■n次元の立方体と直角三角錐(その349)

 (その344)の計算を間違いも発見.s=o

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  Gs(n+1,s+1)=Go(n,s+1)2^s+1

 そこで,Gkの最大値は(k+1)!,最小値はk+1であるから,

  0≦s+1≦Gs≦(s+1)!

  0≦s+1≦Go≦(s+1)!

  1/s!≦Gs/Go≦s!

  Gs/Go=(n,s+1)2^s+1/(n+1,s+1)

 =(n−s)2^s+1/(n+1)

[1](n−s)2^s+1/(n+1)≧1/s!

  s!2^s+1≧(n+1)/(n−s)

  n(s!2^s+1−1)≧s!2^s+1s+1

[2](n−s)2^s+1/(n+1)≦s!

  2^s+1/s!≦(n+1)/(n−s)

  n(2^s+1/s!−1)≦2^s+1/(s−1)!+1

  n(2^s+1−s!)≦2^s+1s+s!

[3]0≦s≦n−1

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【1】n(s!2^s+1−1)≧s!2^s+1s+1

[1]s=1のとき,n≧2

[2]s=2のとき,n≧3

[3]s=3のとき,n≧4

[4]s=4のとき,n≧5

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【2】n(2^s+1−s!)≦2^s+1s+s!

[1]s=1のとき,n≦1

[2]s=2のとき,n≦3

[3]s=3のとき,n≦5

[4]s=4のとき,n≦19

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