王守恩 发表于 2025-5-21 13:15:31

用 x,y,z 来表示无穷级数的通用公式

$\sum _{k=1}^{\infty}{(k+x)(k+y)(k+z)}/{k!}=A*e-B$, x,y,z=正整数。譬如。
x,y,z=1,1,1=15e-1,
x,y,z=1,1,2=20e-2,
x,y,z=1,1,3=25e-3,
x,y,z=1,3,5=61e-15,
x,y,z=1,4,6=85e-24,
x,y,z=2,4,6=121e-48,
x,y,z=3,5,7=211e-105,
x,y,z=6,6,6=365e-216,
x,y,z=7,8,9=748e-504,
......
怎么用 x,y,z 来表示无穷级数的通用公式?谢谢!

iseemu2009 发表于 2025-5-21 13:54:18

这题好难,能有通项公式吗,怀疑。

mathe 发表于 2025-5-21 14:48:03

\(\frac{(k+x)(k+y)(k+z)}{k!}=\frac{k(k-1)(k-2)+(x+y+z+3)k(k-1)+(xy+yz+zx+x+y+z+1)k+(xyz)}{k!}\)
所以答案就应该是\((1+(x+y+z+3)+(xy+yz+zx+x+y+z+1)+xyz)e-xyz\)

王守恩 发表于 2025-5-21 16:12:25

一次到位!!!别指望我,我是想不好的。

$\sum _{k=s}^{\infty}{(k+x)(k+y)(k+z)}/{k!}=A*e-B$,   s,x,y,z=正整数。

怎么用 x,y,z 来表示A,怎么用 x,y,z 来表示B?谢谢!

northwolves 发表于 2025-5-21 21:56:58

$A=(x + 1) (y + 1) (z + 1) + (x + y + z) + 4$
$B=\frac{\lfloor e s!\rfloor -( (s+x+y+z)(s+1)+xy+xz+yz+3)}{(s-1)!}$

王守恩 发表于 2025-5-22 06:34:35

这题目还得改!!!

$\sum _{k=R}^{\infty}{(k+a1)(k+a2)(k+a3)(k+a4)(k+a5)cdotscdots}/{k!}=A*e-B$,R,a1,a2,a3,a4,a5,......=正整数。

怎么用 a1,a2,a3,a4,a5,...... 来表示A,怎么用 a1,a2,a3,a4,a5,...... 来表示B?谢谢!

wayne 发表于 2025-5-22 13:44:14

要引入BellB数.$\sum _{k=0}^{\infty } \frac{k^n}{k!} = e*BellB(n)$,https://oeis.org/A000110

王守恩 发表于 2025-5-27 05:24:14

接6楼。

1, $\sum _{k=R}^{\infty}{k+a1}/{k!}=A*e-B$

$B=\sum _{k=0}^{R-1}{k+a1}/{k!}$

$A=1+a1$

2, $\sum _{k=R}^{\infty}{(k+a1)(k+a2)}/{k!}=A*e-B$

$B=\sum _{k=0}^{R-1}{(k+a1)(k+a2)}/{k!}$

$A=2+(a1+a2)+a1a2$

3, $\sum _{k=R}^{\infty}{(k+a1)(k+a2)(k+a3)}/{k!}=A*e-B$

$B=\sum _{k=0}^{R-1}{(k+a1)(k+a2)(k+a3)}/{k!}$

$A=5+2(a1+a2+a3)+(a1a2+a1a3+a2a3)+a1a2a3$

4, $\sum _{k=R}^{\infty}{(k+a1)(k+a2)(k+a3)(k+a4)}/{k!}=A*e-B$

$B=\sum _{k=0}^{R-1}{(k+a1)(k+a2)(k+a3)(k+a4)}/{k!}$

$A=15+5(a1+a2+a3+a4)+2(a1a2+a1a3+a1a4+a2a3+a2a4+a3a4)+(a1a2a3+a1a2a4+a1a3a4+a2a3a4)+a1a2a3a4$

5, $\sum _{k=R}^{\infty}{(k+a1)(k+a2)(k+a3)(k+a4)(k+a5)}/{k!}=A*e-B$

$B=\sum _{k=0}^{R-1}{(k+a1)(k+a2)(k+a3)(k+a4)(k+a5)}/{k!}$

$A=52 + 15 (a1 + a2 + a3 + a4 + a5) + 5 (a1 a2 + a1 a3 + a1 a4 + a1 a5 + a2 a3 + a2 a4 + a2 a5 + a3 a4 + a3 a5 + a4 a5)
+ 2 (a1 a2 a3 + a1 a2 a4 + a1 a2 a5 + a1 a3 a4 + a1 a3 a5 + a1 a4 a5 + a2 a3 a4 + a2 a3 a5 + a2 a4 a5 + a3 a4 a5) + (a1 a2 a3 a4 + a1 a2 a3 a5 + a1 a2 a4 a5 + a1 a3 a4 a5 + a2 a3 a4 a5) + a1 a2 a3 a4 a5$

6, $\sum _{k=R}^{\infty}{(k+a1)(k+a2)(k+a3)(k+a4)(k+a5)(k+a6)}/{k!}=A*e-B$

$B=\sum _{k=0}^{R-1}{(k+a1)(k+a2)(k+a3)(k+a4)(k+a5)(k+a6)}/{k!}$

$A=203+52(a1 + a2 + a3 + a4 + a5 + a6) + 15 (a1 a2 + a1 a3 + a1 a4 + a1 a5 + a1 a6 + a2 a3 + a2 a4 + a2 a5 + a2 a6 + a3 a4 + a3 a5 + a3 a6 + a4 a5 + a4 a6+a5a6)+5(a1a2a3+a1a2a4$ $+a1a2a5+a1a2a6+a1a3a4+a1a3a5+a1a3a6+a1a4a5+a1a4a6+a1a5a6+a2a3a4+a2a3a5+a2a3a6+a2a4a5+a2a4a6+a2a5a6+a3a4a5+a3a4a6+a3a5a6+a4a5a6) +$
$2 (a1 a2 a3 a4 + a1 a2 a3 a5 + a1 a2 a3 a6 + a1 a2 a4 a5 + a1 a2 a4 a6 + a1 a2 a5 a6 + a1 a3 a4 a5 + a1 a3 a4 a6 + a1 a3 a5 a6 + a1 a4 a5 a6 + a2 a3 a4 a5 + a2 a3 a4 a6 + a2 a3 a5 a6 + a2 a4 a5 a6 + a3 a4 a5 a6)$
$ +(a1 a2 a3 a4 a5 + a1 a2 a3 a4 a6 + a1 a2 a3 a5 a6 + a1 a2 a4 a5 a6 + a1 a3 a4 a5 a6 + a2 a3 a4 a5 a6) + a1 a2 a3 a4 a5 a6$

$A=203+52S1+15S2+5S3+2S4+S5+S6$

7, $\sum _{k=R}^{\infty}{(k+a1)(k+a2)(k+a3)(k+a4)(k+a5)(k+a6)(k+a7)}/{k!}=A*e-B$

$B=\sum _{k=0}^{R-1}{(k+a1)(k+a2)(k+a3)(k+a4)(k+a5)(k+a6)(k+a7)}/{k!}$

$A=877++203S1+52S2+15S3+5S4+2S5+S6+S72$

8, $\sum _{k=R}^{\infty}{(k+a1)(k+a2)(k+a3)(k+a4)(k+a5)(k+a6)(k+a7)(k+a8)}/{k!}=A*e-B$

$B=\sum _{k=0}^{R-1}{(k+a1)(k+a2)(k+a3)(k+a4)(k+a5)(k+a6)(k+a7)(k+a8)}/{k!}$

$A=4140+877S1+203S2+52S3+15S4+5S5+2S6+S7+S8$

9, $\sum _{k=R}^{\infty}{(k+a1)(k+a2)(k+a3)(k+a4)(k+a5)(k+a6)(k+a7)(k+a8)(k+a9)}/{k!}=A*e-B$

$B=\sum _{k=0}^{R-1}{(k+a1)(k+a2)(k+a3)(k+a4)(k+a5)(k+a6)(k+a7)(k+a8)(k+a9)}/{k!}$

$A=21147+4140S1+877S2+203S3+52S4+15S5+5S6+2S7+S8+S9$

{1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, 4213597, 27644437}

王守恩 发表于 2025-5-28 10:07:21

这个B不好搞。我们特别地拉一串数出来熟悉熟悉。

$\sum _{k=1}^{\infty}{k+1}/{k!}=1$

$\sum _{k=2}^{\infty}{k+1}/{k!}=3$

$\sum _{k=3}^{\infty}{(k+1)(k+2)}/{k!}=14$

$\sum _{k=4}^{\infty}{(k+1)(k+2)(k+3)}/{k!}=80$

$\sum _{k=5}^{\infty}{(k+1)(k+2)(k+3)(k+4)}/{k!}=534$

$\sum _{k=6}^{\infty}{(k+1)(k+2)(k+3)(k+4)(k+5)}/{k!}=4102$

$\sum _{k=7}^{\infty}{(k+1)(k+2)(k+3)(k+4)(k+5)(k+6)(k+7)}/{k!}=35916$

$\sum _{k=8}^{\infty}{(k+1)(k+2)(k+3)(k+4)(k+5)(k+6)(k+7)(k+8)}/{k!}=354888$

$\sum _{k=9}^{\infty}{(k+1)(k+2)(k+3)(k+4)(k+5)(k+6)(k+7)(k+8)(k+9)}/{k!}=3915750$

1, 3, 14, 80, 534, 4102, 35916, 354888, 3915750, 47754938, 637840356, 9256590928, 144977618044, 2436460447020, 43719637179224, 834042701945520, 16852447379512710, 359468276129261730, 8070500634880125300,

A377662有这串数——Nov 07 2024——通项公式没我们的好——Table/k!, {k, 0, n}], {n, 0, 19}]

王守恩 发表于 2025-6-1 09:33:02

接楼上。

$\sum _{k=1}^{\infty}{k+1}/{k!}=1$

$\sum _{k=1}^{\infty}{(k+1)(k+1)}/{k!}=1$

$\sum _{k=1}^{\infty}{(k+1)(k+1)(k+2)}/{k!}=2$

$\sum _{k=1}^{\infty}{(k+1)(k+1)(k+2)(k+3)}/{k!}=6$

$\sum _{k=1}^{\infty}{(k+1)(k+1)(k+2)(k+3)(k+5)}/{k!}=30$

$\sum _{k=1}^{\infty}{(k+1)(k+1)(k+2)(k+3)(k+5)(k+8)}/{k!}=240$

$\sum _{k=1}^{\infty}{(k+1)(k+1)(k+2)(k+3)(k+5)(k+8)(k+13)}/{k!}=3120$

$\sum _{k=1}^{\infty}{(k+1)(k+1)(k+2)(k+3)(k+5)(k+8)(k+13)(k+21)}/{k!}=65520$

{1, 1, 2, 6, 30, 240, 3120, 65520, 2227680, 122522400, 10904493600, 1570247078400, 365867569267200, 137932073613734400, 84138564904377984000}

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