A295025
Cubes whose largest digit is 5.
Original entry on oeis.org
125, 512, 125000, 405224, 512000, 531441, 1225043, 5000211, 5545233, 13312053, 43243551, 54010152, 102503232, 115501303, 125000000, 221445125, 320013504, 400315553, 405224000, 512000000, 531441000, 1204550144, 1225043000, 2053225511, 2253243231, 2543302125
Offset: 1
512 is in the sequence because it is a cube, 512 = 8^3, and its largest digit is 5.
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for(n=1,2e8, vecmax(digits(n^3))==5&&print1(n^3,","))
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def ok(cube): return max(str(cube)) == "5"
print([c for c in (i**3 for i in range(1370)) if ok(c)]) # Michael S. Branicky, Dec 05 2021
A294664
Numbers n such that the largest digit of n^3 is 4.
Original entry on oeis.org
7, 68, 70, 324, 680, 700, 3240, 6800, 7000, 7618, 31177, 32400, 52308, 68000, 69314, 70000, 76180, 311770, 324000, 353068, 523080, 680000, 693140, 700000, 756658, 761800, 1039247, 2715974, 2732441, 3117700, 3240000, 3511617, 3530680, 4689368, 5230800, 6800000, 6931400, 7000000
Offset: 1
7 is in the sequence because the largest digit of 7^3 = 343 is 4.
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select(n -> max(convert(n^3,base,10))=4, [$1..10^6]); # Robert Israel, Nov 13 2017
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for(n=1,2e8, vecmax(digits(n^3))==4&&print1(n","))
A295021
Cubes whose largest digit is 6.
Original entry on oeis.org
64, 216, 15625, 46656, 50653, 64000, 132651, 216000, 262144, 456533, 614125, 636056, 1601613, 1643032, 2406104, 2515456, 3112136, 3652264, 6331625, 10360232, 13144256, 15625000, 41063625, 46656000, 50653000, 52313624, 55306341, 56623104, 64000000, 66430125, 100544625
Offset: 1
64 is in the sequence because it is a cube, 64 = 4^3, and its largest digit is 6.
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Select[Range[500]^3,Max[IntegerDigits[#]]==6&] (* Harvey P. Dale, Jun 21 2022 *)
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for(n=1,500, vecmax(digits(n^3))==6 &&print1(n^3,","))
A295022
Cubes whose largest digit is 7.
Original entry on oeis.org
27, 2744, 3375, 12167, 17576, 27000, 157464, 166375, 175616, 250047, 274625, 300763, 474552, 753571, 1157625, 1367631, 1771561, 2000376, 2352637, 2460375, 2571353, 2744000, 3176523, 3375000, 4330747, 4657463, 4741632, 5177717, 5451776, 6644672, 7645373, 11543176, 12167000
Offset: 1
27 is in the sequence because it is a cube, 27 = 3^3, and its largest digit is 7.
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for(n=1,250, vecmax(digits(n^3))==7 &&print1(n^3,","))
A295024
Cubes whose largest digit is 9.
Original entry on oeis.org
729, 2197, 4096, 4913, 6859, 9261, 19683, 21952, 24389, 29791, 35937, 39304, 59319, 68921, 79507, 91125, 97336, 110592, 117649, 185193, 195112, 205379, 226981, 287496, 328509, 357911, 389017, 438976, 493039, 592704, 704969, 729000, 912673, 941192, 970299, 1092727, 1191016
Offset: 1
2197 is in the sequence because it is a cube, 2197 = 13^3, and its largest digit is 9.
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for(n=1,150, vecmax(digits(n^3))==8 &&print1(n^3,","))
A295023
Cubes whose largest digit is 8.
Original entry on oeis.org
8, 1728, 5832, 8000, 10648, 13824, 32768, 42875, 54872, 74088, 85184, 103823, 140608, 148877, 238328, 373248, 421875, 551368, 571787, 658503, 681472, 778688, 804357, 830584, 857375, 884736, 1061208, 1124864, 1481544, 1520875, 1728000, 1815848, 1860867, 2048383, 2628072, 2803221
Offset: 1
8 is in the sequence because it is a cube, 8 = 2^3, and its largest digit is 8.
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for(n=1,200, vecmax(digits(n^3))==8 &&print1(n^3,","))
A294662
Least k > a(n-1) such that k^3 has no digit in common with a(n-1) and a(n+1), a(0)=0.
Original entry on oeis.org
0, 1, 2, 4, 5, 6, 7, 8, 9, 10, 29, 55, 88, 90, 111, 200, 211, 400, 518, 654, 888, 889, 1111, 2825, 3131, 4244, 11111, 28222, 31535, 42449, 53355, 90000, 111181, 590000, 618181, 900000, 1111115, 9000000, 11111115, 60660090, 114144155
Offset: 0
a(3) cannot be 3 because 3^3 = 27 would have the digit '2' in common with a(2) = 2, therefore a(3) = 4, which does not violate this condition.
After a(9) = 10, none of the numbers { 11, ..., 19 } can follow, because they have the digit '1' in common with a(9)^3 = 1000. Numbers { 20, ..., 28 } are excluded because their cube would have a digit '0' or '1' in common with a(9). Therefore, a(10) = 29 which hasn't a digit in common with a(9)^3, nor has 29^3 = 24389 a digit in common with a(9).
a(38) = 11111115 with 11111115^3 = 1371743552812575445875 using all digits except for 0, 6 and 9. So a(39) = 60660090 is possible, with a(39)^3 = 223207688999086038729000 having all digits except for 1, 4 and 5.
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nxt(a,L=oo,D(a)=Set(digits(a)),S=D(a),T=D(a^3))={for(k=a+1,L, #setintersect(D(k),T)||#setintersect(D(k^3),S)||return(k))}
A294662=List(); a=0; until(!a=nxt(a,1e7),write("/tmp/A294662.txt",#A294662," ",a);listput(A294662,a))
Showing 1-7 of 7 results.
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