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A173006号 |
| 基于递归的乘积三角序列:A=5;f(n,a)=(2*a+1)*f(n-1,a)+f(n-2,a) |
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1, 1, 1, 1, 11, 1, 1, 120, 120, 1, 1, 1309, 14280, 1309, 1, 1, 14279, 1699201, 1699201, 14279, 1, 1, 155760, 202190640, 2205562898, 202190640, 155760, 1, 1, 1699081, 24058986960, 2862818956682, 2862818956682, 24058986960, 1699081, 1, 1
(列表;桌子;图表;参考;听;历史;文本;内部格式)
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抵消
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0,5
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评论
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行总和为:
{1, 2, 13, 242, 16900, 3426962, 2610255700, 5773759285448, 47972252879976100,
1157507562695117906888, 104909162208463229766370000,...}.
这个结果似乎将这些新的递归直接连接到q形式。
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链接
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配方奶粉
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a=5;f(n,a)=(2*a+1)*f(n-1,a)+f(n-2,a);
c(n)=如果[n==0,1,乘积[f(i,a),{i,1,n}]];
t(n,m)=c(n)/(c(m)*c(n-m)
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例子
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{1},
{1, 1},
{1, 11, 1},
{1, 120, 120, 1},
{1, 1309, 14280, 1309, 1},
{1, 14279, 1699201, 1699201, 14279, 1},
{1, 155760, 202190640, 2205562898, 202190640, 155760, 1},
{1, 1699081, 24058986960, 2862818956682, 2862818956682, 24058986960, 1699081, 1},
{1, 18534131, 2862817257601, 3715936800366098, 40534653607660438, 3715936800366098, 2862817257601, 18534131, 1},
{1, 202176360, 340651194667560, 4823283104057937603, 573930157592104171920, 573930157592104171920, 4823283104057937603, 340651194667560, 202176360, 1},
{1,2205405829,405346293348182040,6260617753130421176727,8126277060814812179812443,88644086770258081457215920,8126277060814812179812443,6260617753130421176727,405346293348182040,2205405829,1}
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数学
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清除[f,c,a,t];
f[0,a_]:=0;f[1,a_]:=1;
f[n,a_]:=f[n、a]=(2*a+1)*f[n-1,a]-f[n-2,a];
c[n_,a_]:=如果[n==0,1,乘积[f[i,a],{i,1,n}]];
t[n,m,a]:=c[n,a]/(c[m,a]*c[n-m,a〕);
表格[扁平[表格[表格[t[n,m,a],{m,0,n}],{n,0,10}]],{a,1,10}]
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交叉参考
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关键词
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作者
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状态
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经核准的
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