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 21-30 of 36 results. Next

A186416 a(n) = binomial(2n,n)^4/(n+1)^3.

Original entry on oeis.org

1, 2, 48, 2500, 192080, 18670176, 2125170432, 270968717448, 37634544090000, 5588044012339360, 875419364366134016, 143310129125665075392, 24338673855047938317568, 4264316875814353400000000, 767401591466550107174400000, 141345980472409642279275210000, 26569505644587874058090478570000
Offset: 0

Views

Author

Emanuele Munarini, Feb 21 2011

Keywords

Crossrefs

Programs

  • Maple
    A186416 := proc(n) binomial(2*n,n)^4/(n+1)^3 ; end proc: # R. J. Mathar, Feb 23 2011
  • Mathematica
    Table[Binomial[2n,n]^4/(n+1)^3,{n,0,40}]
  • Maxima
    makelist(binomial(2*n,n)^4/(n+1)^3,n,0,40);

Formula

G.f.: 4F3(1/2,1/2,1/2,1/2;2,2,2;256*x), where nFm(...;..;.) denotes a generalized hypergeometric series.
a(n) = (A000108(n))^3*A000984(n). - R. J. Mathar, Feb 23 2011

A186284 Self-convolution square equals A127776.

Original entry on oeis.org

1, 2, 48, 1704, 71490, 3291780, 160844160, 8189867280, 429832053840, 23088359467040, 1263134996327680, 70138971602098560, 3942799810867610280, 223942062435751452240, 12831882367225056387840, 740872398293620831990080
Offset: 0

Views

Author

Paul D. Hanna, Feb 16 2011

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 48*x^2 + 1704*x^3 + 71490*x^4 + 3291780*x^5 +...
Related expansions.
The g.f. of A127776 equals A(x)^2:
A(x)^2 = 1 + 4*x + 100*x^2 + 3600*x^3 + 152100*x^4 + 7033104*x^5 +...+ A004981(n)^2*x^n +...
The g.f. of A002897 equals A(x)^4:
A(x)^4 = 1 + 8*x + 216*x^2 + 8000*x^3 + 343000*x^4 + 16003008*x^5 +...+ A000984(n)^3*x^n +...
The g.f. of A004981 begins:
1/(1-8*x)^(1/4) = 1 + 2*x + 10*x^2 + 60*x^3 + 390*x^4 + 2652*x^5 +...
where A004981(n) = (2^n/n!)*Product_{k=0..n-1} (4k + 1).
The g.f. of A000984 begins:
1/(1-4*x)^(1/2) = 1 + 2*x + 6*x^2 + 20*x^3 + 70*x^4 + 252*x^5 +...
where A000984(n) = (2n)!/(n!)^2 forms the central binomial coefficients.
		

Crossrefs

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Sqrt[Hypergeometric2F1[ 1/4, 1/4, 1, 64*x]], {x, 0, nmax}], x] (* Vaclav Kotesovec, Apr 10 2018 *)
  • PARI
    {a(n)=local(A004981=1/(1-8*x+x*O(x^n))^(1/4),A=sum(m=0,n,polcoeff(A004981,m)^2*x^m+x*O(x^n))^(1/2));polcoeff(A,n)}
    
  • PARI
    {a(n)=local(A000984=1/(1-4*x+x*O(x^n))^(1/2),A=sum(m=0,n,polcoeff(A000984,m)^3*x^m+x*O(x^n))^(1/4));polcoeff(A,n)}

Formula

Self-convolution 4th power equals A002897.
G.f.: sqrt( K(k)/(Pi/2) ) in powers of (kk'/4)^2, where K(k) is complete elliptic integral of first kind evaluated at modulus k. [From a formula by Michael Somos in A002897]
G.f.: sqrt( 1/AGM(1, (1-16x)^(1/2)) ) in powers of x(1-16x) where AGM() is the arithmetic-geometric mean. [From a formula by Michael Somos in A004981]
a(n) ~ Pi^(3/4) * 2^(6*n - 1/2) / (Gamma(1/4)^3 * n^(3/2)). - Vaclav Kotesovec, Apr 10 2018

A224735 G.f.: exp( Sum_{n>=1} binomial(2*n,n)^3 * x^n/n ).

Original entry on oeis.org

1, 8, 140, 3616, 116542, 4316080, 175593800, 7640774080, 349626142909, 16632958651688, 816163494236860, 41069537125459360, 2110206360805542510, 110346590629125981872, 5857345961837113457864, 314962180518584299711424, 17128125582951726423704502, 940726748732537798295599280
Offset: 0

Views

Author

Paul D. Hanna, Apr 16 2013

Keywords

Examples

			G.f.: A(x) = 1 + 8*x + 140*x^2 + 3616*x^3 + 116542*x^4 + 4316080*x^5 +...
where
log(A(x)) = 2^3*x + 6^3*x^2/2 + 20^3*x^3/3 + 70^3*x^4/4 + 252^3*x^5/5 + 924^3*x^6/6 + 3432^3*x^7/7 + 12870^3*x^8/8 +...+ A000984(n)^3*x^n/n +...
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[Exp[8*x*HypergeometricPFQ[{1, 1, 3/2, 3/2, 3/2}, {2, 2, 2, 2}, 64*x]], {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 27 2025 *)
  • PARI
    {a(n)=polcoeff(exp(sum(k=1,n,binomial(2*k,k)^3*x^k/k)+x*O(x^n)),n)}
    for(n=0,20,print1(a(n),", "))

Formula

Logarithmic derivative yields A002897.

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)

A380243 Number of rooted 2n-regular planar maps with 3 vertices.

Original entry on oeis.org

1, 54, 3000, 171500, 10001880, 591666768, 35371207872, 2131746903000, 129299660919000, 7883256659941520, 482689850761774656, 29661047546558142624, 1828220386252351000000, 112982297841774018000000, 6998159395715622920640000, 434337846995341921726638000, 27004842919501042631643927000
Offset: 1

Views

Author

Andrew Howroyd, Jan 22 2025

Keywords

Comments

There are no odd valent regular planar maps with 3 vertices.

Crossrefs

Row n=3 of A380241.

Programs

  • PARI
    a(n) = n*binomial(2*n-1, n)^3

Formula

a(n) = n*binomial(2*n-1, n)^3.
a(n) = n*A002897(n - 1).

A186418 a(n) = binomial(2*n,n)^4/(n + 1)^2.

Original entry on oeis.org

1, 4, 144, 10000, 960400, 112021056, 14876193024, 2167749739584, 338710896810000, 55880440123393600, 9629613008027474176, 1719721549507980904704, 316402760115623198128384, 59700436261400947600000000
Offset: 0

Views

Author

Emanuele Munarini, Feb 21 2011

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Binomial[2n,n]^4/(n+1)^2,{n,0,40}]
  • Maxima
    makelist(binomial(2*n,n)^4/(n+1)^2,n,0,40);

Formula

G.f.: 4F3({1/2,1/2,1/2,1/2},{1,2,2},256x), where 4F3 is a hypergeometric series.

A367177 Triangle read by rows, T(n, k) = [x^k] hypergeom([1/2, -n, -n], [1, 1], 4*x).

Original entry on oeis.org

1, 1, 2, 1, 8, 6, 1, 18, 54, 20, 1, 32, 216, 320, 70, 1, 50, 600, 2000, 1750, 252, 1, 72, 1350, 8000, 15750, 9072, 924, 1, 98, 2646, 24500, 85750, 111132, 45276, 3432, 1, 128, 4704, 62720, 343000, 790272, 724416, 219648, 12870
Offset: 0

Views

Author

Peter Luschny, Nov 07 2023

Keywords

Examples

			Triangle T(n, k) starts:
  [0] 1;
  [1] 1,   2;
  [2] 1,   8,    6;
  [3] 1,  18,   54,     20;
  [4] 1,  32,  216,    320,      70;
  [5] 1,  50,  600,   2000,    1750,     252;
  [6] 1,  72, 1350,   8000,   15750,    9072,     924;
  [7] 1,  98, 2646,  24500,   85750,  111132,   45276,    3432;
  [8] 1, 128, 4704,  62720,  343000,  790272,  724416,  219648,   12870;
  [9] 1, 162, 7776, 141120, 1111320, 4000752, 6519744, 4447872, 1042470, 48620;
		

Crossrefs

Cf. A002893 (row sum), A002897 (central column), A000984 (main diagonal).

Programs

  • Maple
    p := n -> hypergeom([1/2, -n, -n], [1, 1], 4*x):
    T := (n, k) -> coeff(simplify(p(n)), x, k):
    seq(seq(T(n, k), k = 0..n), n = 0..9);

Formula

T(n, k) = binomial(n, k)^2 * binomial(2*k, k).

A186419 a(n) = binomial(2*n,n)^4/(n + 1).

Original entry on oeis.org

1, 8, 432, 40000, 4802000, 672126336, 104133351168, 17341997916672, 3048398071290000, 558804401233936000, 105925743088302215936, 20636658594095770856448, 4113235881503101575668992, 835806107659613266400000000, 172665358079973774114240000000
Offset: 0

Views

Author

Emanuele Munarini, Feb 21 2011

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Binomial[2n,n]^4/(n+1),{n,0,40}]
  • Maxima
    makelist(binomial(2*n,n)^4/(n+1),n,0,12);

Formula

G.f.: 4F3({1/2,1/2,1/2,1/2},{1,1,2},256x), where 4F3 is a hypergeometric series.

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
Previous Showing 21-30 of 36 results. Next