无穷数列的最大公因数

hbghlyj

对整数k>1,设f(k)为最大的正整数n>1,满足$\forall i\in\mathbb N_+:n|(i^k-(i-1)^k-1)(i^k-(i-1)^k+1)(i^k-(i+1)^k-1)(i^k-(i+1)^k+1)$.

In[1]:Table[GCD@@Table[(i^n-(i-1)^n-1)(i^n-(i-1)^n+1)(i^n-(i+1)^n-1)(i^n-(i+1)^n+1),{i,1000}],{n,100}]
Out[1]:
{0,192,8640,3072,100800,2304,987840,12288,129600,768,2439360,9216,119246400,768,8640,49152,29131200,6912,50944320,3072,5227200,768,10664640,36864,119246400,768,77760,3072,84772800,2304,16409574720,196608,12484800,768,141120,27648,58932644097600,768,8640,12288,20502820800,2304,260930880,3072,68558400,768,44533440,147456,1688648270400,768,1045440,3072,283147200,20736,50944320,12288,36331200,768,70176960,9216,51595723265486400,768,25920,786432,29131200,2304,335114982720,3072,43200,768,12296813760,110592,314052060396110400,768,8640,3072,705600,2304,880729920,49152,68564361729600,768,138882240,9216,185429225217600,768,8640,12288,422373067200,6912,41466995317440,3072,95428800,768,20160,589824,324254930126400,768,25920,3072}
In[2]:Tally[%1]
Out[2]:
{{0,1},{192,1},{8640,5},{3072,9},{100800,1},{2304,5},{987840,1},{12288,4},{129600,1},{768,16},{2439360,1},{9216,3},{119246400,2},{49152,2},{29131200,2},{6912,2},{50944320,2},{5227200,1},{10664640,1},{36864,1},{77760,1},{84772800,1},{16409574720,1},{196608,1},{12484800,1},{141120,1},{27648,1},{58932644097600,1},{20502820800,1},{260930880,1},{68558400,1},{44533440,1},{147456,1},{1688648270400,1},{1045440,1},{283147200,1},{20736,1},{36331200,1},{70176960,1},{51595723265486400,1},{25920,2},{786432,1},{335114982720,1},{43200,1},{12296813760,1},{110592,1},{314052060396110400,1},{705600,1},{880729920,1},{68564361729600,1},{138882240,1},{185429225217600,1},{422373067200,1},{41466995317440,1},{95428800,1},{20160,1},{589824,1},{324254930126400,1}}
In[3]:Position[%1, 3072]
Out[3]:{{4},{20},{28},{44},{52},{68},{76},{92},{100}}
从这里可以看出一些规律:作差分得
16,8,16,8,16,8,16,8
这个规律会继续吗?

睡神

回复 1# hbghlyj
很久之前算了一下，没算错的话，应该是100800的所有约数

hbghlyj

回复 1# hbghlyj
按照这个规律
In[4]:n = 16 k + 100; Table[
GCD @@ Table[(i^n - (i - 1)^n - 1) (i^n - (i - 1)^n +
      1) (i^n - (i + 1)^n - 1) (i^n - (i + 1)^n + 1), {i, 5000}], {k,
  50}]
Out[4]:
{3072,9216,3072,3072,27648,3072,3072,9216,3072,3072,9216,3072,3072,248832,3072,3072,9216,3072,3072,9216,3072,3072,27648,3072,3072,9216,3072,3072,9216,3072,3072,27648,3072,3072,9216,3072,3072,9216,3072,3072,82944,3072,3072,9216,3072,3072,9216,3072,3072,27648}
In[5]:Sum[Boole[Not[Divisible[%4[[t]], 3072]]], {t, Length[%4]}]
Out[5]:0
可见这个列表里面都是3072的倍数

hbghlyj

原贴是$\forall i=1,2,\cdots,n$,后来觉得改成任意正整数更加自然,就这样了

睡神

看不懂在说啥的路过…

hbghlyj

回复 2# 睡神
您算的是对的.我把题改成k了.就这样了.\摊手