■単純リー環を使った面数数え上げ(その22)

 パスカルの三角形に似た漸化式,たとえば

  fj^(n)=fj^(n-1)+fj-1^(n-1)

は成立しないだろうか?

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【1】Anのボロノイ細胞の要素数

  fk^(n)=2(2^k+1−1)(n+1,k+2)

  fk^(n-1)=2(2^k+1−1)(n,k+2)

  fk-1^(n-1)=2(2^k−1)(n,k+1)

  (n,k+2)+(n,k+1)=(n+1,k+2)

  fk^(n)=2(2^k+1−1)(n+1,k+2)

=2(2^k+1−1){(n,k+2)+(n,k+1)}

=2(2^k+1−1)(n,k+2)+2(2^k+1−1)(n,k+1)

  2(2^k+1−1)=4(2^k−1)+2

  fk^(n)=2(2^k+1−1)(n+1,k+2)

=fk^(n-1)+2fk-1^(n-1)+2(n,k+1)

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【2】Cnのボロノイ細胞の要素数

  fk^(n)=2^k+1(n,k+1)+2^k+2(n−k−1)(n,k+1)

=2^k+1(2n−2k−1)(n,k+1)

  fk^(n-1)=2^k+1(2n−2k−3)(n−1,k+1)

  fk-1^(n-1)=2^k(2n−2k−1)(n−1,k)

  (n−1,k+1)+(n−1,k)=(n,k+1)

  fk^(n)=2^k+1(2n−2k−1)(n,k+1)

=2^k+1(2n−2k−1){(n−1,k+1)+(n−1,k)}

=2^k+1(2n−2k−1)(n−1,k+1)+2^k+1(2n−2k−1)(n−1,k)

  2^k+1(2n−2k−1)=2^k+1(2n−2k−3)+2^k+2

  fk^(n)=2^k+1(2n−2k−3)(n−1,k+1)+2^k+2(n−1,k+1)+2^k+1(2n−2k−1)(n−1,k)

=fk^(n-1)+2^k+2(n−1,k+1)+2fk-1^(n-1)

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【3】まとめ

 どちらも,余り実用的ではないことがわかる.

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