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%I #18 Jul 19 2016 12:33:14
%S 1257,32769,262176,262208,1081344,4198400,16777217,16809984
%N Numbers D such that D^2 = A^4 + B^5 + C^6 has more than one solution in positive integers (A, B, C).
%C A subsequence of A255830. Sequences A256604 and A256603 are the analog for A180241 and A256091.
%C Terms a(2) - a(8) have Hamming weight 2: 32769 = 2^15 + 1, 262176 = 2^18 + 2^5, 262208 = 2^18 + 2^6, 1081344 = 2^20 + 2^15, 4198400 = 2^22 + 2^12, 16777217 = 2^24 + 1, 16809984 = 2^24 + 2^15.
%C Given D^2 = A^4+B^5+C^6, multiply by u^60, u>1, to get (u^30*D)^2 = (u^15*A)^4 + (u^12*B)^5 + (u^10*C)^6. If D is a solution then so is u^30*D. - _Lars Blomberg_, Apr 26 2015
%C Solutions for a(1)-a(8) as well as some larger terms:
%C ..A1.....B1....C1......A2.....B2....C2..............D
%C ..35......8.....6......32......2.....9...........1257
%C ..16......1....32......16.....64.....1..........32769
%C ..64......4....64.....512......4....16.........262176
%C ...8.....32....64.....512.....32.....4.........262208
%C 1024.....64....64.....512....256....32........1081344
%C .480....240...160....2048....128....16........4198400
%C ...1.....32...256....4096.....32.....1.......16777217
%C 1024.....64...256....4096....256....32.......16809984
%C .512......4..1024...32768......4....64.....1073741856
%C 1024...4096.....8...32768....256.....8.....1073742336
%C 4096...2048..1024...32768...2048...256.....1090519040
%C ...1..16384....64.....512..16384.....1....34359738369
%C 4096..16384....16......64..16384...256....34359742464
%C 4096..16384..1024...32768..16384...256....34376515584
%C .512...2048..4096..262144...2048....64....68719738880
%C ...1....256..8192....1024......1..8192...549755813889
%C 1024...4096..8192...32768....256..8192...549756862464
%C - _Lars Blomberg_, Apr 26 2015
%e (A, B, C) = (32, 2, 9): 32^4 + 2^5 + 9^6 = 1048576 + 32 + 531441 = 1580049 = 1257^2, and
%e (A, B, C) = (35, 8, 6): 35^4 + 8^5 + 6^6 = 1500625 + 32768 + 46656 = 1580049 = 1257^2,
%e so 1257 is a term.
%o (PARI) is_A256652(D,f=-1)={my(C=0,B,D2C6);while(1<D2C6=D^2-C++^6, B=0; while(0<D2C6-B++^5, ispower(D2C6-B^5,4)&&f++&&return(1)))}
%o for(D=2,10^5,is_A256652(D)&&print1(D",")) \\ Converted to integer arithmetic by _M. F. Hasler_, May 01 2015
%Y Cf. A255830, A256603, A256091, A256613, A256604, A180241, A180242.
%K nonn,bref,more
%O 1,1
%A _M. F. Hasler_, Apr 06 2015
%E a(5)-a(8) from _Lars Blomberg_, Apr 26 2015
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