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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A081798 a(n) = Sum_{k = 0..n} C(n,k) * C(n+k,k) * C(n+2*k,k).

Original entry on oeis.org

1, 7, 115, 2371, 54091, 1307377, 32803219, 844910395, 22188235867, 591446519797, 15953338537885, 434479441772845, 11927609772412075, 329653844941016785, 9163407745486783435, 255982736410338609931, 7181987671728091545787
Offset: 0

Views

Author

Emanuele Munarini, Apr 23 2003

Keywords

Comments

a(n) is also a generalization of Delannoy numbers to 3D; i.e. the number of walks from (0,0,0) to (n,n,n) in a 3D square lattice where each step is in the direction of one of (1,0,0), (0,1,0), (0,0,1) and (1,1,1). - Theodore Kolokolnikov, Jul 04 2010
Diagonal of the rational function 1/(1 - x - y - z - x*y*z). - Gheorghe Coserea, Jul 06 2016

Crossrefs

Related to diagonal of rational functions: A268545-A268555.

Programs

  • Maple
    w := proc(i,j,k) option remember; if i=0 and j=0 and k = 0 then 1; elif i<0 or j<0 or k<0 then 0 else w(i-1,j,k)+w(i,j-1,k)+w(i,j,k-1)+w(i-1,j-1,k-1); end: end: for k from 0 to 10 do lprint(w(k,k,k)):end: # Theodore Kolokolnikov, Jul 04 2010
    # second Maple program:
    a:= proc(n) option remember; `if`(n<3, 51*n^2-45*n+1,
         ((3*n-2)*(30*n^2-50*n+13)*a(n-1)+(3*n-1)*(n-2)^2*a(n-3)
         -(9*n^3-30*n^2+29*n-6)*a(n-2))/(n^2*(3*n-4)))
        end:
    seq(a(n), n=0..20); # Alois P. Heinz, Sep 22 2013
  • Mathematica
    f[n_] := Sum[ Binomial[n, k] Binomial[n + k, k] Binomial[n + 2k, k], {k, 0, n}]; Array[f, 17, 0] (* Robert G. Wilson v *)
    CoefficientList[Series[HypergeometricPFQ[{1/3, 2/3}, {1}, 27*x/(1 - x)^3]/(1 - x), {x, 0, 20}], x] (* Vaclav Kotesovec, Jul 07 2016 *)
  • Maxima
    makelist(sum(binomial(n,k)*binomial(n+k,k)*binomial(n+2*k,k),k,0,n),n,0,12);
    
  • PARI
    {a(n)=polcoeff(sum(m=0,n,(3*m)!/m!^3*x^m/(1-x+x*O(x^n))^(3*m+1)), n)} \\ Paul D. Hanna, Sep 22 2013
    
  • PARI
    a(n) = sum(k = 0, n, binomial(n, k) * binomial(n+k, k) * binomial(n+2*k, k)); \\ Michel Marcus, Jan 14 2016

Formula

a(n) = w(n,n,n) where w(i,j,k)=w(i-1,j,k)+w(i,j-1,k)+w(i,j,k-1)+w(i-1,j-1,k-1) and where w(0,0,0)=1 and w(i,j,k)=0 if one of i,j,k is strictly negative. - Theodore Kolokolnikov, Jul 04 2010
G.f.: hypergeom([1/3, 2/3],[1],27*x/(1-x)^3)/(1-x). - Mark van Hoeij, Oct 24 2011
G.f.: Sum_{n>=0} (3*n)!/n!^3 * x^n / (1-x)^(3*n+1). - Paul D. Hanna, Sep 22 2013
a(n) ~ c*d^n/(Pi*n), where d = (3*(292 + 4*sqrt(5))^(2/3) + 132 + 20*(292 + 4*sqrt(5))^(1/3)) / (2*(292 + 4*sqrt(5))^(1/3)) = 29.900786688498085... is the root of the equation -1 + 3*d - 30*d^2 + d^3 = 0 and c = 1/(2*sqrt(((81 - 27*sqrt(5))/2)^(1/3) + 3*((3 + sqrt(5))/2)^(1/3) - 6)) = 0.8959908650405192232... is the root of the equation -1 - 72*c^2 - 1296*c^4 + 1728*c^6 = 0. - Vaclav Kotesovec, Sep 23 2013, updated Jul 07 2016
From Peter Bala, Jan 13 2016: (Start)
a(n) = Sum_{k = 0..n} multinomial(n + 2*k, k, k, k, n - k). Cf. A001850(n) = Sum_{k = 0..n} multinomial(n + k, k, k, n - k).
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 4*x^2 + 42*x^3 + 639*x^4 + 11571*x^5 + ... appears to have integer coefficients. (End)
Conjecture: n^2*(3*n-4)*a(n) -(3*n-2)*(30*n^2-50*n+13)*a(n-1) +(9*n^3-30*n^2+29*n-6)*a(n-2) -(3*n-1)*(n-2)^2*a(n-3)=0. - R. J. Mathar, Apr 15 2016
Conjecture: (n^2)*a(n) +(-28*n^2+24*n-3)*a(n-1) +3*(-19*n^2+78*n-77)*a(n-2) +(5*n-12)*(n-3)*a(n-3) -2*(n-3)^2*a(n-4)=0. - R. J. Mathar, Apr 15 2016
0 = (2*x+1)*(x^3-3*x^2+30*x-1)*x*y'' + (6*x^4-8*x^3+51*x^2+60*x-1)*y' + (x-1)*(2*x^2+2*x-7)*y, where y is g.f. - Gheorghe Coserea, Jul 06 2016

A082488 a(n) = Sum_{k = 0..n} C(n,k) * C(n+k,k) * C(n+2*k,k) * C(n+3*k,k).

Original entry on oeis.org

1, 25, 2641, 392641, 67982041, 12838867105, 2564949195985, 533008982952625, 114035552691160585, 24950692835328410305, 5557138347370070346601, 1255741805437716400557625, 287180884347761929741524361, 66343186345544102086872515761
Offset: 0

Views

Author

Emanuele Munarini, Apr 28 2003

Keywords

Comments

Diagonal of the rational function 1/(1-(x + y + z + w + x*y*z*w)). - Gheorghe Coserea, Jul 15 2016

Examples

			G.f.: A(x) = 1 + 25*x + 2641*x^2 + 392641*x^3 + 67982041*x^4 + 12838867105*x^5 +...
where
A(x) = 1/(1-x) + (4!/1!^4)*x/(1-x)^5 + (8!/2!^4)*x^2/(1-x)^9 + (12!/3!^4)*x^3/(1-x)^13 + (16!/4!^4)*x^4/(1-x)^17 + (20!/5!^4)*x^5/(1-x)^21 +... [Hanna]
Equivalently,
A(x) = 1/(1-x) + 24*x/(1-x)^5 + 2520*x^2/(1-x)^9 + 369600*x^3/(1-x)^13 + 63063000*x^4/(1-x)^17 + 11732745024*x^5/(1-x)^21 +...+ A008977(n)*x^n/(1-x)^(4*n+1) +...
		

Crossrefs

Cf. A081798.
Column k = 4 of A229142.
Related to diagonal of rational functions: A268545-A268555.

Programs

  • End
    
    				
  • Magma
    [&+[Binomial(n, k)*Binomial(n+k, k)*Binomial(n+2*k, k)*Binomial(n+3*k, k): k in [0..n]]: n in [0..20]]; // Vincenzo Librandi, Oct 16 2018
  • Maple
    with(combinat):
    a:= n-> add(multinomial(n+3*k, n-k, k$4), k=0..n):
    seq(a(n), n=0..15);  # Alois P. Heinz, Sep 23 2013
  • Mathematica
    Table[Sum[Binomial[n,k]*Binomial[n+k,k]*Binomial[n+2*k,k]* Binomial[n+3*k,k], {k, 0, n}],{n,0,20}] (* Vaclav Kotesovec, Sep 23 2013 *)
  • PARI
    {a(n)=polcoeff(sum(m=0, n, (4*m)!/m!^4*x^m/(1-x+x*O(x^n))^(4*m+1)), n)}
    for(n=0, 15, print1(a(n), ", "))
    
  • PARI
    {a(n)=sum(k=0, n, binomial(n, k)*binomial(n+k, k)*binomial(n+2*k, k)*binomial(n+3*k, k))}
    for(n=0, 15, print1(a(n), ", "))
    

Formula

G.f.: Sum_{n>=0} (4*n)!/n!^4 * x^n / (1-x)^(4*n+1). - Paul D. Hanna, Sep 22 2013
Recurrence: n^3*(2*n-3)*(4*n-9)*(4*n-5)*a(n) = (4*n-9)*(4*n-3)*(520*n^4 - 1820*n^3 + 2109*n^2 - 905*n + 121)*a(n-1) - (192*n^6 - 1536*n^5 + 4748*n^4 - 7050*n^3 + 5065*n^2 - 1563*n + 171)*a(n-2) + (4*n-1)*(32*n^5 - 296*n^4 + 1040*n^3 - 1689*n^2 + 1209*n - 279)*a(n-3) - (n-3)^3*(2*n-1)*(4*n-5)*(4*n-1)*a(n-4). - Vaclav Kotesovec, Sep 23 2013
a(n) ~ c*d^n/(Pi^(3/2)*n^(3/2)), where d = 65 + 46*sqrt(2) + 2*sqrt(2*(1055 + 746*sqrt(2))) = 259.976980158726979... is the maximal positive root of the equation 1 - 4*d + 6*d^2 - 260*d^3 + d^4 = 0 and c = sqrt(8 + 5*sqrt(2) + sqrt(14*(11 + 8*sqrt(2))))/8 = 0.71529801573844067904424114047445568721... - Vaclav Kotesovec, Sep 23 2013, updated Jul 16 2016
G.f.: hypergeom([1/8, 3/8],[1],256*x/(1-x)^4)^2/(1-x). - Mark van Hoeij, Sep 23 2013
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 13*x^2 + 893*x^3 + 99125*x^4 + 13706093*x^5 + ... appears to have integer coefficients. - Peter Bala, Jan 13 2016
0 = x^2*(3*x+1)^2*(1-260*x+6*x^2-4*x^3+x^4)*y''' + 3*x*(3*x+1)*(1-390*x-378*x^2+8*x^3-15*x^4+6*x^5)*y'' + (1-836*x+133*x^2+768*x^3-69*x^4-60*x^5+63*x^6)*y' + (-25+397*x-378*x^2-6*x^3+3*x^4+9*x^5)*y, where y is the g.f. - Gheorghe Coserea, Jul 15 2016

A206178 a(n) = Sum_{k=0..n} binomial(n,k)^3 * 2^k.

Original entry on oeis.org

1, 3, 21, 171, 1521, 14283, 138909, 1385163, 14072193, 145039923, 1512191781, 15914734443, 168802010001, 1802247516891, 19350710547021, 208783189719531, 2262263134211073, 24604815145831011, 268499713118585781, 2938736789722114731, 32250788066104022961
Offset: 0

Views

Author

Paul D. Hanna, Feb 04 2012

Keywords

Comments

Ignoring initial term, equals the logarithmic derivative of A206177.
Compare to Franel numbers: A000172(n) = Sum_{k=0..n} binomial(n,k)^3.
Diagonal of rational functions 1/(1 - x*y + y*z + 2*x*z - 3*x*y*z), 1/(1 + y + z + x*y + y*z + 2*x*z + 3*x*y*z), 1/(1 - x + 2*z + x*y - y*z - 2*x*z + 3*x*y*z), 1/(1 - x - y - z + x*y + y*z + x*z - 3*x*y*z), 1/(1 - x + y + 2*z - x*y + 2*y*z - 2*x*z - 3*x*y*z). - Gheorghe Coserea, Jul 03 2018

Examples

			L.g.f.: L(x) = 3*x + 21*x^2/2 + 171*x^3/3 + 1521*x^4/4 + 14283*x^5/5 +...
Exponentiation equals the g.f. of A206177:
exp(L(x)) = 1 + 3*x + 15*x^2 + 93*x^3 + 657*x^4 + 5067*x^5 + 41579*x^6 +...
		

Crossrefs

Related to diagonal of rational functions: A268545-A268555.

Programs

  • Mathematica
    Flatten[{1,RecurrenceTable[{(n+3)^2*(3*n+4)*a[n+3]-3*(9*n^3+57*n^2+116*n+74)*a[n+2]-3*(27*n^3+144*n^2+252*n+145)*a[n+1]-27*(3*n+7)*(n+1)^2*a[n]==0, a[1]==3, a[2]==21, a[3]==171},a,{n,1,20}]}] (* Vaclav Kotesovec, Sep 11 2012 *)
    Table[HypergeometricPFQ[{-n, -n, -n}, {1, 1}, -2], {n, 0, 20}] (* Jean-François Alcover, Oct 25 2019 *)
  • PARI
    {a(n)=sum(k=0,n,binomial(n,k)^3*2^k)}
    
  • Sage
    A206178 = lambda n: hypergeometric([-n,-n,-n], [1,1], -2)
    [Integer(A206178(n).n(100)) for n in (0..20)] # Peter Luschny, Sep 23 2014

Formula

a(2*3^n) == 3 (mod 9) for n>=0; a(n) == 0 (mod 9) if n/2 > 1 is not a power of 3.
Recurrence: (n+3)^2*(3*n+4)*a(n+3) - 3*(9*n^3+57*n^2+116*n+74)*a(n+2) - 3*(27*n^3+144*n^2+252*n+145)*a(n+1) - 27*(3*n+7)*(n+1)^2*a(n) = 0. - Vaclav Kotesovec, Sep 11 2012
a(n) ~ (1 + 2^(1/3))^(3*n + 2) / (2^(4/3)*sqrt(3)*Pi*n). - Vaclav Kotesovec, Sep 19 2012, simplified Apr 24 2025
G.f.: hypergeom([1/3, 2/3],[1],54*x^2/(1-3*x)^3)/(1-3*x). - Mark van Hoeij, May 02 2013
a(n) = hypergeom([-n,-n,-n],[1,1], -2). - Peter Luschny, Sep 23 2014
G.f. y=A(x) satisfies: 0 = x*(3*x + 2)*(27*x^3 + 27*x^2 + 9*x - 1)*y'' + (243*x^4 + 378*x^3 + 189*x^2 + 36*x - 2)*y' + 3*(x + 1)*(27*x^2 + 12*x + 2)*y. - Gheorghe Coserea, Jul 01 2018

Extensions

Minor edits by Vaclav Kotesovec, Mar 31 2014

A275051 Expansion of 3F2([1/9, 4/9, 5/9], [1/3,1], 729*x).

Original entry on oeis.org

1, 60, 20475, 9373650, 4881796920, 2734407111744, 1605040007778900, 973419698810097000, 604759111060745718000, 382741738086972337402560, 245810413547242455520545552, 159759730493918131135425965280, 104861901534978616465850670348000
Offset: 0

Views

Author

Gheorghe Coserea, Jul 19 2016

Keywords

Comments

"One may consider the following conjecture: all the irreducible factors of the minimal order linear differential operator annihilating a diagonal of a rational function should be homomorphic to their adjoint (possibly on an algebraic extension). [...]
"If our conjecture above was correct, this would be a way to show that the series cannot be the diagonal of a rational function." (See Boukraa link.)

Examples

			1 + 60*x + 20475*x^2 + 9373650*x^3 + ...
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[HypergeometricPFQ[{1/9, 4/9, 5/9}, {1/3,1}, 729*x], {x, 0, 15}], x] (* Vaclav Kotesovec, Jul 28 2016 *)
    a[n_] := FullSimplify[(729^n Cos[Pi/18] Gamma[1/3] Gamma[1/9 + n] Gamma[4/9 + n] Gamma[5/9 + n])/(Pi Gamma[1/9] Gamma[1/3 + n] n!^2)] (* Benedict W. J. Irwin, Aug 05 2016 *)
  • PARI
    \\ system("wget http://www.jjj.de/pari/hypergeom.gpi");
    read("hypergeom.gpi");
    N = 21; x = 'x + O('x^N);
    Vec(hypergeom_sym([1/9, 4/9, 5/9], [1/3,1], 729*x, N))
    
  • PARI
    my(x = 'x + O('x^20)); Vec(hypergeom([1/9, 4/9, 5/9], [1/3,1], 729*x)) \\ Michel Marcus, Apr 11 2023

Formula

G.f.: hypergeom([1/9, 4/9, 5/9], [1/3,1], 729*x).
From Vaclav Kotesovec, Jul 28 2016: (Start)
Recurrence: n^2*(3*n - 2)*a(n) = 3*(9*n - 8)*(9*n - 5)*(9*n - 4)*a(n-1).
a(n) ~ Gamma(1/3) * cos(Pi/18) * 3^(6*n) / (Pi * Gamma(1/9) * n^(11/9)).
(End)
a(n) = 729^n*cos(Pi/18)*Gamma(1/3)*Gamma(1/9+n)*Gamma(4/9+n)*Gamma(5/9+n) /(Pi*Gamma(1/9)*Gamma(1/3+n)*n!^2). - Benedict W. J. Irwin, Aug 05 2016
From Karol A. Penson, Apr 11 2023: (Start)
a(n) = Integral_{x=0..729} x^n*W(x), where
W(x) = W1(x) + W2(x) + W3(x), and
W1(x) = (2*cos(Pi/18)*3^(1/3)*2^(4/9)*sqrt(Pi)*Gamma(13/18)*hypergeom([1/9, 1/9, 7/9], [5/9, 2/3], x/729))/(9*Gamma(2/3)^2*Gamma(1/9)*Gamma(8/9)^2*x^(8/9));
W2(x) = cos(Pi/18)*2^(1/9)*Gamma(2/9)*Gamma(1/18)*hypergeom([4/9, 4/9, 10/9], [8/9, 4/3], x/729)/(162*Gamma(2/3)*Gamma(1/9)*Pi^(3/2)*x^(5/9));
W3(x) = cos(Pi/18)*3^(1/6)*2^(4/9)*Gamma(5/18)*Gamma(-1/18)*hypergeom([5/9, 5/9, 11/9], [10/9, 13/9], x/729)/(324*Gamma(2/3)*Gamma(1/9)*Pi*Gamma(4/9)*x^(4/9)).
This integral representation is unique as W(x) is the solution of the Hausdorff power moment problem. Using only the definition of a(n), W(x) can be proven to be positive. W(x) is singular at x = 0, with the singularity x^(-4/9), and for x > 0 is monotonically decreasing to zero at x = 729. At x = 729 the first derivative of W(x) is infinite. (End)

A275054 G.f.: 3F2([1/9, 2/9, 8/9], [2/3,1], 729 x).

Original entry on oeis.org

1, 24, 6732, 2771340, 1342525275, 711891288108, 399866544799722, 233750557331494632, 140707672445849703480, 86621407014527646518400, 54278825541246092520182592, 34504174655166790354911360048, 22195631874904018057471849288020, 14421008706115620277976088538033200
Offset: 0

Views

Author

Gheorghe Coserea, Jul 20 2016

Keywords

Comments

"Other hypergeometric 'blind spots' for Christol’s conjecture" - (see Bostan link).

Examples

			1 + 24*x + 6732*x^2 + 2771340*x^3 + ...
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[HypergeometricPFQ[{1/9, 2/9, 8/9}, {2/3, 1}, 729 x], {x, 0, 13}], x] (* Michael De Vlieger, Jul 26 2016 *)
    a[n_] := FullSimplify[(729^n Gamma[2/3] Gamma[1/9 + n] Gamma[2/9 + n] Gamma[8/9 + n] Sin[Pi/9])/(Pi (n!)^2 Gamma[2/9] Gamma[2/3 + n])] (* Benedict W. J. Irwin, Aug 05 2016 *)
  • PARI
    \\ system("wget http://www.jjj.de/pari/hypergeom.gpi");
    read("hypergeom.gpi");
    N = 12; x = 'x + O('x^N);
    Vec(hypergeom([1/9, 2/9, 8/9], [2/3,1], 729*x, N))

Formula

G.f.: hypergeom([1/9, 2/9, 8/9], [2/3,1], 729*x).
From Vaclav Kotesovec, Jul 28 2016: (Start)
Recurrence: n^2*(3*n - 1)*a(n) = 3*(9*n - 8)*(9*n - 7)*(9*n - 1)*a(n-1).
a(n) ~ 2 * sin(Pi/9) * 3^(6*n - 1/2) / (Gamma(1/3) * Gamma(2/9) * n^(13/9)).
(End)
a(n) = 729^n*Gamma(2/3)*Gamma(1/9+n)*Gamma(2/9+n)*Gamma(8/9+n)*Sin(Pi/9)/(Pi*(n!)^2*Gamma(2/9)*Gamma(2/3+n)). - Benedict W. J. Irwin, Aug 05 2016

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

A243949 Squares of the central Delannoy numbers: a(n) = A001850(n)^2.

Original entry on oeis.org

1, 9, 169, 3969, 103041, 2832489, 80802121, 2365752321, 70611901441, 2139090528969, 65568745087209, 2029206892664961, 63300531617048961, 1987912809986437161, 62787371136571152009, 1992942254830520803329, 63531842302018973818881, 2033004661359005674887561
Offset: 0

Views

Author

Paul D. Hanna, Aug 17 2014

Keywords

Comments

In general, we have the binomial identity:
if b(n) = Sum_{k=0..n} t^k * C(2*k, k) * C(n+k, n-k), then b(n)^2 = Sum_{k=0..n} (t^2+t)^k * C(2*k, k)^2 * C(n+k, n-k), where the g.f. of b(n) is 1/sqrt(1 - (4*t+2)*x + x^2), and the g.f. of b(n)^2 is 1 / AGM(1-x, sqrt((1+x)^2 - (4*t+2)^2*x)), where AGM(x,y) = AGM((x+y)/2,sqrt(x*y)) is the arithmetic-geometric mean.
Note that the g.f. of A001850 is 1/sqrt(1 - 6*x + x^2).
Limit_{n -> oo} a(n+1)/a(n) = (3 + 2*sqrt(2))^2 = 17 + 12*sqrt(2).
From Gheorghe Coserea, Jul 05 2016: (Start)
Diagonal of the rational function 1/(1 - x - y - z - x*y + x*z - y*z - x*y*z).
Annihilating differential operator: x*(x-1)*(x+1)*(x^2-34*x+1)*Dx^2 + (3*x^4-66*x^3-70*x^2+70*x-1)*Dx + x^3-7*x^2-35*x+9.
(End).
The sequence b(n) mentioned above is the sequence of shifted Legendre polynomials P(n,2*t + 1) (see A063007). See Zudilin for a g.f. for the sequence b(n)^2. - Peter Bala, Mar 02 2017

Examples

			G.f.: A(x) = 1 + 9*x + 169*x^2 + 3969*x^3 + 103041*x^4 + 2832489*x^5 +...
		

Crossrefs

Sequences of the form LegendreP(n, 2*m+1)^2: A000012 (m=0), this sequence (m=1), A243943 (m=2), A243944 (m=3), A243007 (m=4).
Related to diagonal of rational functions: A268545 - A268555.

Programs

  • Magma
    [Evaluate(LegendrePolynomial(n), 3)^2 : n in [0..40]]; // G. C. Greubel, May 17 2023
    
  • Mathematica
    Table[Sum[2^k *Binomial[2*k, k]^2 *Binomial[n+k, n-k], {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Aug 18 2014 *)
    a[n_]:= HypergeometricPFQ[{1/2, -n, n+1}, {1, 1}, -8];
    Table[a[n], {n, 0, 17}] (* Peter Luschny, Mar 14 2018 *)
    LegendreP[Range[0, 30], 3]^2 (* G. C. Greubel, May 17 2023 *)
  • PARI
    {a(n) = sum(k=0, n, 2^k * binomial(2*k, k)^2 * binomial(n+k, n-k) )}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    {a(n)=polcoeff( 1 / agm(1-x, sqrt((1+x)^2 - 36*x +x*O(x^n))), n)}
    for(n=0, 20, print1(a(n), ", "))
    
  • Python
    from math import comb
    def A243949(n): return sum(comb(n,k)*comb(n+k,k) for k in range(n+1))**2 # Chai Wah Wu, Mar 23 2023
    
  • SageMath
    [gen_legendre_P(n,0,3)^2 for n in range(41)] # G. C. Greubel, May 17 2023

Formula

G.f.: 1 / AGM(1-x, sqrt(1-34*x+x^2)). - Paul D. Hanna, Aug 30 2014
a(n) = Sum_{k=0..n} 2^k * C(2*k, k)^2 * C(n+k, n-k).
a(n)^(1/2) = Sum_{k=0..n} C(2*k, k) * C(n+k, n-k).
Recurrence: n^2*(2*n-3)*a(n) = (2*n-1)*(35*n^2 - 70*n + 26)*a(n-1) - (2*n-3)*(35*n^2 - 70*n + 26)*a(n-2) + (n-2)^2*(2*n-1)*a(n-3). - Vaclav Kotesovec, Aug 18 2014
a(n) ~ (4 + 3*sqrt(2)) * (3 + 2*sqrt(2))^(2*n) / (8*Pi*n). - Vaclav Kotesovec, Aug 18 2014
From Gheorghe Coserea, Jul 05 2016: (Start)
G.f.: hypergeom([1/12, 5/12],[1],27648*x^4*(x^2-34*x+1)*(x-1)^2/(1-36*x+134*x^2-36*x^3+x^4)^3)/(1-36*x+134*x^2-36*x^3+x^4)^(1/4).
0 = x*(x-1)*(x+1)*(x^2-34*x+1)*y'' + (3*x^4-66*x^3-70*x^2+70*x-1)*y' + (x^3-7*x^2-35*x+9)*y, where y is g.f.
(End)
a(n) = Sum_{k = 0..n} 4^k*binomial(n+k,2*k)^2*binomial(2*k,k). - Peter Bala, Mar 02 2017
a(n) = hypergeom([1/2, -n, n + 1], [1, 1], -8). - Peter Luschny, Mar 14 2018
G.f.: Sum_{n >= 0} (2^n)*binomial(2*n,n)^2 *x^n/(1-x)^(2*n+1). - Peter Bala, Feb 07 2022

A078678 Number of binary strings with n 1's and n 0's avoiding zigzags, that is avoiding the substrings 101 and 010.

Original entry on oeis.org

1, 2, 4, 8, 18, 42, 100, 242, 592, 1460, 3624, 9042, 22656, 56970, 143688, 363348, 920886, 2338566, 5949148, 15157874, 38674978, 98803052, 252701484, 646990518, 1658066668, 4252908542, 10917422860, 28046438252, 72099983802, 185469011130, 477383400300
Offset: 0

Views

Author

Emanuele Munarini, Dec 17 2002

Keywords

Comments

Also number of Grand Dyck paths of length 2*n with no zigzags, that is, with no factors UDU or DUD. - Emanuele Munarini, Jul 07 2011

Examples

			For n = 2 : 0011, 0110, 1001, 1100.
For n = 3 : 000111, 011001, 100011, 110001, 001110, 011100, 100110, 111000.
		

Crossrefs

Cf. A003440.
Main diagonal of array A099172.
Related to diagonal of rational functions: A268545-A268555.

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<5, [1, 2, 4, 8, 18][n+1],
         (2*n*a(n-1)+(n-2)*a(n-2)+(2*n-8)*a(n-3)-(n-4)*a(n-4))/n)
        end:
    seq(a(n), n=0..40);  # Alois P. Heinz, Feb 13 2020
  • Mathematica
    Table[SeriesCoefficient[Series[Sqrt[(1 + x + x^2)/(1 - 3 x + x^2)], {x, 0, n}], n], {n, 0, 40}]
  • Maxima
    a(n):=coeff(taylor((1+x+x^2)/sqrt(1-2*x-x^2-2*x^3+x^4),x,0,n),x,n);
    makelist(a(n),n,0,12); /* Emanuele Munarini, Jul 07 2011 */
    
  • PARI
    my(x='x+O('x^99)); Vec(((1+x+x^2)/(1-3*x+x^2))^(1/2)) \\ Altug Alkan, Jul 18 2016

Formula

G.f.: sqrt( ( 1 + x + x^2 ) / ( 1 - 3*x + x^2 ) ).
a(n) = Sum_{k=0..n+floor(n/2)} binomial( n - k + 2*floor(k/3), floor(k/3) )^2.
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k,k)^2*( 2*n^2 - 6*n*k + 6*k^2 )/(n-k)^2, n > 0.
a(n) ~ 2 * ((3+sqrt(5))/2)^n / (5^(1/4)*sqrt(Pi*n)). - Vaclav Kotesovec, Mar 21 2014
a(n) = [x^n y^n](1+x*y+x^2*y^2)/(1-x-y+x*y-x^2*y^2). - Gheorghe Coserea, Jul 18 2016
D-finite with recurrence: n*a(n) -2*n*a(n-1) +(-n+2)*a(n-2) +2*(-n+4)*a(n-3) +(n-4)*a(n-4)=0. [Doslic] - R. J. Mathar, Jun 21 2018

A124435 Number of effective multiple alignments of three equal-length sequences.

Original entry on oeis.org

1, 5, 67, 1109, 20251, 391355, 7847155, 161476565, 3387271675, 72114452255, 1553475100717, 33786532319435, 740681494769659, 16346552430326123, 362830907979309067, 8093356178498583509, 181311959402343288955, 4077310062938894133623, 91999289732199733092601
Offset: 0

Views

Author

Lee A. Newberg, Dec 15 2006

Keywords

Comments

This counts effective alignments rather than standard alignments, so that for example the following two alignments are equivalent:
-A A-
-T T-
C- -C
See Dress, Morgenstern and Stoye for more information.

Examples

			a(1) = 5 because the five alignments are
  A--   A-   A-   A-   A
  -C-   C-   -C   -C   C
  --T   -T   T-   -T   T
		

Crossrefs

Programs

  • Maple
    G := series( hypergeom([1/3, 2/3],[1],27*x/(1+x)^3)/(1+x), x=0, 31);
    seq(coeff(G,x,i),i=0..30);  # Mark van Hoeij, Dec 20 2013
  • Mathematica
    a[n_] := Sum[(-1)^(n-k) Binomial[n, k] Binomial[n+2k, n] Binomial[2k, k], {k, 0, n}];
    Table[a[n], {n, 0, 18}] (* Jean-François Alcover, Sep 18 2018, after Wadim Zudilin *)
  • 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] = polcoef(a[n], n-1)));
      return(a);
    };
    x='x; y='y; z='z; diag(1/(1 - x - y - z + x*y*z), 19)
    
  • PARI
    \\ system("wget http://www.jjj.de/pari/hypergeom.gpi");
    read("hypergeom.gpi");
    N = 20; x = 'x + O('x^N);
    Vec(hypergeom([1/3, 2/3],[1],27*x/(1+x)^3, N)/(1+x)) \\ Gheorghe Coserea, Jul 06 2016

Formula

The recurrence is three-dimensional with the order of the three parameters immaterial. That is, a(i,j,k)=a(i,k,j)=a(j,i,k)=a(j,k,i)=a(k,i,j)=a(k,j,i). a(i, j, 0) = (i+j)! / i! / j! a(i, j, k) = a(i-1,j,k) + a(i,j-1,k) + a(i,j,k-1) - a(i-1,j-1,k-1).
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n,k)*binomial(n+2*k,n)*binomial(2*k,k). - Wadim Zudilin, Nov 26 2015
Diagonal of 1/(1 - x - y - z + x*y*z). - Mark van Hoeij, Dec 20 2013
G.f.: hypergeom([1/3, 2/3],[1],27*x/(1+x)^3)/(1+x). - Mark van Hoeij, Dec 20 2013
(3*n-1)*(n+1)^2*a(n+1)-(3*n+1)*(24*n^2+8*n-5)*a(n)+(9*n^3-3*n^2-4*n+2)*a(n-1)+(3*n+2)*(n-1)^2*a(n-2)=0. - Robert Israel, Nov 26 2015
0 = (2*x-1)*(x^3+3*x^2-24*x+1)*x*y'' + (6*x^4+8*x^3-57*x^2+48*x-1)*y' + (x+1)*(2*x^2-2*x+5)*y, where y is g.f. - Gheorghe Coserea, Jul 06 2016
From Peter Bala, Mar 16 2023: (Start)
(3*n - 4)*n^2*a(n) = (3*n - 2)*(24*n^2 - 40*n + 11)*a(n-1) - (9*n^3 - 30*n^2 + 29*n - 6)*a(n-2) - (3*n - 1)*(n - 2)^2*a(n-3) with a(0) = 1, a(1) = 5 and a(2) = 67.
Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for positive integers n and r and all primes p >= 5. (End)

Extensions

More terms from Mark van Hoeij, Dec 21 2013

A268549 Diagonal of (1 - 9 x y)/((1 - 3 y - 2 x + 3 y^2 + 9 x^2 y) * (1 - u - z) * (1 - v - w)).

Original entry on oeis.org

1, 12, 648, 50400, 4630500, 468087984, 50345463168, 5655718328832, 656151696743400, 78036148295820000, 9465472643689782720, 1166663950520357802240, 145719568153188579382560, 18405635030728188793200000
Offset: 0

Views

Author

N. J. A. Sloane, Feb 29 2016

Keywords

Comments

"The corresponding (order-three) linear differential operator is not homomorphic to its adjoint, even with an algebraic extension." (see A. Bostan link) - Gheorghe Coserea, Aug 15 2016

Examples

			1 + 12*x + 648*x^2 + 50400*x^3 + ...
		

Crossrefs

Programs

  • Maple
    A268549 := proc(n)
        (1-9*x*y)/(1-3*y-2*x+3*y^2+9*x^2*y)/(1-u-z)/(1-v-w) ;
        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(A268549(n),n=0..40) ; # R. J. Mathar, Mar 11 2016
    series(hypergeom([1/3, 1/2, 1/2], [1, 1], 144*x), x=0, 14); # Gheorghe Coserea, Aug 15 2016
  • Mathematica
    FullSimplify[Table[3^(2*n)*(2*n)!^2*Gamma[n + 1/3]/(Gamma[1/3]*(n!)^5), {n, 0, 15}]] (* Vaclav Kotesovec, Jul 01 2016 *)

Formula

a(n) = [(xyzuvw)^n] (1 - 9*x*y)/((1 - 3*y - 2*x + 3*y^2 + 9*x^2*y) * (1 - u - z) * (1 - v - w)).
D-finite with recurrence: n^3*a(n) -12*(3*n-2)*(-1+2*n)^2*a(n-1)=0. - R. J. Mathar, Mar 11 2016 [follows from the hypergeometric g.f. below - Georg Fischer, Jul 30 2022]
From Vaclav Kotesovec, Jul 01 2016: (Start)
a(n) = 3^(2*n) * (2*n)!^2 * Gamma(n + 1/3) / (Gamma(1/3) * (n!)^5).
a(n) ~ 12^(2*n)/(Gamma(1/3)*Pi*n^(5/3)).
(End)
From Gheorghe Coserea, Aug 16 2016: (Start)
a(n) = [(xyzuv)^n] 1/((1 - x + 3*y - 27*x*y^3 - 27*x*y^2 - 9*x*y + 3*y^2) * (1 - u - v - u*z - v*z)).
G.f.: hypergeom([1/3, 1/2, 1/2], [1, 1], 144*x).
(End)
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