■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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