A130694 -id:A130694 - cp's OEIS frontend

A043537 Number of distinct base-10 digits of n.

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2

Offset: 1

Clark Kimberling

a(A000079(A130694(n))) = 10. - Reinhard Zumkeller, Jul 29 2007
a(A000290(A016070(n))) = 2. - Reinhard Zumkeller, Aug 05 2010
a(n) = 10 for almost all n. - Charles R Greathouse IV, Oct 02 2013

R. Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A047726, A118668, A119797, A119798, A119799, A119823, A038378, A076493, A055642, A178788.

import Data.List (nub)
a043537 = length . nub . show  -- Reinhard Zumkeller, Apr 16 2011

A043537 := proc(n)
        convert(convert(n,base,10),set) ;
        nops(%) ;
end proc: # R. J. Mathar, Dec 22 2012

Count[DigitCount@ #, n_ /; n > 0] & /@ Range@ 120 (* Michael De Vlieger, Aug 29 2015 *)

a(n)=#vecsort(digits(n),,8) \\ Charles R Greathouse IV, Oct 02 2013

def A043537(n): return len(set(str(n))) # Dimiter Skordev, Oct 02 2021

A130696 Numbers k such that 2^k does not contain all ten decimal digits.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 69, 71, 72, 73, 74, 75, 76, 77, 78, 80, 81, 83, 85, 86, 90, 91, 92, 93, 99, 102, 107, 108, 153, 168

Offset: 1

Greg Dresden, Jul 10 2007

It is believed that 168 is the last number in this list; 2^168 is a 51-digit number that contains all the digits except (oddly enough) 2.

There are no more terms less than 10^10. - David Radcliffe, Apr 11 2019

			20 is in this list because 2^20 = 1048576, which doesn't contain all ten digits.
68 is the first number not in this list; 2^68 = 295147905179352825856 and this contains all ten digits.

Cf. A007377, A062518, A090493.

Complement of A130694.

A2 := {}; Do[If[Length[Union[ IntegerDigits[2^ n]]] != 10, A2 = Join[A2, {n}]], {n, 1, 3000}]; Print[A2]
Select[Range[10^6]-1,MemberQ[DigitCount[2^#],0]&] (* Hans Rudolf Widmer, Jun 23 2021 *)

hasalldigits(n) = #vecsort(digits(n), , 8)==10
is(n) = !hasalldigits(2^n) \\ Felix Fröhlich, Apr 11 2019

print([n for n in range(1000) if len(set(str(2**n))) < 10]) # David Radcliffe, Apr 11 2019

a(1) = 0 prepended by David Radcliffe, Apr 11 2019

A090493 Least k such that n^k contains all the digits from 0 through 9, or 0 if no such k exists.

Original entry on oeis.org

0, 68, 39, 34, 19, 20, 18, 28, 24, 0, 23, 22, 22, 21, 12, 17, 14, 21, 17, 51, 17, 18, 14, 19, 11, 18, 13, 11, 12, 39, 11, 14, 16, 14, 19, 10, 13, 14, 17, 34, 11, 17, 13, 16, 15, 11, 12, 12, 9, 18, 16, 11, 13, 10, 12, 7, 13, 11, 11, 20, 14, 18, 13, 14, 10, 13, 10, 9, 11, 18, 15

Offset: 1

Amarnath Murthy, Dec 03 2003

Note that the values of n for which a(n) = 1 have density 1.
Is it known that a(n)=0 only for n a power of 10? - Christopher J. Smyth, Aug 21 2014
a(n) >= ceiling(log_n(10)*9), whenever a(n)>0. This is because in order for an integer to have 10 digits its base-10 magnitude must be at least 9. - Ely Golden, Sep 06 2017

			a(5)=19: 5^19 = 19073486328125.

Seiichi Manyama, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe)
Ely Golden, Python program to generate a(n)

Cf. A020665, A062518, A171102.

Exponents of powers of k that contain all ten decimal digits: A130694 (k=2), A236673 (k=3), A284670 (k=5), A284672 (k=7).

a:= proc(n) local k;
   if n = 10^ilog10(n) then return 0 fi;
   for k from 1 do
     if nops(convert(convert(n^k,base,10),set))=10 then return k fi
   od
end proc:
seq(a(n),n=1..100); # Robert Israel, Aug 20 2014

Table[If[IntegerQ@ Log10[n], 0, SelectFirst[Range[#, # + 100] &@ Ceiling[9 Log[n, 10]], NoneTrue[DigitCount[n^#], # == 0 &] &]], {n, 71}] (* Michael De Vlieger, Sep 06 2017 *)

a(n) = if (n == 10^valuation(n, 10), return (0)); k=1; while(#vecsort(digits(n^k),,8)!=10, k++); k; \\ Michel Marcus, Aug 20 2014

def a(n):
  s = str(n)
  if n == 1 or (s.count('0')==len(s)-1 and s.startswith('1')):
    return 0
  k = 1
  count = 0
  while count != 10:
    count = 0
    for i in range(10):
      if str(n**k).count(str(i)) == 0:
        count += 1
        break
    if count:
      k += 1
    else:
      return k
n = 1
while n < 100:
  print(a(n), end=', ')
  n += 1
# Derek Orr, Aug 20 2014

a(10^e) = 0; a(m^e) = a(m)/e for e dividing a(m). - Reinhard Zumkeller, Dec 06 2004

More terms from Reinhard Zumkeller, Dec 06 2004

Corrected a(15), a(17), a(38), a(48), a(56) and a(65). (For each of these terms, the only 1 in n^k is the first digit.) - Jon E. Schoenfield, Sep 20 2008

A236673 Exponents of powers of 3 that contain all ten decimal digits.

Original entry on oeis.org

39, 45, 47, 48, 53, 57, 60, 61, 62, 63, 64, 65, 67, 69, 70, 71, 72, 73, 74, 76, 77, 78, 79, 80, 82, 83, 85, 86, 87, 88, 89, 90, 92, 93, 94, 95, 96, 97, 98, 99, 102, 103, 105, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123

Offset: 1

Derek Orr, Jan 29 2014

It is conjectured that after a(43), a(n) = n + 63 (i.e., natural numbers beginning with 107).

Complement of A236674.

			3^53 = 19383245667680019896796723, which contains two 1's, two 2's, three 3's, one 4, one 5, five 6's, three 7's, three 8's, four 9's and two 0's, so 53 is in the sequence.
3^57 = 1570042899082081611640534563, which contains four 1's, two 2's, two 3's, three 4's, three 5's, three 6's, one 7, three 8's, two 9's and five 0's.
58 is not in the sequence because there are no 5's in 3^58 = 4710128697246244834921603689.

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

Cf. A130694, A236674.

Select[Range[0, 200], Union[IntegerDigits[3^#]] == Range[0, 9] &] (* T. D. Noe, Jan 29 2014 *)

def PanDig(x):
  a = '1234567890'
  for n in range(10**3):
    count = 0
    for i in a:
      if str(x**n).count(i) > 0:
        count += 1
      else:
        break
    if count == len(a):
      print(n)

A284670 Exponents of powers of 5 that contain all ten decimal digits.

Original entry on oeis.org

19, 22, 26, 27, 28, 29, 31, 34, 35, 37, 39, 40, 41, 49, 50, 51, 52, 56, 57, 59, 60, 61, 62, 63, 64, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104

Offset: 1

Seiichi Manyama, Apr 01 2017

It is conjectured that a(n) = n + 40 for n > 25.

			5^19 = 19073486328125, which contains two 1's, two 2's, two 3's, one 4, one 5, one 6, one 7, two 8's, one 9 and one 0, so 19 is in the sequence.

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

Cf. A130694 (k=2), A236673 (k=3), this sequence (k=5).

Select[Range[110], Union[IntegerDigits[5^#]] == Range[0, 9] &] (* Indranil Ghosh, Apr 01 2017 *)

isok(n) = #vecsort(digits(5^n),,8) == 10; \\ Michel Marcus, Apr 01 2017

from sympy.ntheory.factor_ import digits
r10 = set(range(10))
print([n for n in range(1, 111) if set(sorted(digits(5**n)[1:])) == r10]) # Indranil Ghosh, Apr 01 2017

A215830 Numbers k such that 2^k contains each decimal digit at least twice.

Original entry on oeis.org

88, 104, 113, 114, 116, 117, 118, 119, 120, 125, 126, 131, 133, 134, 136, 140, 141, 142, 144, 145, 146, 147, 148, 150, 155, 156, 157, 159, 160, 161, 162, 163, 164, 165, 166, 170, 171, 172, 175, 177, 178, 179, 180, 181, 182, 185, 186, 187, 188, 189, 190, 191, 192

Offset: 1

M. F. Hasler, Aug 25 2012

A subsequence of A130694.
Motivated by J. Merickel's question about the least power of 2 in which all digits 0-9 occur a prime number of times. The first 4 terms of this sequence are all such that this is the case for all but one digit; see Examples.
Beyond 184, the numbers 195-197 and 229 are the only exponents < 10^4 which are not in this sequence. Is 229 the largest such number?

			Digit counts for 2^n:
.
  n\d| 0  1  2  3  4  5  6  7  8  9
  ---+-----------------------------
   88| 5  2  2  2  3  3  2  2  4* 2
  104| 5  3  6* 2  3  2  5  2  2  2
  113| 5  3  3  2  3  5  4* 3  2  5
  114| 3  7  2  5  4* 2  2  2  5  3
.
*nonprime counts

J. Merickel, Are these new questions?, Yahoo! group "primenumbers", Aug 24 2012.
James Merickel, Are these new questions?, message 24399 in primenumbers Yahoo group, Aug 24, 2012.

PARI
```
is_A215830(n)={my(c=vector(10),N=[1<1}
```

A284672 Exponents of powers of 7 that contain all ten decimal digits.

Original entry on oeis.org

18, 27, 34, 36, 37, 38, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100

Offset: 1

Seiichi Manyama, Apr 01 2017

nonn

It is conjectured that a(n) = n + 34 for n > 27.

			7^18 = 1628413597910449, which contains three 1's, one 2, one 3, three 4's, one 5, one 6, one 7, one 8, three 9's and one 0, so 18 is in the sequence.

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

Cf. A130694 (k=2), A236673 (k=3), A284670 (k=5), this sequence (k=7).

G.f.: x*(-x^28 + x^27 - x^7 + x^6 - x^4 - 5*x^3 - 2*x^2 - 9*x + 18)/(x - 1)^2 (as conjectured in comments). - Chai Wah Wu, Jan 26 2020

A347164 Positive integers k such that the decimal representation of 2^k ends with some permutation of the string "0123456789".

Original entry on oeis.org

7386, 11061, 15176, 16054, 19950, 24493, 26749, 29160, 33902, 42207, 43013, 44233, 45627, 52235, 54727, 56186, 59228, 59229, 59230, 60883, 62823, 63468, 65404, 69960, 71225, 71804, 75176, 78392, 89416, 96576, 96682, 97723, 98085, 98561, 102735, 104125, 105301, 105760

Offset: 1

Dimiter Skordev, Aug 20 2021

If k is a term of the sequence then some nonzero digit must occur more than once in the decimal representation of 2^k because 1+2+3+4+5+6+7+8+9=45, and 2^k is not divisible by 9. Thus 2^k>10^10 and therefore k>33 for any term k.
A positive integer k is a term of the sequence iff a decimal representation of the remainder of 2^k modulo 10^10 (possibly containing a leading zero) is a permutation of the string "0123456789".
Let d = 7812500 = 4*5^9 = phi(5^10) where phi is Euler's totient function. The remainders of the powers of 2 modulo 10^10 form an eventually periodic sequence with period d: if k >= 10 then 2^(k+d) - 2^k is divisible by 10^10 since 2^(k+d) - 2^k = 2^k*(2^d-1) and 10^10 = 2^10*5^10. Hence if k >= 10 then k + d is a term iff k is a term.
Actually the above equivalence holds for all positive integers k because neither k nor k + d is a term of the sequence for k < 10 (the decimal representations of the numbers 2^(k + d) with k = 1, 2, ..., 9 end, respectively, with the following strings: 3574218752, 7148437504, 4296875008, 8593750016, 7187500032, 4375000064, 8750000128, 7500000256, 5000000512).
There are 2795 terms not exceeding d. The last of them is 7808304, with decimal representation of the corresponding power of 2 ending with 9745238016.

			7386, 11061 and 15176 are in the sequence because the decimal representations of the corresponding powers of 2 end with 9815307264, 4706813952 and 0294875136, respectively.

Dimiter Skordev, Table of n, a(n) for n = 1..6000

Cf. A090493, A291926. Subsequence of A130694.

q:= n-> (l-> is({l[], `if`(nops(l)<10, 0, [][])}=
    {$0..9}))(convert(2&^n mod 10^10, base, 10)):
select(q, [$1..120000])[];  # Alois P. Heinz, Aug 23 2021

Select[Range[10^5],Union[If[Length[s=IntegerDigits@PowerMod[2,#,10^10]]==9,Join[{0},s],s]]==0~Range~9&] (* Giorgos Kalogeropoulos, Sep 03 2021 *)

isok(k) = my(d=digits(lift(Mod(2, 10^10)^k))); if (#d<10, d = concat(d, 0)); #Set(d) == 10; \\ Michel Marcus, Oct 01 2021

k,r,n=1,2,1
while n<=6000:
    s,t=set(),r
    for i in range(10):
        s.add(t%10)
        t=t//10
    if len(s)==10:
        print(n,k)
        n=n+1
    k,r=k+1,2*r%10**10

a(n+c) = a(n) + d with c=2795 and d as above.

A352040 a(n) is the least number k such that 2^k contains each of the 10 digits at least n times.

Original entry on oeis.org

68, 88, 119, 200, 209, 246, 291, 318, 396, 398, 443, 443, 495, 586, 592, 622, 646, 707, 758, 758, 813, 866, 875, 903, 923, 1001, 1022, 1022, 1105, 1111, 1111, 1231, 1243, 1245, 1327, 1342, 1419, 1453, 1453, 1454, 1534, 1536, 1537, 1626, 1676, 1699, 1699, 1763

Offset: 1

Amiram Eldar, Apr 16 2022

a(1)-a(8) were given in the solution to Problem 410 (Crux Mathematicorum, 1979), but a(2) = 170 is wrong.

a(1) was calculated by Rudolph Ondrejka in 1976.

			a(1) = 68 since 2^68 = 295147905179352825856 is the least power of 2 that contains all the 10 digits at least once.

Amiram Eldar, Table of n, a(n) for n = 1..10000
James Gary Propp, Problem 410, Crux Mathematicorum, Vol. 5, No. 1 (1979), p. 17; Solution to Problem 410, ibid., Vol. 5, No. 10 (1979), pp. 296-299.

Cf. A000079, A007377, A020862, A130694, A130696.

a:= proc(n) option remember; local k; for k from a(n-1)
      while min((p-> seq(coeff(p, x, j), j=0..9))(add(
            x^i, i=convert(2^k, base, 10))))Alois P. Heinz, Apr 22 2022

s = Table[Min[DigitCount[2^n, 10, Range[0, 9]]], {n, 1, 2500}]; Table[FirstPosition[s, _?(# >= n &)], {n, 1, Max[s]}] // Flatten

from sympy import ceiling, log
def A352040(n):
    k = 10*n-1+int(ceiling((10*n-1)*log(5,2)))
    s = str(c := 2**k)
    while any(s.count(d) < n for d in '0123456789'):
        c *= 2
        k += 1
        s = str(c)
    return k   # Chai Wah Wu, Apr 16 2022

Conjecture: a(n) ~ c*n, where c = 10*log_2(10) = 33.21928... .

a(n) >= (10n-1)*log_2(10), i.e., c = 10*log_2(10) is a lower bound on the asymptotic growth rate. - Chai Wah Wu, Apr 16 2022

A382681 Conjecturally, the numbers k (not multiples of 5) such that for all x >= 0, k*2^x has a '0' in its decimal expansion.

Original entry on oeis.org

7501221, 7508793, 10006109, 10625334, 12970254, 15002442, 15017586, 15685077, 17975049, 20012218, 20752359, 21250668, 22500771, 23501007, 24625029, 24875024, 25033207, 25034183, 25034771, 25940508, 29003907, 29057504, 29450021, 29590047, 29625044, 29850293, 30004884, 30035172, 30175941

Offset: 1

Brian Kehrig, Jun 02 2025

If k is a multiple of 5, then k*2^x always ends with '0' for x >= 1. We exclude these trivial cases.
If k is in this sequence, then so is k*2^x for all x >= 1.
All terms up to 30175941 have been tested up to x=10^6.

			7501221*2^0 = 7501221 contains a '0'
7501221*2^1 = 15002442 contains a '0'
7501221*2^2 = 30004884 contains a '0'
7501221*2^3 = 60009768 contains a '0'
7501221*2^4 = 120019536 contains a '0'
7501221*2^5 = 240039072 contains a '0'
...
Conjecturally, all further numbers of the form 7501221*2^k also contain '0'. Thus, 7501221 is in the sequence.

Cf. A007377, A027870, A130694.

cp's OEIS Frontend

A043537 Number of distinct base-10 digits of n.

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A130696 Numbers k such that 2^k does not contain all ten decimal digits.

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A090493 Least k such that n^k contains all the digits from 0 through 9, or 0 if no such k exists.

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A236673 Exponents of powers of 3 that contain all ten decimal digits.

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A284670 Exponents of powers of 5 that contain all ten decimal digits.

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A215830 Numbers k such that 2^k contains each decimal digit at least twice.

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A284672 Exponents of powers of 7 that contain all ten decimal digits.

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