Three positive integers 2520 < b < c constitute three consecutive terms of a
harmonic sequence and, b divides c.
Find the total count of pairs (b,c) for which this is possible.
(In reply to
Analytic solution by Steve Herman)
Indeed I neglected the part about b dividing c.
4 kill "conscnt.txt"
5 open "conscnt.txt" for output as #2
10 for C=1 to 10000000
20 F=1//2520+1//C
30 B=2//F
40 if B=int(B) then if C @ B=0 then print B,C,C/B:print #2,B,C,C/B:Ct=Ct+1
50 next
60 print Ct:print #2,Ct
b c c/b
2520 2520 1.0
3780 7560 2.0
4200 12600 3.0
4410 17640 4.0
4536 22680 5.0
4620 27720 6.0
4680 32760 7.0
4725 37800 8.0
4760 42840 9.0
4788 47880 10.0
4830 57960 12.0
4860 68040 14.0
4872 73080 15.0
4900 88200 18.0
4914 98280 20.0
4920 103320 21.0
4935 118440 24.0
4950 138600 28.0
4956 148680 30.0
4968 173880 35.0
4970 178920 36.0
4977 199080 40.0
4980 209160 42.0
4984 224280 45.0
4995 279720 56.0
4998 299880 60.0
5000 315000 63.0
5004 350280 70.0
5005 360360 72.0
5010 420840 84.0
5012 451080 90.0
5016 526680 105.0
5019 602280 120.0
5020 632520 126.0
5022 703080 140.0
5025 844200 168.0
5026 904680 180.0
5028 1055880 210.0
5030 1267560 252.0
5031 1408680 280.0
5032 1585080 315.0
5033 1811880 360.0
5034 2114280 420.0
5035 2537640 504.0
5036 3172680 630.0
5037 4231080 840.0
5038 6347880 1260.0
47 is the revised count
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Posted by Charlie
on 2015-04-24 21:49:59 |
revised program
