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 1-2 of 2 results.

A220267 Main diagonal of triangle A220265.

Original entry on oeis.org

1, 9, 177, 9800, 1131750, 225598527, 69153712446, 30211650109440, 17832410391617082, 13670065258130703125, 13203133188522251881137, 15685246720965582029887488, 22477297594725738707224270105, 38231902029181930196176183861755, 76144787589417130318451646093750000
Offset: 1

Views

Author

Paul D. Hanna, Dec 09 2012

Keywords

Comments

G.f. of row n of triangle A220265 equals: Sum_{k=0..n^2-n+1} A220265(n,k)*y^k = (2*(1+y)^n - 1)*((1+y)^n - 1)^(n-1)/y^(n-1) for n>=1.

Examples

			 Irregular triangle A220265 begins:
1, 2;
2, 9, 8, 2;
9, 72, 177, 222, 163, 72, 18, 2;
64, 800, 3696, 9800, 17408, 22284, 21340, 15554, 8652, 3633, 1120, 240, 32, 2;
625, 11250, 82500, 365000, 1131750, 2654250, 4922750, 7425000, 9274150, 9704600, 8566200, 6398000, 4042345, 2152890, 959690, 354020, 106251, 25300, 4600, 600, 50, 2; ...
		

Crossrefs

Programs

  • PARI
    {a(n)=polcoeff((2*(1+x)^n-1)*((1+x)^n-1)^(n-1)/x^(n-1), n-1)}
    for(n=1, 20, print1(a(n), ", "))

Formula

a(n) = A220265(n,n-1) for n>=1.

A220266 G.f.: Sum_{n>=1} (2*(1+x)^n - 1) * ((1+x)^n - 1)^(n-1).

Original entry on oeis.org

1, 4, 18, 144, 1604, 22944, 400624, 8259680, 196358760, 5287879092, 159094582274, 5288950560768, 192527721428892, 7616404083126180, 325361411700398046, 14926683772801407168, 731947910056020737036, 38204289826040411251632, 2114787166947079113869760
Offset: 0

Views

Author

Paul D. Hanna, Dec 09 2012

Keywords

Comments

Compare the g.f. of this sequence to the identity (when G(x) = 1+x):
1 = Sum_{n>=1} (2*G(x)^n - 1) * (1 - G(x)^n)^(n-1) for all G(x) such that G(0)=1.

Examples

			G.f.: A(x) = 1 + 4*x + 18*x^2 + 144*x^3 + 1604*x^4 + 22944*x^5 +...
where
A(x) = (1+2*x) + (1+4*x+2*x^2)*(2*x+x^2) + (1+6*x+6*x^2+2*x^3)*(3*x+3*x^2+x^3)^2 + (1+8*x+12*x^2+8*x^3+2*x^4)*(4*x+6*x^2+4*x^3+x^4)^3 +...
Compare the g.f. to the identity:
1 = (1+2*x) - (1+4*x+2*x^2)*(2*x+x^2) + (1+6*x+6*x^2+2*x^3)*(3*x+3*x^2+x^3)^2 - (1+8*x+12*x^2+8*x^3+2*x^4)*(4*x+6*x^2+4*x^3+x^4)^3 +-...
		

Crossrefs

Programs

  • PARI
    {a(n)=polcoeff(sum(m=1,n+1,(2*(1+x)^m - 1) * ((1+x)^m - 1 +x*O(x^n))^(m-1)),n)}
    for(n=0,20,print1(a(n),", "))
    
  • PARI
    /* As Row Sums of Triangle A220265: */
    {A220265(n,k)=polcoeff((2*(1+x)^n-1)*((1+x)^n-1)^(n-1)/x^(n-1),k)}
    {a(n)=sum(k=0,n,A220265(n-k+1,k))}
    for(n=0,20,print1(a(n),", "))
    
  • PARI
    {a(n)=polcoeff(1+sum(m=1,n\2+1,2*(2*(1+x)^(2*m) - 1) * ((1+x)^(2*m) - 1 +x*O(x^n))^(2*m-1)),n)}
    for(n=0,20,print1(a(n),", "))
    
  • PARI
    {a(n)=polcoeff(-1+sum(m=0,n\2,2*(2*(1+x)^(2*m+1) - 1) * ((1+x)^(2*m+1) - 1 +x*O(x^n))^(2*m)),n)}
    for(n=0,20,print1(a(n),", "))

Formula

Equals the antidiagonal sums of triangle A220265:
a(n) = Sum_{k=0..n} A220265(n-k+1,k) for n>=0.
G.f.: 1 + Sum_{n>=1} 2*(2*(1+x)^(2*n) - 1) * ((1+x)^(2*n) - 1)^(2*n-1).
G.f.: -1 + Sum_{n>=0} 2*(2*(1+x)^(2*n+1) - 1) * ((1+x)^(2*n+1) - 1)^(2*n).
Showing 1-2 of 2 results.