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.

A107051 Numerators of coefficients that satisfy: 4^n = Sum_{k=0..n} c(k)*x^k for n>=0, where c(k) = a(k)/A107052(k).

Original entry on oeis.org

1, 3, 9, 5, 127, 124273, 385829, 70009765747, 220026935042111, 59574747365570286907, 113453152114585319883313, 4471148647570383262775217527741887
Offset: 0

Views

Author

Paul D. Hanna, May 10 2005

Keywords

Comments

Sum_{k>=0} a(k)/A107052(k) = 8.2025187671791748426202820386803825244610468145759213023...

Examples

			4^0 = 1;
4^1 = 1 + (3)*1;
4^2 = 1 + (3)*2 + (9/4)*2^2;
4^3 = 1 + (3)*3 + (9/4)*3^2 + (5/4)*3^3;
4^4 = 1 + (3)*4 + (9/4)*4^2 + (5/4)*4^3 + (127/256)*4^4.
Initial coefficients are:
A107051/A107052 = {1, 3, 9/4, 5/4, 127/256, 124273/800000,
385829/9600000, 70009765747/7906012800000,
220026935042111/129532113715200000, ...}.
		

Crossrefs

Programs

  • PARI
    {a(n)=numerator(sum(k=0,n,4^k*(matrix(n+1,n+1,r,c,if(r>=c,(r-1)^(c-1)))^-1)[n+1,k+1]))}

Formula

a(n)/A107052(n) = Sum_{k=0..n} T(n, k)*4^k where T(n, k) = A107045(n, k)/A107046(n, k) = [A079901^-1](n, k) (matrix inverse of A079901).