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POWER CHAIN. (Posted on 2010-07-22) Difficulty: 4 of 5
{75, 100, 125} is an example of an arithmetic progression of positive integers such that the n-th term is a perfect n-th power.
Find a longer sequence with this feature.

What is the longest you can get?
P.S. Trival solution(d=0) excluded.

No Solution Yet Submitted by Ady TZIDON    
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Some Thoughts computer results | Comment 5 of 6 |
 50       dim h(12)
100       FOR sr = 1 TO 1000
110         sq = sr * sr
120         cr0 = -INT(-(sq ^ (1 / 3)))
130         FOR cr = cr0 TO cr0 + 1000
140           cu = cr * cr * cr
150           diff = cu - sq
160           IF diff > 0 THEN
170           : pwr = 3: rslt = cu
180           : repeat
190           :   pwr = pwr + 1
200           :   rslt = rslt + diff
210           :   root = INT(rslt ^ (1 / pwr) + .5)
220           :   rchk = 1
230           :   good = 0
240           :   FOR i = 1 TO pwr
250           :    rchk = rchk * root
260           :   NEXT
270           :   IF rchk = rslt THEN good = 1:endif
280           :   IF good THEN h(pwr) = rslt:endif
290           : until good = 0
300           : pwr = pwr - 1
310           : IF pwr >= max AND pwr > 3 THEN
320           :   PRINT pwr, sq; cu;
330           :   FOR i = 4 TO pwr
340           :     PRINT h(i);
350           :   NEXT
360           :   PRINT
370           : END IF
380          :END IF
390         NEXT
400       NEXT

finds 4 as the highest power within its range of up to 10,000 for the square root of the square term, and up to 10,000 for the difference in the arithmetic sequence:

run
 4       90000  125000  160000
 4       1194649  48627125  96059601
 4       15816529  1532808577  3049800625
 4       44289025  8703679193  17363069361
 4       73530625  1838265625  3603000625
OK

These represent only the 2nd, 3rd and 4th powers. But all except the first shown would result in the first term being negative, thus disqualifying the whole series.

Thus the best progression found by the program is {55000, 90000, 125000, 160000}.

Now that I have seen Justin's series, with negative d, a variation allowing such is:

  50       dim H(12)
 100       for Sr=1 to 400
 110         Sq=Sr*Sr
 120         Cr0=-int(-(Sq^(1/3)))
 130         for Cr=1 to 1000
 140           Cu=Cr*Cr*Cr
 150           Diff=Cu-Sq
 160           if Diff<>0 then
 170           :Pwr=3:Rslt=Cu
 180           :repeat
 190           :Pwr=Pwr+1
 200           :Rslt=Rslt+Diff
 210           :Root=int(abs(Rslt)^(1/Pwr)+0.5)
 220           :Rchk=1
 230           :Good=0
 240           :for I=1 to Pwr
 250           :Rchk=Rchk*Root
 260           :next
 270           :if Rchk=Rslt then Good=1:endif
 280           :if Good then H(Pwr)=Rslt:endif
 290           :until Good=0
 300           :Pwr=Pwr-1
 310           :if Pwr>=Max and Pwr>3 then
 320           :print Pwr,Sq;Cu;
 330           :for I=4 to Pwr
 340           :print H(I);
 350           :next
 360           :print
 370           :endif
 380          :endif
 390         next
 400       next

which finds

run
 4       16  8  0
 4       169  125  81
 4       1024  512  0
 4       11664  5832  0
 4       30625  15625  625
 4       33856  17576  1296
 4       65536  32768  0
 4       90000  125000  160000
 4       123201  68921  14641
 4       123904  64000  4096
 

with the first power extrapolated back from the 2nd, 3rd and 4th powers shown. Justin's solution is the second row. A ones with 0 as the 4th power fail to satisfy the requirement for positive integers.


  Posted by Charlie on 2010-07-22 16:13:40
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