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 POWER CHAIN. (Posted on 2010-07-22)
{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 No Rating

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 computer results | Comment 5 of 6 |
` 50       dim h(12)100       FOR sr = 1 TO 1000110         sq = sr * sr120         cr0 = -INT(-(sq ^ (1 / 3)))130         FOR cr = cr0 TO cr0 + 1000140           cu = cr * cr * cr150           diff = cu - sq160           IF diff > 0 THEN170           : pwr = 3: rslt = cu180           : repeat190           :   pwr = pwr + 1200           :   rslt = rslt + diff210           :   root = INT(rslt ^ (1 / pwr) + .5)220           :   rchk = 1230           :   good = 0240           :   FOR i = 1 TO pwr250           :    rchk = rchk * root260           :   NEXT270           :   IF rchk = rslt THEN good = 1:endif280           :   IF good THEN h(pwr) = rslt:endif290           : until good = 0300           : pwr = pwr - 1310           : IF pwr >= max AND pwr > 3 THEN320           :   PRINT pwr, sq; cu;330           :   FOR i = 4 TO pwr340           :     PRINT h(i);350           :   NEXT360           :   PRINT370           : END IF380          :END IF390         NEXT400       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  3603000625OK`

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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