A119735 -id:A119735 - cp's OEIS frontend

A029776 Digits of n appear in n^3.

0, 1, 4, 5, 6, 9, 10, 11, 12, 21, 24, 25, 29, 32, 33, 34, 39, 40, 44, 49, 50, 51, 54, 55, 56, 59, 60, 61, 64, 65, 66, 67, 71, 72, 73, 75, 76, 88, 90, 97, 99, 100, 101, 102, 106, 109, 110, 111, 112, 114, 115, 116, 119, 120, 124, 125, 129, 137, 151, 153, 176

Offset: 1

Patrick De Geest

			66 is in the list because 66^3 = 287496 and 6 appears in 287496.
153 is in the list because 153^3 = 3581577 and 1, 5, 3 appear in 3581577.

Robert Israel, Table of n, a(n) for n = 1..10000

Contains A119735.

[n: n in [0..200] | Intseq(n) subset Intseq(n^3)]; // Bruno Berselli, Aug 01 2013

filter:= n -> convert(convert(n,base,10),set) subset convert(convert(n^3,base,10),set):select(filter, [$0..200]); # Robert Israel, Mar 18 2020

isok(m) = my(d=Set(digits(m)), ddd=Set(digits(m^3))); setintersect(d, ddd) == d; \\ Michel Marcus, Mar 18 2020

from itertools import count, islice
def A029776_gen(startvalue=0): # generator of terms >= startvalue
    return filter(lambda n:set(str(n)) <= set(str(n**3)), count(max(startvalue,0)))
A029776_list = list(islice(A029776_gen(),20)) # Chai Wah Wu, Apr 03 2023

Offset changed to 1 by Robert Israel, Mar 18 2020

A235807 Numbers n such that n^3 has one or more occurrences of exactly five different digits.

Original entry on oeis.org

22, 24, 27, 29, 32, 35, 38, 41, 47, 48, 49, 51, 52, 54, 55, 57, 61, 63, 65, 71, 72, 82, 85, 87, 89, 94, 96, 102, 103, 104, 105, 108, 109, 119, 120, 123, 125, 126, 127, 130, 133, 134, 136, 137, 138, 141, 143, 144, 149, 152, 153, 154, 155, 158, 162, 165, 167

Offset: 1

Colin Barker, Jan 19 2014

			22 is in the sequence because 22^3 = 10648, which contains exactly five different digits: 0, 1, 4, 6, 8.
87 is in the sequence because 87^3 = 658503, which contains exactly five different digits: 0, 3, 5, 6, 8.

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

Cf. A030292, A155146, A155147, A235808-A235811, A119735.

[n: n in [0..200] | #Set(Intseq(n^3)) eq 5]; // Bruno Berselli, Jan 19 2014

Select[Range[200], Length[Union[IntegerDigits[#^3]]] == 5 &] (* Bruno Berselli, Jan 19 2014 *)

s=[]; for(n=1, 200, if(#vecsort(eval(Vec(Str(n^3))),,8)==5, s=concat(s, n))); s

A235807_list, m = [], [6, -6, 1, 0]
for n in range(1,10**5+1):
    for i in range(3):
        m[i+1] += m[i]
    if len(set(str(m[-1]))) == 5:
        A235807_list.append(n) # Chai Wah Wu, Nov 05 2014

A235811 Numbers n such that n^3 has one or more occurrences of exactly nine different digits.

Original entry on oeis.org

1018, 1028, 1112, 1452, 1475, 1484, 1531, 1706, 1721, 1733, 1818, 1844, 1895, 1903, 2008, 2033, 2208, 2214, 2217, 2223, 2257, 2274, 2277, 2327, 2329, 2336, 2354, 2394, 2403, 2524, 2525, 2589, 2647, 2686, 2691, 2694, 2727, 2733, 2784, 2842, 2866, 2884, 2890

Offset: 1

Colin Barker, Jan 19 2014

			1018 is in the sequence because 1018^3 = 1054977832, which contains exactly nine different digits.

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

Cf. A030292, A155146, A155147, A235807-A235810, A119735.

Select[Range[3000],Count[DigitCount[#^3],0]==1&] (* Harvey P. Dale, Dec 17 2021 *)

s=[]; for(n=1, 3000, if(#vecsort(eval(Vec(Str(n^3))),,8)==9, s=concat(s, n))); s

A235811_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]))) == 9:
        A235811_list.append(n) # Chai Wah Wu, Nov 05 2014

A363905 Numbers whose square and cube taken together contain each decimal digit.

Original entry on oeis.org

69, 128, 203, 302, 327, 366, 398, 467, 542, 591, 593, 598, 633, 643, 669, 690, 747, 759, 903, 923, 943, 1016, 1018, 1027, 1028, 1043, 1086, 1112, 1182, 1194, 1199, 1233, 1278, 1280, 1282, 1328, 1336, 1364, 1396, 1419, 1459, 1463, 1467, 1472, 1475

Offset: 1

M. F. Hasler, Jun 27 2023

The first term, a(1) = 69, is the only number for which the square and the cube together contain each decimal digit 0 to 9 exactly once.

a(820) = 6534 is the only number of which the square and cube taken together contain each digit 0 to 9 exactly twice.

			69^2 = 4761, 69^3 = 328509, which together contain each digit 0-9 exactly once.

Robert G. Wilson v, Table of n, a(n) for n = 1..10000
Harold Suarez, Interesting..., Number Theory group on LinkedIn, June 2023.

Cf. A036744, A054038, A071519 and A156977 for "pandigital" squares.

Cf. A119735: Numbers n such that every digit occurs at least once in n^3.

fQ[n_] := Union[ Join[ IntegerDigits[n^2], IntegerDigits[n^3]]] == {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}; Select[Range@1500, fQ] (* Robert G. Wilson v, Jun 27 2023 *)

is(k)=#setunion(Set(digits(k^2)),Set(digits(k^3)))>9
select(is,[1..9999])

from itertools import count, islice
def A363905_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n:len(set(str(n**2))|set(str(n**3)))==10,count(max(startvalue,1)))
A363905_list = list(islice(A363905_gen(),20)) # Chai Wah Wu, Jun 27 2023

A121321 Numbers k such that every digit occurs at least once in k^4.

Original entry on oeis.org

763, 767, 1066, 1088, 1206, 1304, 1425, 1557, 1561, 1634, 1653, 1712, 1739, 1782, 1818, 1839, 1866, 1878, 1975, 2032, 2045, 2055, 2134, 2192, 2232, 2233, 2299, 2318, 2339, 2347, 2358, 2425, 2441, 2489, 2509, 2511, 2575, 2646, 2682, 2692

Offset: 1

Tanya Khovanova, Aug 25 2006

			763^4 = 338920744561 contains every digits at least once.

Lars Blomberg, Table of n, a(n) for n = 1..10000

Cf. A119735 (in k^3), A054038 (in k^2).

[ n: n in [0..2695] | Seqset(Intseq(n^4)) eq {0..9} ]; // Bruno Berselli, May 17 2011

Select[Range[2692],ContainsAll[IntegerDigits[#^4],Range[0,9]]&] (* James C. McMahon, Oct 16 2024 *)

PARI
```
isok(k) = #Set(digits(k^4)) == 10;
```

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

Original entry on oeis.org

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

A235809 Numbers k such that k^3 has one or more occurrences of exactly seven different digits.

Original entry on oeis.org

135, 145, 203, 221, 223, 225, 227, 233, 243, 244, 245, 247, 249, 254, 257, 265, 272, 273, 275, 276, 299, 313, 327, 329, 334, 338, 341, 345, 347, 352, 365, 366, 368, 382, 384, 388, 393, 395, 398, 403, 405, 409, 411, 412, 434, 439, 447, 452, 455, 473, 486, 493

Offset: 1

Colin Barker, Jan 19 2014

			135 is in the sequence because 135^3 = 2460375, which contains exactly seven different digits.

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

Cf. A030292, A155146, A155147, A235807, A235808, A235810, A235811, A119735.

[n: n in [0..1200] | #Set(Intseq(n^3)) eq 7]; // Vincenzo Librandi, Nov 07 2014

Select[Range[500], Length[Union[IntegerDigits[#^3]]]==7&] (* Vincenzo Librandi, Nov 07 2014 *)

s=[]; for(n=1, 600, if(#vecsort(eval(Vec(Str(n^3))),,8)==7, s=concat(s, n))); s

for(n=0,10^3,if(#Set(digits(n^3))==7,print1(n,", "))); \\ Joerg Arndt, Nov 10 2014

from itertools import count, islice
def A235809gen(): return filter(lambda n:len(set(str(n**3))) == 7,count(0))
A235809_list = list(islice(A235809gen(),26)) # Chai Wah Wu, Dec 23 2021

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

A363927 Numbers N such that in the concatenation of N^2 and N^3, each of the 10 decimal digits appears equally often.

Original entry on oeis.org

69, 6534, 497375, 539019, 543447, 586476, 589629, 601575, 646479, 858609, 895688, 959097, 46839081, 47469378, 47693199, 47760623, 47841576, 48038964, 48527792, 48733506, 48886836, 48965892, 49229103, 49397283, 49594832, 49670616, 50013116, 50247423, 50359157

Offset: 1

Charles R Greathouse IV and M. F. Hasler, Jun 28 2023

a(3) = 497375 and a(11) = 895688 are the only terms < 10^6 that are not divisible by 3.
Each term has an even number of decimal digits, k, and a corresponding value between 10^(k-1)*100^(1/3) and 10^k. - Michael S. Branicky, Jun 29 2023
Indeed, the number of digits of concat(N^2, N^3) is floor(2*L + 1) + floor(3*L + 1) where L = log_10(N). This is a multiple of 10 iff L mod 2 is in the interval [5/3, 2), which means that N is in the above range for some even k. - M. F. Hasler, Jul 02 2023

Michael S. Branicky, Table of n, a(n) for n = 1..10000

Cf. A363905, A363909: concat(n^2, n^3) has each digit at least once / twice.
Cf. A171102: pandigital numbers.
Cf. A036744, A054038, A071519 and A156977 for "pandigital squares".
Cf. A119735: n^3 is pandigital.

fQ[n_] := Length@ Union[ Count[ Sort[ Join[ IntegerDigits[n^2], IntegerDigits[n^3]]], #] & /@ Range[0, 9]] == 1; Select[ Range@ 52000000, fQ] (* Robert G. Wilson v, Jul 01 2023 *)

is(n)={my(v=concat(digits(n^2),digits(n^3)), c=#v); c%10==0 && vecsort(v)==[0..c-1]\(c\10)}
for(n=1,1e6, is(n)&& print1(n","))

a(13) and beyond from Michael S. Branicky, Jun 28 2023

A121322 Numbers m such that m^5 contains every digit at least once.

Original entry on oeis.org

309, 418, 462, 474, 575, 635, 662, 699, 702, 713, 737, 746, 747, 748, 765, 771, 795, 838, 875, 876, 892, 897, 943, 945, 976, 1009, 1012, 1018, 1033, 1072, 1104, 1107, 1137, 1143, 1149, 1167, 1174, 1183, 1187, 1195, 1203, 1233, 1248, 1249, 1269, 1292

Offset: 1

Tanya Khovanova, Aug 25 2006

			309^5 = 2817036000549 contains every digit at least once.

T. D. Noe, Table of n, a(n) for n = 1..10000

Cf. A054038 (with m^2), A119735 (with m^3).

Select[Range@1500, 1+Union@IntegerDigits@(#^5)==Range@10&] (* Hans Rudolf Widmer, Nov 02 2021 *)

isok(m) = #Set(digits(m^5)) == 10; \\ Michel Marcus, Nov 02 2021

def ok(n): return len(set(str(n**5))) == 10
print([m for m in range(1293) if ok(m)]) # Michael S. Branicky, Nov 02 2021

cp's OEIS Frontend

A029776 Digits of n appear in n^3.

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A235807 Numbers n such that n^3 has one or more occurrences of exactly five different digits.

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A235811 Numbers n such that n^3 has one or more occurrences of exactly nine different digits.

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A363905 Numbers whose square and cube taken together contain each decimal digit.

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A121321 Numbers k such that every digit occurs at least once in k^4.

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

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Magma

Mathematica

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A235809 Numbers k such that k^3 has one or more occurrences of exactly seven different digits.

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