(In reply to
computer exploration -- far from a proof by Charlie)
5 point 255:dim Term(20)
10 for N=2 to 10
20 Prod=1
25 erase Term()
30 dim Term(N)
40 for I=1 to N
50 Term(I)=sqrt(I)
60 next
70 for F=0 to int(2^N-0.5)
80 Tot=0
90 Bs="":F1=F
100 while F1>0
110 D=F1@2:F1=F1\2
120 Bs=cutspc(str(D))+Bs
130 wend
140 Bs=right("00000000000"+Bs,N)
150 for I=1 to len(Bs)
160 if mid(Bs,I,1)="0" then
170 :Tot=Tot-Term(I)
180 :else
190 :Tot=Tot+Term(I)
200 :endif
210 next
220 Prod=Prod*Tot
230 if N=3 then print using(10,15),Tot
240 next F
250 print N,int(Prod+0.5)
260 next N
2 1
3 64
4 4096
5 23323703841
6 63703464216016403230349121
7 316699666163357097153212433469030615484754548657341071360000
8 122650145964677485280855470979489605455779538020711656443170899516425755208153578926158095907728782598859083391629280893573529600000000
Then there's an overflow.
Rounding was used as there were long strings of 9's after the decimal point in each case.
|
Posted by Charlie
on 2015-09-20 08:23:36 |