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.

A105557 Row sums of triangle A105556, in which column n equals the row sums of A001263^n, which is the n-th matrix power of the Narayana triangle A001263.

Original entry on oeis.org

1, 2, 4, 10, 32, 128, 626, 3681, 25574, 206402, 1908996, 20024149, 236142157, 3106393358, 45265833590, 726249472784, 12761749378320, 244453274012442, 5082582988294164, 114258645210526486, 2767462674168199303
Offset: 0

Views

Author

Paul D. Hanna, Apr 14 2005

Keywords

Crossrefs

Programs

  • PARI
    {a(n)=local(N=matrix(n+1,n+1,m,j,if(m>=j, binomial(m-1,j-1)*binomial(m,j-1)/j))); sum(k=0,n,sum(j=0,n-k,(N^k)[n-k+1,j+1]))}

A105558 Central terms in even-indexed rows of triangle A105556 and thus equals the n-th row sum of the n-th matrix power of the Narayana triangle A001263.

Original entry on oeis.org

1, 2, 12, 148, 3105, 99156, 4481449, 272312216, 21414443481, 2116193061340, 256712977920256, 37506637787774112, 6496315164318118165, 1316230822119433518312, 308426950979497974254310
Offset: 0

Views

Author

Paul D. Hanna, Apr 14 2005

Keywords

Comments

Each term a(n) is divisible by (n+1) for all n>=0.

Examples

			Terms a(n) divided by (n+1) begin:
1,1,4,37,621,16526,640207,34039027,2379382609,211619306134,...
Contribution from _Paul D. Hanna_, Jan 31 2009: (Start)
G.f.: A(x) = 1 + 2*x + 12*x^2/3 + 148*x^3/18 + 3105*x^4/180 +...+ a(n)*x^n/[n!*(n+1)!/2^n] +...
G.f.: A(x) = d/dx x*F(x) where F(x) = B(x*F(x)) and:
F(x) = 1 + x + 4*x^2/3 + 37*x^3/18 + 621*x^4/180 + 16526*x^5/2700 +...+ A155926(n)*x^n/[n!*(n+1)!/2^n] +...
B(x) = 1 + x + x^2/3 + x^3/18 + x^4/180 +...+ x^n/[n!*(n+1)!/2^n] +... (End)
		

Crossrefs

Programs

  • PARI
    a(n)=local(N=matrix(n+1,n+1,m,j,if(m>=j, binomial(m-1,j-1)*binomial(m,j-1)/j))); sum(j=0,n,(N^n)[n+1,j+1])
    for(n=0,20,print1(a(n),", "))
    
  • PARI
    a(n)=local(F=sum(k=0,n,x^k/(k!*(k+1)!/2^k))+x*O(x^n));polcoeff(deriv(serreverse(x/F)),n)*n!*(n+1)!/2^n
    for(n=0,20,print1(a(n),", ")) \\ Paul D. Hanna, Jan 31 2009

Formula

Contribution from Paul D. Hanna, Jan 31 2009: (Start)
a(n) = (n+1)*A155926(n) for n>=0.
G.f.: A(x) = d/dx x*F(x) where F(x) = B(x*F(x)) and F(x) = Sum_{n>=0} A155926(n)*x^n/[n!*(n+1)!/2^n] with B(x) = Sum_{n>=0} x^n/[n!*(n+1)!/2^n] and A(x) = Sum_{n>=0} a(n)*x^n/[n!*(n+1)!/2^n]. (End)

A155926 G.f. satisfies: A(x) = B(x*A(x)) where A(x) = Sum_{n>=0} a(n)*x^n/[n!*(n+1)!/2^n] and B(x) = Sum_{n>=0} x^n/[n!*(n+1)!/2^n].

Original entry on oeis.org

1, 1, 4, 37, 621, 16526, 640207, 34039027, 2379382609, 211619306134, 23337543447296, 3125553148981176, 499716551101393705, 94016487294245251308, 20561796731966531616954, 5172827581575899147920471
Offset: 0

Views

Author

Paul D. Hanna, Jan 30 2009

Keywords

Examples

			G.f.: A(x) = 1 + x + 4*x^2/3 + 37*x^3/18 + 621*x^4/180 + 16526*x^5/2700 +...+ a(n)*x^n/[n!*(n+1)!/2^n] +...
B(x) = 1 + x + 1/3*x^2 + 1/18*x^3 + 1/180*x^4 +...+ x^n/[n!*(n+1)!/2^n] +... where
A(x) = B(x*A(x)) and B(x) = A(x/B(x)) ;
1/B(x) = 1 - x + 2*x^2/3 - 7*x^3/18 + 39*x^4/180 - 321*x^5/2700 +...+ (-1)^n*A103365(n)*x^n/[n!*(n+1)!/2^n] +...
Also, A(x) = C(x*A(x)^2) where:
C(x) = 1 + x - 2*x^2/3 + 19*x^3/18 - 379*x^4/180 + 12726*x^5/2700 +...+ A155927(n)*x^(n+1)/[n!*(n+1)!/2^n] +...
A(x)^2 = 1 + 2*x + 11*x^2/3 + 122*x^3/18 + 2302*x^4/180 + 66482*x^5/2700 +...
		

Crossrefs

Programs

  • PARI
    {a(n)=local(F=sum(k=0,n,x^k/(k!*(k+1)!/2^k))+x*O(x^n));polcoeff(serreverse(x/F)/x,n)*n!*(n+1)!/2^n}
    
  • PARI
    {a(n)=local(N=matrix(n+1, n+1, m, j, if(m>=j, binomial(m-1, j-1)*binomial(m, j-1)/j))); sum(j=0, n, (N^n)[n+1, j+1])/(n+1)}

Formula

a(n) = A105558(n)/(n+1) = A105556(2n,n)/(n+1) = [N^(n+1)](n+1,1)/(n+1) for n>=0, where N^(n+1) is the (n+1)-th matrix power of the Narayana triangle N=A001263.
G.f.: A(x) = Series_Reversion[x/B(x)]/x where B(x) = A(x/B(x)) = Sum_{n>=0} x^n/[n!*(n+1)!/2^n].
G.f. satisfies: A(x) = C(x*A(x)^2) and C(x) = A(x/C(x)^2) where C(x) is the g.f. of A155927.
Showing 1-3 of 3 results.