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1973年
斐波那契三乘积三角形:c(n)=如果[n==0,1,乘积[Fibonacci[(i-1)]*Fibonaci[i]*Fiponacci[i+1],{i,2,n}];t(n,m)=c(n)/(c(m)*c(n-m))
0
1, 1, 1, 1, 2, 1, 1, 6, 6, 1, 1, 30, 90, 30, 1, 1, 120, 1800, 1800, 120, 1, 1, 520, 31200, 156000, 31200, 520, 1, 1, 2184, 567840, 11356800, 11356800, 567840, 2184, 1, 1, 9282, 10135944, 878448480, 3513793920, 878448480, 10135944, 9282, 1, 1, 39270
抵消
0,5
评论
行总和为:
{1, 2, 4, 14, 152, 3842, 219442, 23853650, 5290981334, 2432822210282,
2340979254062902,...}.
配方奶粉
c(n)=如果[n==0,1,乘积[Fibonacci[(i-1)]*Fibonaci[i]*Fiponacci[i+1],{i,2,n}];
t(n,m)=c(n)/(c(m)*c(n-m))
例子
{1},
{1, 1},
{1, 2, 1},
{1, 6, 6, 1},
{1, 30, 90, 30, 1},
{1, 120, 1800, 1800, 120, 1},
{1, 520, 31200, 156000, 31200, 520, 1},
{1, 2184, 567840, 11356800, 11356800, 567840, 2184, 1},
{1, 9282, 10135944, 878448480, 3513793920, 878448480, 10135944, 9282, 1},
{1, 39270, 182252070, 66339753480, 1149889060320, 1149889060320, 66339753480, 182252070, 39270, 1},
{1, 166430, 3267853050, 5055368668350, 368030839055880, 1594800302575480, 368030839055880, 5055368668350, 3267853050, 166430, 1}
数学
c[n_]=如果[n==0,1,乘积[Fibonacci[(i-1)]*Fibonaci[i]*Fiponacci[i+1],{i,2,n}];
t[n,m]=c[n]/(c[m]*c[n-m]);
表[表[t[n,m],{m,0,n}],{n,0,10}];
压扁[%]
关键词
非n,,未经编辑的
作者
罗杰·L·巴古拉2010年3月1日
状态
经核准的