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A171636号 |
| 按行读取的表。n=m和降次幂的Lommel多项式L(n,m,z)=(Gamma(n+m)/(Gamma(n)*(z/2)^m))*超几何([(1-m)/2,-m/2],[n,-m,1-n-m],z^2)的系数。 |
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2, 24, 0, 1, 480, 0, 16, 13440, 0, 360, 0, 1, 483840, 0, 10752, 0, 42, 21288960, 0, 403200, 0, 1728, 0, 1, 1107025920, 0, 18247680, 0, 79200, 0, 80, 66421555200, 0, 968647680, 0, 4118400, 0, 5280, 0, 1, 4516665753600, 0, 59041382400, 0, 242161920
(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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1,1
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评论
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Lommel多项式是有理函数,而不是多项式。
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链接
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例子
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{2},
{24, 0, 1},
{480, 0, 16},
{13440, 0, 360, 0, 1},
{483840, 0, 10752, 0, 42},
{21288960, 0, 403200, 0, 1728, 0, 1},
{1107025920, 0, 18247680, 0, 79200, 0, 80},
{66421555200, 0, 968647680, 0, 4118400, 0, 5280, 0, 1},
{4516665753600, 0, 59041382400, 0, 242161920, 0, 349440, 0, 130},
{343266597273600, 0, 4064999178240, 0, 15968010240, 0, 24460800, 0, 12600, 0, 1}
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MAPLE公司
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L:=(n,m,z)->(γ(n+m)/(γ(n)*(z/2)^m))*超几何([(1-m)/2,-m/2],
[n,-m,1-n-m],z^2);
对于从1到10的n,做L(n,n,1/z):转换(级数(%,z,12),多项式):
lprint(seq(系数(展开(%),z,n-k),k=0。。n-爱尔兰(n,2)):od:
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数学
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Lommel[m_,n_,z_]:=(Gamma[n+m]/(Gamma[n]((z/2))^m))超几何PFQ[{((1-m))/2,(-m)/2},{n,(/m),1-n-m},z^2]
表[系数列表[展开[Lommel[n,n,x]*x^n],x],{n,1,10}]
压扁[%]
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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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