(1 + y + z + 2*x*y + y*z + x*z + x*y*z), 1/(1 + y + 2*z + x*y + 2*y*z + x*z + x*y*z), 1/(1 + x + y - 2*z - y*z - 2*x*z + x*y*z). - Gheorghe Coserea, Jul 03 2018

Crossrefs

Programs

  • Mathematica
    gf = Hypergeometric2F1[1/12, 5/12, 1, 13824*x^5*(1 - 19*x + 27*x^2 + x^3) / (1 - 20*x + 54*x^2 + 28*x^3 + x^4)^3]/(1 - 20*x + 54*x^2 + 28*x^3 + x^4)^(1/4);
    CoefficientList[gf + O[x]^20, x] (* Jean-François Alcover, Dec 01 2017 *)
    a[n_] := HypergeometricPFQ[{-n, -n, n + 1}, {1, 1}, 2];
    Table[a[n], {n, 0, 19}]  (* Peter Luschny, Nov 15 2021 *)
  • PARI
    \\ system("wget http://www.jjj.de/pari/hypergeom.gpi");
    read("hypergeom.gpi");
    N = 20; x = 'x + O('x^N);
    Vec(hypergeom([1/12, 5/12],[1],13824*x^5*(1-19*x+27*x^2+x^3)/(1-20*x+54*x^2+28*x^3+x^4)^3, N)/(1-20*x+54*x^2+28*x^3+x^4)^(1/4))
    
  • PARI
    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] = polcoeff(a[n], n-1)));
      return(a);
    };
    diag(1/(1 - x - y - z - x*y + x*z + y*z - x*y*z), 20)
    \\ test: diag(1/(1 - x - y - z - x*y + x*z + y*z - x*y*z)) == diag(1/(1 + x + y - 2*z - y*z - 2*x*z + x*y*z))
    \\ Gheorghe Coserea, Jul 03 2018

Formula

G.f.: hypergeom([1/12, 5/12],[1],13824*x^5*(1-19*x+27*x^2+x^3)/(1-20*x+54*x^2+28*x^3+x^4)^3)/(1-20*x+54*x^2+28*x^3+x^4)^(1/4).
0 = x*(4*x-5)*(x^3+27*x^2-19*x+1)*y'' + (12*x^4+196*x^3-481*x^2+190*x-5)*y' + (4*x^3+18*x^2-95*x+25)*y, where y is the g.f.
a(n) = Sum_{k=0..n} binomial(n,k)^2 * binomial(n+k,k) * 2^k. - Ilya Gutkovskiy, Nov 15 2021
a(n) = hypergeom([-n, -n, n + 1], [1, 1], 2). - Peter Luschny, Nov 15 2021
Recurrence: n^2*(37*n - 62)*a(n) = (703*n^3 - 1881*n^2 + 1383*n - 330)*a(n-1) - (999*n^3 - 3672*n^2 + 4267*n - 1510)*a(n-2) - (n-2)^2*(37*n - 25)*a(n-3). - Vaclav Kotesovec, Nov 15 2021
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