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.

Showing 1-4 of 4 results.

A362733 a(n) = [x^n] F(x)^n, where F(x) = exp( Sum_{k >= 1} A362732(k)*x^k/k ).

Original entry on oeis.org

1, 6, 234, 10428, 492522, 24033006, 1197423396, 60530725380, 3092592004074, 159295600885794, 8258018380659234, 430335300869496072, 22521831447746893092, 1182951246247357578348, 62325193477833011143260, 3292376206935392483917428, 174323297281680647978503146, 9248680725006429075147528150
Offset: 0

Views

Author

Peter Bala, May 06 2023

Keywords

Crossrefs

Programs

  • Maple
    E(n,x) := series(exp(n*add((3*k)!/k!^3*x^k/k, k = 1..20)), x, 21):
    A362732(n) := coeftayl(E(n,x), x = 0, n):
    F(n,x) := series(exp(n*add(A362732(k)*x^k/k, k = 1..20)), x, 21):
    seq(coeftayl(F(n,x), x = 0, n), n = 0..20);
    # alternative program
    G(n,x) := series(exp(n*add((3*k)!/k!^3*x^(2*k)/k, k = 1..40)), x, 41):
    seq((1/2)*coeftayl(G(2*n,x), x = 0, 2*n), n = 1..20); # Peter Bala, Oct 27 2024

Formula

a(n*p^r) == a(n*p^(r-1)) (mod p^r) (Gauss congruence) holds for all primes p and positive integers n and r.
Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for all primes p >= 3 and positive integers n and r.
For n >= 1, a(n) = (1/2) * [x^n] G(x)^(2*n), where G(x) = exp(Sum_{k >= 1} A006480(k)*x^k/k) is the g.f. of A229451. - Peter Bala, Oct 27 2024

A229451 G.f.: exp( Sum_{n>=1} (3*n)!/n!^3 * x^n/n ).

Original entry on oeis.org

1, 6, 63, 866, 13899, 246366, 4676768, 93322596, 1934035965, 41286407510, 902562584556, 20119266633060, 455832458083577, 10470568749165246, 243361203186769659, 5714294570067499930, 135377464019074334826, 3232534121305720233264, 77726654423445817800164
Offset: 0

Views

Author

Paul D. Hanna, Sep 23 2013

Keywords

Comments

The sixth root of the o.g.f. A(x)^(1/6) = 1 + x + 8*x^2 + 101*x^3 + 1569*x^4 + 27445*x^5 + ... appears to have integer coefficients. See A229452. More generally, if A(m,x) := exp( Sum_{n >= 1} (m*n)!/n!^m * x^n/n ), m = 1,2,3,..., then it can be shown that the expansion of A(m,x) has integer coefficients. We conjecture that the expansion of A(m,x)^(1/m!) also has integer coefficients. - Peter Bala, Feb 16 2020

Examples

			G.f.: A(x) = 1 + 6*x + 63*x^2 + 866*x^3 + 13899*x^4 + 246366*x^5 +...,
where
log(A(x)) = 6*x + 90*x^2/2 + 1680*x^3/3 + 34650*x^4/4 + 756756*x^5/5 +...+ A006480(n)*x^n/n + ....
		

Crossrefs

Cf. A229452, A006480 (de Bruijn's S(3,n)), A061401, A333042, A333043, A370288, A362732, A370289, A370293.

Programs

  • Mathematica
    CoefficientList[Series[Exp[6*x*HypergeometricPFQ[{1,1,4/3,5/3},{2,2,2},27*x]],{x,0,20}],x] (* Vaclav Kotesovec, Dec 25 2013 *)
  • PARI
    {a(n)=polcoeff(exp(sum(k=1,n,(3*k)!/k!^3*x^k/k) +x*O(x^n)),n)}
    for(n=0,25,print1(a(n),", "))

Formula

a(n) ~ c * 3^(3*n)/n^2, where c = 2^11 * 3^(7/2) * Pi^5 * A370293^6 = 0.406436497... - Vaclav Kotesovec, Dec 25 2013, updated Feb 14 2024
a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} A006480(k) * a(n-k). - Seiichi Manyama, Feb 09 2024
From Peter Bala, Oct 24 2024: (Start)
Series reversion of x*A(-x) = x + 6*x^2 + 9*x^3 + 56*x^4 - 300*x^5 + 3942*x^6 - ... is the g.f. of A061401.
The g.f. A(x) satisfies [x^n] A(x)^n = A362732(n). (End)

A184423 a(n) = (2*n)!*(3*n)!/n!^5.

Original entry on oeis.org

1, 12, 540, 33600, 2425500, 190702512, 15849497664, 1369618398720, 121821136479900, 11079206239530000, 1025579963180813040, 96310511463483233280, 9152842704012278107200, 878622906816654279840000
Offset: 0

Views

Author

Paul D. Hanna, Jan 13 2011

Keywords

Comments

Denoted by h_3[n] by T. Piezas III. He also gives formulas for 1/Pi such as 1/Pi = 2 * Sum_{n>=0} a(n) * (-1)^n * (51*n + 7) / (12^3)^(n + 1/2). - Michael Somos, May 31 2012
Diagonal of the rational function R(x,y,z,w) = 1/(1-(w*y+w*z+x+y+z)). - Gheorghe Coserea, Jul 15 2016
From Peter Bala, Jun 28 2023: (Start)
The supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for primes p >= 5 and positive integers n and r. This follows from Meštrović equation 39, since a(n) = binomial(3*n,n) * binomial(2*n,n)^2.
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 primes p >= 5 and positive integers n and r. Cf. A362730 and A362732. (End)

Examples

			G.f.: A(x) = 1 + 12*x + 540*x^2 + 33600*x^3 + 2425500*x^4 +...
G.f. of A184424 equals A(x)^(1/2):
A(x)^(1/2) = 1 + 6*x + 252*x^2 + 15288*x^3 + 1089270*x^4 + 84963060*x^5 +...+ [(3^n/n!^2)*Product_{k=1..n} (6*k-4)*(6*k-5)]*x^n +...
		

Crossrefs

Related to diagonal of rational functions: A268545-A268555.

Programs

  • Mathematica
    Table[((2n)!(3n)!)/(n!)^5,{n,0,20}] (* Harvey P. Dale, Dec 18 2018 *)
  • PARI
    {a(n)=(3*n)!*(2*n)!/n!^5}
    
  • PARI
    {a(n)=polcoeff(sum(m=0,n,3^m*prod(k=1,m,(6*k-4)*(6*k-5))/m!^2*x^m+x*O(x^n))^2,n)}

Formula

Self-convolution of A184424:
a(n) = Sum_{k=0..n} A184424(k)*A184424(n-k) where A184424(n) = (3^n/n!^2)*Product_{k=1..n} (6*k-4)*(6*k-5).
a(n) = 6 * (2*n - 1) * (3*n - 1) * (3*n - 2) / n^3 * a(n-1) if n>0. - Michael Somos, May 31 2012
0 = (x^2-108*x^3)*y''' + (3*x-486*x^2)*y''+ (1-348*x)*y' - 12*y, where y is g.f. - Gheorghe Coserea, Jul 15 2016
a(n) ~ 3^(1/2)/(2*Pi^(3/2)) * n^(-3/2) * 108^n. - Ilya Gutkovskiy, Jul 15 2016
a(n) = C(2*n,n)^2 * C(3*n,n) = ( [x^n](1 + x)^(2*n) )^2 * ( [x^n](1 + x)^(3*n) ) = [x^n]( F(x)^(12*n) ), where [x^n] is the coefficient extraction operator and where F(x) = 1 + x + 11*x^2 + 350*x^3 + 15293*x^4 + 794433*x^5 + 45958617*x^6 + ... appears to have integral coefficients. Cf. A000897 and A001451. - Peter Bala, Dec 30 2019

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)
Showing 1-4 of 4 results.