A235809 -id:A235809 - cp's OEIS frontend

A235808 Numbers k such that k^3 has one or more occurrences of exactly six different digits.

59, 66, 69, 73, 75, 76, 84, 88, 93, 97, 107, 112, 113, 115, 116, 118, 124, 128, 129, 131, 139, 147, 148, 151, 156, 159, 161, 166, 168, 169, 174, 178, 181, 183, 184, 187, 189, 193, 194, 196, 207, 219, 226, 232, 234, 235, 236, 238, 240, 241, 246, 253, 255, 262

Offset: 1

Colin Barker, Jan 19 2014

			59 is in the sequence because 59^3 = 205379, which contains exactly six different digits: 0, 2, 3, 5, 7, 9.
107 is in the sequence because 107^3 = 1225043, which contains exactly six different digits: 0, 1, 2, 3, 4, 5.

Colin Barker, Table of n, a(n) for n = 1..1000

Cf. A030292, A155146, A155147, A235807, A235809-A235811, A119735.

[n: n in [0..300] | #Set(Intseq(n^3)) eq 6]; // Bruno Berselli, Jan 19 2014

Select[Range[300], Length[Union[IntegerDigits[#^3]]] == 6 &] (* Bruno Berselli, Jan 19 2014 *)
Select[Range[300],Count[DigitCount[#^3],0]==4&] (* Harvey P. Dale, May 28 2025 *)

s=[]; for(n=1, 300, if(#vecsort(eval(Vec(Str(n^3))),,8)==6, s=concat(s, n))); s

A235808_list, m = [], [6, -6, 1, 0]
for n in range(1,10**4+1):
    for i in range(3):
        m[i+1] += m[i]
    if len(set(str(m[-1]))) == 6:
        A235808_list.append(n) # Chai Wah Wu, Nov 05 2014

A235810 Numbers n such that n^3 has one or more occurrences of exactly eight different digits.

Original entry on oeis.org

289, 297, 302, 319, 467, 494, 515, 557, 562, 595, 621, 623, 676, 682, 709, 712, 721, 862, 887, 909, 939, 945, 949, 963, 984, 987, 1012, 1015, 1016, 1025, 1029, 1043, 1049, 1065, 1075, 1087, 1104, 1106, 1107, 1114, 1118, 1132, 1137, 1154, 1161, 1167, 1178

Offset: 1

Colin Barker, Jan 19 2014

			289 is in the sequence because 289^3 = 24137569, which contains exactly eight different digits.

Colin Barker, Table of n, a(n) for n = 1..1000

Cf. A030292, A155146, A155147, A235807-A235809, A235811, A119735.

[n: n in [0..1200] | #Set(Intseq(n^3)) eq 8]; // Vincenzo Librandi, Nov 07 2014

s=[]; for(n=1, 1500, if(#vecsort(eval(Vec(Str(n^3))),,8)==8, s=concat(s, n))); s

A235810_list, m = [], [6, -6, 1, 0]
for n in range(1,10**3+1):
    for i in range(3):
        m[i+1] += m[i]
    if len(set(str(m[-1]))) == 8:
        A235810_list.append(n) # Chai Wah Wu, Nov 05 2014

cp's OEIS Frontend

A235808 Numbers k such that k^3 has one or more occurrences of exactly six different digits.

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