The speed with which a series can be developed can be accellerated by repeated application of the mapping:

- U(a,b)F ⇒ U(a,U
_{2})[F^{2}-2]

*base-2*accelleration (because it uses U2). Let's look at U(1,1)5 - the terius of 5 - for an example:

1 | 1 | 4 | 19 | 91 | 436 | 2089 | 10009 | 47956 | 229771 | 1100899 | 5274724 | 25272721 | 121088881 | 580171684 | 2779769539 | 13318676011 | 63813610516 | 305749376569 | 1464933272329 | 7018916985076 |

U_{0} | U_{1} | U_{2} | U_{3} | U_{4} | U_{5} | U_{6} | U_{7} | U_{8} | U_{9} | U_{10} | U_{11} | U_{12} | U_{13} | U_{14} | U_{15} | U_{16} | U_{17} | U_{18} | U_{19} | U_{20} |

After the mapping U(1,1)5 ⇒ U(1,4)23 we look at the latter:

1 | 4 | 91 | 2089 | 47956 | 1100899 | 25272721 | 580171684 | 13318676011 | 305749376569 | 7018916985076 |

U_{0} | U_{1} | U_{2} | U_{3} | U_{4} | U_{5} | U_{6} | U_{7} | U_{8} | U_{9} | U_{10} |

After the mapping U(1,4)23 ⇒ U(1,91)527 we look at the latter:

1 | 91 | 47956 | 25272721 | 13318676011 | 7018916985076 |

U_{0} | U_{1} | U_{2} | U_{3} | U_{4} | U_{5} |

After the mapping U(1,91)527 ⇒ U(1,47956)277727 we look at the latter:

1 | 47956 | 13318676011 |

U_{0} | U_{1} | U_{2} |

This, then, is the general idea base-2. It zaps down the parent series with steps that are powers of 2: 2, 4, 8, 16, ...

We can do it base-3, but that's just like base-2 really. It employs the mapping:

- U(a,b)F ⇒ U(a,U
_{3})[F^{3}-3F]

__Matrices of the Factor base-n__

| Of course we need the coefficients matrix of the factor F up to 'base-n' so here is how to go about that: In this matrix every column consists of the subsequent differences of the next column. Using this property the matrix can be extended indefinitely. The base defines the highest power of the factor F. Powers decrease with steps of 2. The general term of the polynom base-n could be constructed, but considering the size of the factors that can be reached by extending the matrix, this would seem a bit premature. |

| In a non-zero-series U(a,b,c)F the factor F follows the above matrix just as in a zero-series. The constant c develops as c*(matrix over 1), and that one looks like this: Note that each column consists if the subsequent differences, not so much of the next column, but of the one next to that. Using this property the matrix can be extended indefinitely. The power of the factor F is one less than the base. Powers decrease with steps of 1. |

I hate to work without examples, so let's have a look at U(1,1,2)3 and accelerate it base-5.

Here's the parent series:

1 | 1 | 4 | 13 | 37 | 100 | 265 | 697 | 1828 | 4789 | 12541 | 32836 | 85969 | 225073 | 589252 | 1542685 | 4038805 | 10573732 | 27682393 | 72473449 | 189737956 | 496740421 | 1300483309 | 3404709508 | 8913645217 | 23336226145 | 61095033220 | 159948873517 |

U_{0} | U_{1} | U_{2} | U_{3} | U_{4} | U_{5} | U_{6} | U_{7} | U_{8} | U_{9} | U_{10} | U_{11} | U_{12} | U_{13} | U_{14} | U_{15} | U_{16} | U_{17} | U_{18} | U_{19} | U_{20} | U_{21} | U_{22} | U_{23} | U_{24} | U_{25} | U_{26} | U_{27} |

^{5}- 5F

^{3}+ 5F.

The matrix over c renders c*(F

^{4}+ 2F

^{3}- F

^{2}- 2F + 1).

For F=3 and c=2 this adds up to a new factor 123 and a new constant 242 respectively.

It would seem that the mapping we're after is: U(1,1,2)3 ⇒ U(1,100,242)123

After this mapping we look at the latter again:

1 | 100 | 12541 | 1542685 | 189737956 | 23336226145 |

U_{0} | U_{1} | U_{2} | U_{3} | U_{4} | U_{5} |

Bingo!

Let's repeat the mapping base-5.

For F=123 and c=242 this adds up to a new factor 28143753123 and a new constant 56287506242 respectively.

It would seem that the mapping we're after is: U(1,100,242)123 ⇒ U(1,23336226145,56287506242)28143753123

After this mapping we look at the latter again:

1 | 23336226145 | 656768987503665507076 |

U_{0} | U_{1} | U_{2} |

How about that :)

U

_{1}and U

_{2}are of course U

_{25}and U

_{50}of the parent series. Repeat the base-5 mapping and U

_{1}and U

_{2}will be U

_{125}and U

_{250}, repeat it again and you'll get U

_{625}and U

_{1250}and so on.

An awesome tool!