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.

A107049 Numerators of coefficients that satisfy: 3^n = Sum_{k=0..n} c(k)*x^k for n>=0, where c(k) = a(k)/A107050(k).

Original entry on oeis.org

1, 2, 1, 11, 101, 71723, 1462111, 194269981673, 224103520039487, 14876670160046176873, 20871062802926443547323, 606768727432357137728440774281877, 97827345788163051844748893917483101
Offset: 0

Views

Author

Paul D. Hanna, May 10 2005

Keywords

Comments

Sum_{k>=0} a(k)/A107050(k) = 4.5568226185870666883519278484116281050682807568451524897...

Examples

			3^0 = 1;
3^1 = 1 + (2)*1;
3^2 = 1 + (2)*2 + (1)*2^2;
3^3 = 1 + (2)*3 + (1)*3^2 + (11/27)*3^3;
3^4 = 1 + (2)*4 + (1)*4^2 + (11/27)*4^3 + (101/864)*4^4.
Initial coefficients are:
A107049/A107050 = {1, 2, 1, 11/27, 101/864, 71723/2700000,
1462111/291600000, 194269981673/240145138800000,
224103520039487/1967268977049600000, ...}.
		

Crossrefs

Programs

  • PARI
    {a(n)=numerator(sum(k=0,n,3^k*(matrix(n+1,n+1,r,c,if(r>=c,(r-1)^(c-1)))^-1)[n+1,k+1]))}

Formula

a(n)/A107050(n) = Sum_{k=0..n} T(n, k)*3^k where T(n, k) = A107045(n, k)/A107046(n, k) = [A079901^-1](n, k) (matrix inverse of A079901).