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.

A227845 G.f.: Sum_{n>=0} x^n / (1-x)^(2*n+1) * [ Sum_{k=0..n} C(n,k)^2*x^k ]^2.

Original entry on oeis.org

1, 2, 7, 28, 125, 590, 2891, 14536, 74497, 387450, 2038743, 10830148, 57986773, 312542678, 1694166275, 9228580464, 50486521785, 277239830210, 1527533993871, 8441627856300, 46776754474709, 259830443968046, 1446468759734131, 8068688342238328, 45091854560015025, 252423540736438890
Offset: 0

Views

Author

Paul D. Hanna, Aug 01 2013

Keywords

Comments

Equals antidiagonal sums of table A143007.

Examples

			G.f.: A(x) = 1 + 2*x + 7*x^2 + 28*x^3 + 125*x^4 + 590*x^5 + 2891*x^6 +...
where
A(x) = 1/(1-x) + x/(1-x)^3 * (1+x)^2 + x^2/(1-x)^5*(1 + 2^2*x + x^2)^2
+ x^3/(1-x)^7 * (1 + 3^2*x + 3^2*x^2 + x^3)^2
+ x^4/(1-x)^9 * (1 + 4^2*x + 6^2*x^2 + 4^2*x^3 + x^4)^2
+ x^5/(1-x)^11 * (1 + 5^2*x + 10^2*x^2 + 10^2*x^3 + 5^2*x^4 + x^5)^2
+ x^6/(1-x)^13 * (1 + 6^2*x + 15^2*x^2 + 20^2*x^3 + 15^2*x^4 + 6^2*x^5 + x^6)^2 +...
We can also express the g.f. by the binomial series identity:
A(x) = 1 + x*(1 + (1+x)) + x^2*(1 + 2^2*(1+x) + (1+2^2*x+x^2))
+ x^3*(1 + 3^2*(1+x) + 3^2*(1+2^2*x+x^2) + (1+3^2*x+3^2*x^2+x^3))
+ x^4*(1 + 4^2*(1+x) + 6^2*(1+2^2*x+x^2) + 4^2*(1+3^2*x+3^2*x^2+x^3) + (1+4^2*x+6^2*x^2+4^2*x^3+x^4))
+ x^5*(1 + 5^2*(1+x) + 10^2*(1+2^2*x+x^2) + 10^2*(1+3^2*x+3^2*x^2+x^3) + 5^2*(1+4^2*x+6^2*x^2+4^2*x^3+x^4) + (1+5^2*x+10^2*x^2+10^2*x^3+5^2*x^4+x^5)) +...
The square-root of the g.f. is an integer series:
A(x)^(1/2) = 1 + x + 3*x^2 + 11*x^3 + 47*x^4 + 215*x^5 + 1029*x^6 +...+ A227846(n)*x^n +...
The g.f. also satisfies A(x) = F(x*A(x))^2 and F(x)^2 = A(x/F(x)^2) where
F(x) = 1 + x + x^2 + x^4 - 2*x^5 - 4*x^6 - 7*x^8 + 20*x^9 + 42*x^10 + 84*x^12 - 272*x^13 - 584*x^14 - 1239*x^16 +...+ A258053(n)*x^n +...
such that A258053(4*n+3) = 0 for n>=0.
		

Crossrefs

Programs

  • Maple
    U := proc(n) options remember;
            if n < 1 then 1
            elif n = 1 then 2
            elif n = 2 then 7/2
            else
                    (2*(3*n^2-3*n+1)*U(n-2) - (n-1)^2*U(n-4))/n^2
            fi
    end:
    seq(U(n)*U(n-1), n=0..25); # Mark van Hoeij, Jul 10 2024
  • Mathematica
    Table[Sum[Sum[Binomial[n-k,j]^2*Binomial[j,k]^2,{j,k,n-k}],{k,0,Floor[n/2]}],{n,0,20}] (* Vaclav Kotesovec, Jul 05 2014 *)
  • PARI
    /* From definition: */
    {a(n)=local(A=1); A=sum(m=0, n, x^m/(1-x)^(2*m+1)*sum(k=0, m, binomial(m, k)^2*x^k)^2+x*O(x^n)); polcoeff(A, n)}
    for(n=0,30,print1(a(n),", "))
    
  • PARI
    /* From alternate g.f.: */
    {a(n)=polcoeff(sum(m=0,n,x^m*sum(k=0,m,binomial(m,k)^2*sum(j=0,k,binomial(k,j)^2*x^j)+x*O(x^n))),n)}
    for(n=0, 30, print1(a(n), ", "))
    
  • PARI
    /* From formula for a(n): */
    {a(n)=sum(k=0,n\2,sum(j=k,n-k,binomial(n-k,j)^2*binomial(j,k)^2))}
    for(n=0,30,print1(a(n),", "))
    
  • PARI
    /* From g.f.: 1/AGM((1+x)^2, 1-6*x+x^2) */
    {a(n)=local(A);A = 1 / agm((1+x)^2, 1-6*x+x^2 +x*O(x^n));polcoeff(A,n)}
    for(n=0,30,print1(a(n),", "))

Formula

G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} binomial(n,k)^2 * Sum_{j=0..k} binomial(k,j)^2 * x^j.
a(n) = Sum_{k=0..[n/2]} Sum_{j=k..n-k} binomial(n-k,j)^2 * binomial(j,k)^2.
Recurrence: n^2*a(n) = 2*(3*n^2 - 3*n + 1)*a(n-1) - 2*(3*n^2 - 9*n + 7)*a(n-3) + (n-2)^2*a(n-4). - Vaclav Kotesovec, Jul 05 2014
a(n) ~ (3+2*sqrt(2))^(n+1) / (4*Pi*n). - Vaclav Kotesovec, Jul 05 2014
G.f.: 1 / AGM((1+x)^2, 1 - 6*x + x^2), where AGM(x,y) = AGM((x+y)/2,sqrt(x*y)) denotes the arithmetic-geometric mean. - Paul D. Hanna, Jul 31 2014
G.f. satisfies: A(x) = F(x*A(x))^2, where F(x) is the g.f. of A258053. - Paul D. Hanna, May 17 2015
G.f.: hypergeom([1/2, 1/2], [1], -16*x^2/((x+1)^2*(x^2-6*x+1)))/((x+1)*sqrt(x^2-6*x+1)). - Mark van Hoeij, Jul 08 2024
a(n) = U(n)*U(n-1) where the sequences U(-1),U(1),U(3),... and U(0),U(2),U(4),... satisfy a second order recurrence n^2*U(n) = 2*(3*n^2-3*n+1)*U(n-2) - (n-1)^2*U(n-4) with initial terms U(-1), U(1)=2 and U(0)=1, U(2)=7/2. - Mark van Hoeij, Jul 10 2024

Extensions

Name changed by Paul D. Hanna, Sep 07 2014