如何证明？

王守恩

\(f(n)=\big(1+x+x^2+x^3+x^4\big)^n=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+\cdots+a_{4n}x^{4n}\)

\(S(0)=a_{0}+a_{5}+a_{10}+a_{15}+a_{20}+a_{25}+\cdots\)

\(S(1)=a_{1}+a_{6}+a_{11}+a_{16}+a_{21}+a_{26}+\cdots\)

\(S(2)=a_{2}+a_{7}+a_{12}+a_{17}+a_{22}+a_{27}+\cdots\)

\(S(3)=a_{3}+a_{8}+a_{13}+a_{18}+a_{23}+a_{28}+\cdots\)

\(S(4)=a_{4}+a_{9}+a_{14}+a_{19}+a_{24}+a_{29}+\cdots\)

求证：\(S(0)=S(1)=S(2)=S(3)=S(4)=5^{n-1}\)

f(1)=1, 1, 1, 1, 1,
f(2)=1, 2, 3, 4, 5, 4, 3, 2, 1,
f(3)=1, 3, 6, 10, 15, 18, 19, 18, 15, 10, 6, 3, 1,
f(4)=1, 4, 10, 20, 35, 52, 68, 80, 85, 80, 68, 52, 35, 20, 10, 4, 1,
f(5)=1, 5, 15, 35, 70, 121, 185, 255, 320, 365, 381, 365, 320, 255, 185, 121, 70, 35, 15, 5, 1,
f(6)=1,6,21,56,126,246,426,666,951,1246,1506,1686,1751,1686,1506,1246,951,666,426,246,126,56,21,6,1,
......

yigo

\(x^5-1=0\)的5个根分别为\(1,ω,ω^2,ω^3,ω^4\)，带入\(f(n)\),可得`s_0,s_1,s_2,s_3,s_4`的五元一次方程:\[\begin{split}
f(1)&=s_0&+&s_1&+&s_2&+&s_3&+&s_4&=5^n\\
f(ω)&=s_0&+ω&s_1&+ω^2&s_2&+ω^3&s_3&+ω^4&s_4&=0\\
f(ω^2)&=s_0&+ω^2&s_1&+ω^4&s_2&+ω&s_3&+ω^3&s_4&=0\\
f(ω^3)&=s_0&+ω^3&s_1&+ω&s_2&+ω^4&s_3&+ω^2&s_4&=0\\
f(ω^4)&=s_0&+ω^4&s_1&+ω^3&s_2&+ω^2&s_3&+ω&s_4&=0\\
\end{split}\]上述方程有一个显然的解`s_0=s_1=s_2=s_3=s_4=5^{n-1}`.
方程的系数行列式\[\begin{Vmatrix}
1&1&1&1&1\\
1&ω&ω^2&ω^3&ω^4\\
1&ω^2&ω^4&ω&ω^3\\
1&ω^3&ω&ω^4&ω^2\\
1&ω^4&ω^3&ω^2&ω
\end{Vmatrix}\]是一个范德蒙行列式，值等于`\D\prod_{0≤i<j≤4}(ω_j-ω_i)≠0`, 因为多项式`f(x)`无重根。
所以方程有唯一解，就是那个显然的解。

王守恩

这应该是 "显然" 的结论呀！诸位！分享分享你的想法。

\(f(n)=(1+x+x^2+x^3+x^4)^n=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+\cdots+a_{4n}x^{4n}\)

1, 1, 1, 1, 1,
1, 2, 3, 4, 5, 4, 3, 2, 1,
1, 3, 6, 10, 15, 18, 19, 18, 15, 10, 6, 3, 1,
1, 4, 10, 20, 35, 52, 68, 80, 85, 80, 68, 52, 35, 20, 10, 4, 1,
1, 5, 15, 35, 70, 121, 185, 255, 320, 365, 381, 365, 320, 255, 185, 121, 70, 35, 15, 5, 1,
1,6,21,56,126,246,426,666,951,1246,1506,1686,1751,1686,1506,1246,951,666,426,246,126,56,21,6,1,
......
我不知道怎么把这一串数分成5串, 再加起来。
如果我们只是把每串最大的数特别地列出来。OEIS--A324595-- 2024 年 1 月 15 日
1, 5, 19, 85, 381, 1751, 8135, 38173, 180415, 857695, 4096830, 19645975, 94523729, 456079769,
2206005414, 10693086637, 51930129399, 252617434619, 1230714593340, 6003931991895,  ......

1. Table[SeriesCoefficient[(-1+QPochhammer[-1,Sqrt[x]]/2)^n,{x,0,n}],{n,0,25}]

复制代码

mathe

直接计算也可以，看起来稍微容易懂一些。
设\((1+x+x^2+x^3+x^4)^{n-1}=a_0+a_1x+a_2x^2+\cdots+a_{4(n-1)}x^{4(n-1)}\)
\((1+x+x^2+x^3+x^4)^n=b_0+b_1x+b_2x^2+\cdots+b_{4n}x^{4n}\)
为了方便再记\(a_{-4}=a_{-3}=a_{-2}=a_{-1}=a_{4(n-1)+1}=a_{4(n-1)+2}=a_{4(n-1)+3}=a_{4n}=0\)
于是\(b_k=a_k+a_{k-1}+a_{k-2}+a_{k-3}+a_{k-4}\)
\(b_0+b_5+b_{10}+...=a_{-4}+a_{-3}+a_{-2}+a_{-1}+a_0 +a_1+a_2+a_3+a_4+a_5+....=\sum_{k=0}^{4(n-1)}a_k=(1+1+1+1+1)^{n-1}=5^{n-1}\)
同理另外4个和的结果也相同

王守恩

mathe 发表于 2024-2-2 17:01
直接计算也可以，看起来稍微容易懂一些。
设\((1+x+x^2+x^3+x^4)^{n-1}=a_0+a_1x+a_2x^2+\cdots+a_{4(n-1)} ...

谢谢 mathe ！这些都没问题吗？后面的都没问题吗？谢谢 mathe ！

\((1+x)^n=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+\cdots+a_{n}x^{n}\)
\(S(0)=a_{0}+a_{2}+a_{4}+a_{6}+a_{8}+a_{10}+\cdots=2^{n-1}\)
\(S(1)=a_{1}+a_{3}+a_{5}+a_{7}+a_{9}+a_{11}+\cdots=2^{n-1}\)

\((1+x+x^2)^n=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+\cdots+a_{2n}x^{2n}\)
\(S(0)=a_{0}+a_{3}+a_{6}+a_{09}+a_{12}+a_{15}+\cdots=3^{n-1}\)
\(S(1)=a_{1}+a_{4}+a_{7}+a_{10}+a_{13}+a_{16}+\cdots=3^{n-1}\)
\(S(2)=a_{2}+a_{5}+a_{8}+a_{11}+a_{14}+a_{17}+\cdots=3^{n-1}\)

\((1+x+x^2+x^3)^n=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+\cdots+a_{3n}x^{3n}\)
\(S(0)=a_{0}+a_{4}+a_{08}+a_{12}+a_{16}+a_{20}+\cdots=4^{n-1}\)
\(S(1)=a_{1}+a_{5}+a_{09}+a_{13}+a_{17}+a_{21}+\cdots=4^{n-1}\)
\(S(2)=a_{2}+a_{6}+a_{10}+a_{14}+a_{18}+a_{22}+\cdots=4^{n-1}\)
\(S(3)=a_{3}+a_{7}+a_{11}+a_{15}+a_{19}+a_{23}+\cdots=4^{n-1}\)

\((1+x+x^2+x^3+x^4)^n=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+\cdots+a_{4n}x^{4n}\)
\(S(0)=a_{0}+a_{5}+a_{10}+a_{15}+a_{20}+a_{25}+\cdots=5^{n-1}\)
\(S(1)=a_{1}+a_{6}+a_{11}+a_{16}+a_{21}+a_{26}+\cdots=5^{n-1}\)
\(S(2)=a_{2}+a_{7}+a_{12}+a_{17}+a_{22}+a_{27}+\cdots=5^{n-1}\)
\(S(3)=a_{3}+a_{8}+a_{13}+a_{18}+a_{23}+a_{28}+\cdots=5^{n-1}\)
\(S(4)=a_{4}+a_{9}+a_{14}+a_{19}+a_{24}+a_{29}+\cdots=5^{n-1}\)
..........