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 1-3 of 3 results.

A217615 G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} binomial(n,k)^2 * x^k*(1-x)^(n-k).

Original entry on oeis.org

1, 1, 1, 3, 5, 7, 15, 29, 49, 95, 187, 345, 659, 1289, 2465, 4739, 9237, 17911, 34715, 67705, 132063, 257477, 503309, 984983, 1927895, 3778017, 7411237, 14544967, 28565661, 56144615, 110406527, 217225533, 427636561, 842256047, 1659600955, 3271579689, 6451913519
Offset: 0

Views

Author

Paul D. Hanna, Oct 09 2012

Keywords

Comments

Radius of convergence of g.f. is r = 1/2.
More generally, given
A(x) = Sum_{n>=1} x^n * Sum_{k=0..n} binomial(n,k)^2 * x^k*(t-x)^(n-k),
then A(x) = 1/sqrt( (1 - t*x + 2*x^2)^2 - 4*x^2 )
and the radius of convergence r satisfies: (1-r)^2 = r*(t-r) for t > 0.
a(n) is the number of (2k-1)-element subsets of {1, 2, ..., n+1} whose k-th smallest (i.e., k-th largest) element equals 2k-1. - Darij Grinberg, Oct 09 2019

Examples

			G.f.: A(x) = 1 + x + x^2 + 3*x^3 + 5*x^4 + 7*x^5 + 15*x^6 + 29*x^7 + 49*x^8 + ...
where the g.f. equals the series:
A(x) = 1 +
  x*((1-x) + x) +
  x^2*((1-x)^2 + 2^2*x*(1-x) + x^2) +
  x^3*((1-x)^3 + 3^2*x*(1-x)^2 + 3^2*x^2*(1-x) + x^3) +
  x^4*((1-x)^4 + 4^2*x*(1-x)^3 + 6^2*x^2*(1-x)^2 + 4^2*x^3*(1-x) + x^4) +
  x^5*((1-x)^5 + 5^2*x*(1-x)^4 + 10^2*x^2*(1-x)^3 + 10^2*x^3*(1-x)^2 + 5^2*x^4*(1-x) + x^5) + ...
		

Crossrefs

Programs

  • Maple
    a := n -> `if`(n < 4, [1, 1, 1, 3][n+1], hypergeom([1/2, (1-n)/3, (2-n)/3, -n/3], [1, (1-n)/2, -n/2], -27)):
    seq(simplify(a(n)), n=0..36); # Peter Luschny, Oct 09 2019
  • Mathematica
    CoefficientList[Series[1/Sqrt[(1-x+2*x^2)^2-4*x^2], {x, 0, 20}], x] (* Vaclav Kotesovec, Sep 16 2013 *)
  • PARI
    {a(n)=polcoeff(sum(m=0, n+1, x^m*sum(k=0, m, binomial(m, k)^2*x^k*(1-x)^(m-k) + x*O(x^n))), n)}
    for(n=0,40,print1(a(n),", "))
    
  • PARI
    a(n)={sum(k=0, n\2, binomial(2*k, k) * binomial(n-2*k, k))} \\ Andrew Howroyd, Oct 09 2019

Formula

G.f.: A(x) = 1 / sqrt( (1 - x + 2*x^2)^2 - 4*x^2 ).
G.f.: A(x) = 1 / sqrt( (1-x)*(1-2*x)*(1+x+2*x^2) ).
G.f. satisfies: A(x) = (1 + 2*x^2*Sum_{n>=0} A000108(n)*(-x*A(x))^(2*n)) / (1-x+2*x^2) where A000108(n) = binomial(2*n,n)/(n+1) forms the Catalan numbers.
a(n) ~ 2^n/sqrt(Pi*n). - Vaclav Kotesovec, Sep 16 2013
a(n) = Sum_{k=0..floor(n/2)} binomial(2*k, k) * binomial(n-2*k, k). - Darij Grinberg, Oct 09 2019
a(n) = hypergeom([1/2,(1-n)/3,(2-n)/3, -n/3], [1, (1-n)/2, -n/2], -27) for n >= 4. - Peter Luschny, Oct 09 2019

A217461 G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} binomial(n,k)^2 * x^k*(4-x)^(n-k).

Original entry on oeis.org

1, 4, 16, 72, 350, 1768, 9120, 47696, 251974, 1341560, 7186912, 38694000, 209187884, 1134838736, 6174666560, 33681995936, 184138474182, 1008642036184, 5534504908640, 30415064058160, 167378028670788, 922251663816368, 5087347689155264, 28091877168106592
Offset: 0

Views

Author

Paul D. Hanna, Oct 10 2012

Keywords

Comments

Radius of convergence of g.f. is r = (3-sqrt(7))/2 = 0.1771243444...
More generally, given
A(x) = Sum_{n>=1} x^n * Sum_{k=0..n} binomial(n,k)^2 * x^k*(t-x)^(n-k),
then A(x) = 1 / sqrt( (1 - t*x + 2*x^2)^2 - 4*x^2 )
and the radius of convergence r satisfies: (1-r)^2 = r*(t-r) for t>0.

Examples

			G.f.: A(x) = 1 + 4*x + 16*x^2 + 72*x^3 + 350*x^4 + 1768*x^5 +...
where the g.f. equals the series:
A(x) = 1 +
x*((4-x) + x) +
x^2*((4-x)^2 + 2^2*x*(4-x) + x^2) +
x^3*((4-x)^3 + 3^2*x*(4-x)^2 + 3^2*x^2*(4-x) + x^3) +
x^4*((4-x)^4 + 4^2*x*(4-x)^3 + 6^2*x^2*(4-x)^2 + 4^2*x^3*(4-x) + x^4) +
x^5*((4-x)^5 + 5^2*x*(4-x)^4 + 10^2*x^2*(4-x)^3 + 10^2*x^3*(4-x)^2 + 5^2*x^4*(4-x) + x^5) +...
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[1/Sqrt[(1-2*x+2*x^2)*(1-6*x+2*x^2)], {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 24 2012 *)
  • PARI
    {a(n)=polcoeff(sum(m=0, n+1, x^m*sum(k=0, m, binomial(m, k)^2*x^k*(4-x)^(m-k) + x*O(x^n))), n)}
    for(n=0, 40, print1(a(n), ", "))

Formula

G.f.: A(x) = 1 / sqrt( (1 - 4*x + 2*x^2)^2 - 4*x^2 ).
G.f.: A(x) = 1 / sqrt( (1-2*x+2*x^2)*(1-6*x+2*x^2) ).
G.f. satisfies: A(x) = [1 + 2*x^2*Sum_{n>=0} A000108(n)*(-x*A(x))^(2*n)] / (1-4*x+2*x^2) where A000108(n) = binomial(2*n,n)/(n+1) forms the Catalan numbers.
Recurrence: n*a(n) = 4*(2*n-1)*a(n-1) - 16*(n-1)*a(n-2) + 8*(2*n-3)*a(n-3) - 4*(n-2)*a(n-4). - Vaclav Kotesovec, Oct 24 2012
a(n) ~ sqrt(147+56*sqrt(7))*(3+sqrt(7))^n/(14*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 24 2012

A217617 G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} binomial(n,k)^2 * x^k*(3-x)^(n-k).

Original entry on oeis.org

1, 3, 9, 33, 133, 549, 2295, 9711, 41505, 178749, 774387, 3370995, 14733043, 64608555, 284143257, 1252749777, 5535201733, 24503713893, 108659076723, 482566381299, 2146042722591, 9555487997247, 42594294578949, 190060286569677, 848858809506279, 3794468370955587
Offset: 0

Views

Author

Paul D. Hanna, Oct 09 2012

Keywords

Comments

Radius of convergence of g.f. is r = (5-sqrt(17))/4 = 0.21922359...
More generally, given
A(x) = Sum_{n>=1} x^n * Sum_{k=0..n} binomial(n,k)^2 * x^k*(t-x)^(n-k),
then A(x) = 1 / sqrt( (1 - t*x + 2*x^2)^2 - 4*x^2 )
and the radius of convergence r satisfies: (1-r)^2 = r*(t-r) for t>0.

Examples

			G.f.: A(x) = 1 + 3*x + 9*x^2 + 33*x^3 + 133*x^4 + 549*x^5 + 2295*x^6 +...
where the g.f. equals the series:
A(x) = 1 +
x*((3-x) + x) +
x^2*((3-x)^2 + 2^2*x*(3-x) + x^2) +
x^3*((3-x)^3 + 3^2*x*(3-x)^2 + 3^2*x^2*(3-x) + x^3) +
x^4*((3-x)^4 + 4^2*x*(3-x)^3 + 6^2*x^2*(3-x)^2 + 4^2*x^3*(3-x) + x^4) +
x^5*((3-x)^5 + 5^2*x*(3-x)^4 + 10^2*x^2*(3-x)^3 + 10^2*x^3*(3-x)^2 + 5^2*x^4*(3-x) + x^5) +...
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[1/Sqrt[(1-3*x+2*x^2)^2-4*x^2], {x, 0, 20}], x] (* Vaclav Kotesovec, Sep 16 2013 *)
  • PARI
    {a(n)=polcoeff(sum(m=0, n+1, x^m*sum(k=0, m, binomial(m, k)^2*x^k*(3-x)^(m-k) + x*O(x^n))), n)}
    for(n=0,40,print1(a(n),", "))

Formula

G.f.: A(x) = 1 / sqrt( (1 - 3*x + 2*x^2)^2 - 4*x^2 ).
G.f.: A(x) = 1 / sqrt( (1-x+2*x^2)*(1-5*x+2*x^2) ).
G.f. satisfies: A(x) = [1 + 2*x^2*Sum_{n>=0} A000108(n)*(-x*A(x))^(2*n)] / (1-3*x+2*x^2) where A000108(n) = binomial(2*n,n)/(n+1) forms the Catalan
numbers.
Recurrence: n*a(n) = 3*(2*n-1)*a(n-1) - 9*(n-1)*a(n-2) + 6*(2*n-3)*a(n-3) - 4*(n-2)*a(n-4). - Vaclav Kotesovec, Sep 16 2013
a(n) ~ 2*((5+sqrt(17))/2)^n/sqrt((42*sqrt(17)-170)*Pi*n). - Vaclav Kotesovec, Sep 16 2013
Showing 1-3 of 3 results.