The prime factors of 30 are {2,3,5}
The sum of the reciprocals is 1/2+1/3+1/5=31/30
The product of the reciprocals is 1/2*1/3*1/5=1/30
The difference of the two is 31/30-1/30=1
There aren't many numbers where the final value of this procedure is a whole number.
Find more.
Other tasks:
Show that they cannot be semi-prime.
Show they must be square-free.
10 for N=2 to 35000
20 Sum=0:Prod=1
30 N2=N
40 while N2>1
50 F=prmdiv(N2)
60 N2=N2//F
70 Sum=Sum+1//F
80 Prod=Prod//F
90 wend
100 Diff=Sum-Prod
110 if Diff=int(Diff) and N<>prmdiv(N) then print N
120 next N
finds only 30, 858 and 1722.
The "and N<>prmdiv(N)" is to weed out results that are prime, which derive from all prime n.
If these are the only three, then the "other tasks" follow immediately.
Here are a few more from Sloane, factored into primes:
30 2 3 5
858 2 3 11 13
1722 2 3 7 41
66198 2 3 11 17 59
2214408306 2 3 11 23 31 47057
24423128562 2 3 7 43 3041 4447
432749205173838 2 3 7 59 163 1381 775807
14737133470010574 2 3 7 71 103 67213 713863
550843391309130318 2 3 7 71 103 61559 29133437
244197000982499715087866346 2 3 11 23 31 47137
from
10 open "a007850.txt" for input as #1
15 open "giugax.txt" for output as #2
20 while eof(1)=0
30 input #1,G
35 Giuga=val(G)
40 print Giuga;" ";
41 print #2,Giuga;" ";
50 while Giuga>1
60 Pf=prmdiv(Giuga):Giuga=Giuga//Pf
70 print Pf;
71 print #2,Pf;
80 wend
85 print
86 print #2,
90 wend
100 close #2
Sloane had a couple more but they exceeded the limitations of UBASIC's prmdiv function.
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Posted by Charlie
on 2016-07-05 14:43:03 |
program; research; program
