A137214 -id:A137214 - cp's OEIS frontend

A020666 a(n)^n is the least n-th power containing every digit.

1023456789, 32043, 2326, 763, 309, 159, 56, 104, 49, 36, 25, 15, 25, 17, 17, 15, 16, 7, 5, 6, 6, 5, 11, 9, 14, 5, 5, 5, 5, 9, 5, 8, 11, 4, 4, 6, 5, 7, 3, 5, 4, 4, 6, 4, 3, 6, 3, 3, 4, 4, 5, 4, 3, 6, 4, 4, 3, 4, 4, 3, 3, 3, 3, 3, 3, 4, 3, 2, 3, 2, 3, 3, 3, 3, 4, 3, 3, 3, 2, 3, 4, 2, 3, 2, 3, 3, 2, 2, 2

Offset: 1

David W. Wilson

It is extremely probable that a(n) = 2 for all n >= 169.

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Cf. A020667, A112388, A137214.

def a(n):
    if n == 1: return 1023456789
    an = 2
    while not(len(set(str(an**n))) == 10): an += 1
    return an
print([a(n) for n in range(1, 90)]) # Michael S. Branicky, Jul 04 2021

A020667 Least n-th power containing every digit.

Original entry on oeis.org

1023456789, 1026753849, 12584301976, 338920744561, 2817036000549, 16157819263041, 1727094849536, 13685690504052736, 1628413597910449, 3656158440062976, 2384185791015625, 129746337890625, 1490116119384765625, 168377826559400929, 2862423051509815793

Offset: 1

David W. Wilson

It is extremely probable that a(n) = 2^n for all n >= 169.

Seiichi Manyama, Table of n, a(n) for n = 1..3321

Cf. A020666, A137214.

def a(n):
    if n == 1: return 1023456789
    a020667n = 2
    while not(len(set(str(a020667n**n))) == 10): a020667n += 1
    return a020667n**n
print([a(n) for n in range(1, 16)]) # Michael S. Branicky, Jul 04 2021

a(n) = A020666(n)^n. - Seiichi Manyama, Dec 05 2024

A112388 a(n) is the smallest prime p such that p^n contains every digit.

Original entry on oeis.org

10123457689, 101723, 5437, 2339, 1009, 257, 139, 173, 83, 67, 31, 29, 37, 17, 17, 47, 19, 7, 5, 23, 23, 5, 11, 11, 17, 5, 5, 5, 5, 11, 5, 11, 11, 5, 5, 7, 5, 7, 3, 5, 5, 7, 7, 7, 3, 7, 3, 3, 5, 5, 5, 5, 3, 7, 7, 5, 3, 7, 5, 3, 3, 3, 3, 3, 3, 5, 3, 2, 3, 2, 3, 3, 3, 3, 5, 3, 3, 3, 2, 3, 5, 2

Offset: 1

Tanya Khovanova, Dec 05 2005

Conjecture: a(n)=2 for all n>168. Checked up to n = 20000. - Robert Israel, Aug 28 2020

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

Cf. A020666, A020665, A007377, A137214.

f:= proc(n) local k;
  k:= 1:
  do k:= nextprime(k);
    if convert(convert(k^n,base,10),set) = {$0..9} then return k fi
  od
end proc:
f(1):= 10123457689:
map(f, [$1..100]); # Robert Israel, Aug 28 2020

f[n_] := Block[{k = 1}, While[ Union@IntegerDigits[ Prime[k]^n] != {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, k++ ]; Prime[k]]; Array[f, 82] (* Robert G. Wilson v, Dec 06 2005 *)

from sympy import nextprime
def a(n):
    if n == 1: return 10123457689
    p = 2
    while not(len(set(str(p**n))) == 10): p = nextprime(p)
    return p
print([a(n) for n in range(1, 83)]) # Michael S. Branicky, Jul 04 2021

More terms from Robert G. Wilson v, Dec 06 2005

A240069 The number n^k has all 10 decimal digits starting at k = a(n), or a(n) = 0 if 10 digits are not possible.

Original entry on oeis.org

0, 169, 107, 85, 66, 65, 62, 57, 54, 0, 42, 52, 38, 35, 35, 43, 28, 26, 45, 169, 30, 25, 51, 24, 30, 32, 29, 29, 46, 107, 29, 19, 25, 35, 19, 33, 26, 18, 42, 85, 24, 20, 21, 30, 40, 33, 16, 30, 17, 66, 30, 30, 31, 19, 18, 34, 20, 32, 28, 65, 27, 20, 25, 29, 18, 16

Offset: 1

T. D. Noe, Apr 01 2014

It appears that numbers of the form 2 * 10^i have the longest period, 169.

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A137214 (number of distinct decimal digits in 2^n).

mx = 1000; Table[s = Table[Length[Union[IntegerDigits[n^k]]], {k, 0, mx}]; pos = Position[s, 10]; If[pos == {}, 0, 1 + mx - Position[Differences[Reverse[s]], _?(# != 0 &)][[1, 1]]], {n, 100}]

A309432 Number of distinct digits in decimal representation of n^2.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 2, 3, 3, 3, 3, 2, 2, 2, 3, 3, 3, 2, 3, 3, 3, 2, 3, 4, 4, 3, 3, 4, 4, 2, 3, 3, 3, 4, 4, 4, 3, 3, 3, 4, 4, 3, 4, 4, 4, 4, 4, 3, 4, 3, 4, 3, 4, 3, 3, 4, 3, 4, 3, 3, 4, 3, 4, 4, 4, 4, 3, 3, 3, 4, 4, 3, 3, 4, 3, 4, 3, 4

Offset: 0

Hauke Löffler, Aug 01 2019

			a(0) = 1 because 0^2 = 0 has 1 distinct digit (0).
a(5) = 2 because 5^2 = 25 has 2 distinct digits (2, 5).
a(10) = 2 because 10^2 = 100 has 2 distinct digits (0, 1).

Michael De Vlieger, Table of n, a(n) for n = 0..10000

Cf. A000290 (squares), A043537, A054039, A137214, A185679, A309431.

Array[Count[DigitCount[#^2], ?(# > 0 &)] &, 105, 0] (* _Michael De Vlieger, Jun 22 2022 *)

a(n) = if(n==0, 1, #Set(digits(n^2))); \\ Jinyuan Wang, Aug 02 2019, Jun 22 2022 [a(0)=1 corrected by Georg Fischer, Jun 22 2022]

def a(n): return len(set(str(n*n)))
print([a(n) for n in range(87)]) # Michael S. Branicky, Jun 22 2022

[len(set(Integer((n^2)).digits(padto=1))) for n in range(25) ]

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A020666 a(n)^n is the least n-th power containing every digit.

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A020667 Least n-th power containing every digit.

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A112388 a(n) is the smallest prime p such that p^n contains every digit.

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A240069 The number n^k has all 10 decimal digits starting at k = a(n), or a(n) = 0 if 10 digits are not possible.

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A309432 Number of distinct digits in decimal representation of n^2.

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