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 Consecutive Count (Posted on 2015-04-23)
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.

 No Solution Yet Submitted by K Sengupta No Rating

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 revised program Comment 4 of 4 |
(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

 Posted by Charlie on 2015-04-24 21:49:59

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