問題來自羣友:
從1開始,二進制取反然後拼接到後面,重複這一操作,得到
1
10
1001
10010110
1001011001101001
10010110011010010110100110010110
1001011001101001011010011001011001101001100101101001011001101001
......
自然而然想到,從前面增加0.
,這個小數會趨近怎樣一個數呢?
通項公式我沒想到好的方法寫出來,於是掏出麥醬:
RSolve[{
a[1] == 1/2,
a[n] == a[n - 1] + (1 - a[n - 1] - 2^-2^(n - 1))*2^-2^(n - 1)}
, a[n], n] // FullSimplify
得到
……看這樣子應該是個超越數,我肯定算不出,不用算了,前200位:
0.58754596635989240221663863174154471691052162554423044242662058465120640763421741103619545951378786660374363355346186251257838287480639347986900318747408590610118831494175859485120507626316228231836793