A154566 -id:A154566 - cp's OEIS frontend

A154532 a(n) is the largest 10-digit number whose n-th power contains each digit (0-9) n times, or -1 no such number exists.

9876543210, 9994363488, 9999257781, 9999112926, 9995722269, 9999409158, 9998033316, 9993870774, 9986053188, 9964052493, 9975246786, 9966918135, 9938689137, 9998781633, 9813743148, 9970902252, 9740383767, 9829440591, 9873773268, 9985819785, 9766102146, 9863817738

Offset: 1

Zhining Yang, Jan 11 2009

A number with 10*n digits may have all ten digits (0-9) repeated n times. The probability of this is (10n)!/((n!)^10 * 10^((10*n)-10^(10*n-1)). There are 10^10-10^(10-1/n)) numbers which are n-th powers of 10-digit numbers. So there may exist Count=(10n)!*(10^10-10^(10-1/n)))/((n!)^10 * 10^((10*n)-10^(2*n-1)) numbers with the desired property.

From a(23) to a(110) the only terms which exist are a(24)=9793730157, a(26)=9347769564, a(35)=9959167017, and a(38)=9501874278. (The other values of a(n) are -1.) - Zhining Yang, Oct 05 2022

			a(18) = 9829440591, so each digit (0-9) appears 18 times in the decimal expansion of 9829440591^18.

northwolves et al., Ten Digit Numbers
Zhining Yang, Largest Ten Digit Powers

Cf. A010784, A078255, A154566.

def flag(p, n):
    b = True
    for i in range(10):
        if not p.count(str(i)) == n:
            b = False
            break
    return b
def a(n):
    for i in range(10 ** 10 - 1, 3 * int(10 ** (10 - 1 / n) / 3), -3):
        p = str(i ** n)
        if flag(p, n) == True:
            return i
            break
for i in range(1, 23):
    print(i, a(i))  # Zhining Yang, Oct 10 2022

def flag(p, n):
    return all(p.count(d) == n for d in "0123456789")
def a(n):
    return next(i for i in range(10**10-1,3*int(10**(10-1/n)/3), -3) if flag(str(i**n), n))
for i in range(2, 23):
    print(i,a(i))  # Michael_S._Branicky, Oct 10 2022

Edited by N. J. A. Sloane, Jan 12 2009
a(19)-a(22) from Zhining Yang, Oct 05 2022
Definition revised by N. J. A. Sloane, Nov 22 2022

A154871 Numbers n such that n^6 contains every digit exactly 4 times.

Original entry on oeis.org

3470187, 3554463, 3887058, 4328241, 4497738

Offset: 1

Zhining Yang, Jan 16 2009

a(1) is also the number of A074205

			3887058^6=3449261536602702380309782557611971988544, which contains 4 times of each digit 0-9. Total 5 terms

A054038 A074205 A154532 A154566 A154818

A154873 Numbers n such that n^5 contains every digit exactly 3 times.

Original entry on oeis.org

643905, 680061, 720558, 775113, 840501, 878613, 984927

Offset: 1

Zhining Yang, Jan 16 2009

a(5) is also the number of A074205

			840501^5=419460598737334268928156702501, which contains exactly 3 times of each digit 0-9. Total 7 terms

A054038 A074205 A154532 A154566 A154818

Select[Range[630972,999978],Union[DigitCount[#^5]]=={3}&] (* Harvey P. Dale, May 01 2021 *)

A154874 Numbers k such that k^3 contains every digit exactly twice.

Original entry on oeis.org

2158479, 2190762, 2205528, 2219322, 2301615, 2330397, 2336268, 2345811, 2358828, 2359026, 2367609, 2388534, 2389119, 2389638, 2397132, 2428986, 2504736, 2524974, 2536152, 2583258, 2590125, 2607222, 2620827, 2622012, 2647866, 2649369, 2658636, 2671593

Offset: 1

Zhining Yang, Jan 16 2009

This sequence has 138 terms.

			2358828^3 = 13124683009764879552, which contains each digit 0..9 exactly twice.

Nathaniel Johnston, Table of n, a(n) for n = 1..138 (full sequence)

Cf. A054038, A074205, A154532, A154566, A154871, A154873.

lim:=floor((10^20)^(1/3)): for j from ceil((10^19)^(1/3)) to lim do d:=convert(j^3,base,10): doubdig:=true: for k from 0 to 9 do if(numboccur(d,k)<>2)then doubdig:=false:break: fi: od: if(doubdig)then print(j); fi: od: # Nathaniel Johnston, May 28 2011

With[{cmin=Ceiling[Surd[10^19,3]],cmax=Floor[Surd[10^20,3]]},Select[ Range[ cmin, cmax], Union[ DigitCount[#^3]]=={2}&]] (* Harvey P. Dale, Nov 17 2018 *)

A154818 Numbers k such that k^4 contains every digit exactly twice.

Original entry on oeis.org

69636, 70215, 77058, 80892

Offset: 1

Zhining Yang, Jan 15 2009

77058^4 = 35259076387041812496, which contains 2 of each digit 0-9. There are just 4 terms.

Cf. A054038, A074205, A154532, A154566.

Keywords fini and full added. - R. J. Mathar, Jan 17 2009

A357755 Number of solutions for a 10-digit number whose n-th power contains each digit (0-9) exactly n times.

Original entry on oeis.org

3265920, 468372, 65663, 15487, 5020, 1930, 855, 417, 246, 114, 97, 45, 33, 24, 20, 18, 7, 6, 1, 3, 2, 3, 0, 1, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1

Offset: 1

Zhining Yang, Nov 26 2022

A number with 10*n digits may have all ten digits (0-9) repeated n times. The probability of this is (10*n)!/((n!)^10 * (10^(10*n)-10^(10*n-1))). There are 10^10-10^(10-1/n) numbers which are n-th powers of 10-digit numbers. So there may exist Count = (10*n)!*(10^10-10^(10-1/n))/((n!)^10 * (10^(10*n)-10^(10*n-1))) numbers with the desired property.

No solutions were found for n = 39 to 1000.

			a(20) = 3 because there are 3 10-digit numbers (8951993472, 9921107394, and 9985819785) whose 20th power contains each digit (0-9) 20 times.

Cf. A010784, A078255, A154532, A154566.

def flag(p, n):
    return all(p.count(d) == n for d in "0123456789")
def a(n):
    num=0
    for i in range(10**10-1, 3*int(10**(10-1/n)/3), -3):
        if flag(str(i**n), n):
            num+=1
    return(num)

A154875 Numbers k such that k^4 contains every digit exactly 3 times.

Original entry on oeis.org

17824719, 17940018, 18027474, 18197931, 18326025, 18798396, 18915888, 18929424, 19027455, 19149462, 19180275, 19196064, 19235673, 19311297, 19322913, 19324275, 19328322, 19455918, 19522575, 19757886, 19793664

Offset: 1

Zhining Yang, Jan 16 2009

			22807116 ^ 4 = 270571148920443982076865351936, which contains exactly 3 times of each digit 0-9.

Harvey P. Dale, Table of n, a(n) for n = 1..172

Cf. A054038, A074205, A154532, A154566, A154871, A154873, A154874.

Select[Range[17824000,31608000],Union[Tally[IntegerDigits[#^4]][[All,2]]]=={3}&] (* Harvey P. Dale, Dec 24 2016 *)

is(n) = my(v=vector(10), d=digits(n^4)); if(#d!=30,return(0)); for(i=1, #d, v[d[i]+1]++; if(v[d[i]+1] > 3, return(0))); 1 \\ David A. Corneth, Aug 19 2025

A217535 Least number having in its decimal representation each digit n times.

Original entry on oeis.org

1023456789, 10012233445566778899, 100011222333444555666777888999, 1000011122223333444455556666777788889999, 10000011112222233333444445555566666777778888899999, 100000011111222222333333444444555555666666777777888888999999

Offset: 1

M. F. Hasler, Oct 05 2012

Alois P. Heinz, Table of n, a(n) for n = 1..100

Cf. A050278, A051018, A049363, A074205, A154566, A204047, A020666, A020667, A061604, A180489.

A subsequence of A171102.

a:= n-> parse(cat(1,0$n,1$(n-1),seq(i$n, i=2..9))):
seq(a(n), n=1..10);  # Alois P. Heinz, Jun 25 2017

A217535(n)=sum(d=1,9,10^(n-(d==1))\9*d*10^(n*(9-d)))+10^(10*n-1)

a(n) ~ 10^(10n-1). See PARI code for an exact formula.

A370667 Largest pandigital number whose n-th power contains each digit (0-9) exactly n times.

Original entry on oeis.org

9876543210, 9876124053, 9863527104, 9846032571, 9847103256, 9247560381

Offset: 1

Zhining Yang, Mar 13 2024

If an n-th power of a pandigital number k contains each digit (0-9) exactly n times, it implies that 10^(10 - 1/n) <= 9876543210, so n <= 185. It's easy to verify that no solutions exist for n=7 to 185.

			a(4) = 9846032571 because it is the largest 10-digit number that contains each digit (0-9) exactly once and its 4th power 9398208429603554221689707364750715341681 contains each digit (0-9) exactly 4 times.

Cf. A050278, A199630, A199631, A199632, A199633, A078255, A154532, A154566, A357755, A365144.

s=FromDigits/@Permutations[Range[0,9]];For[n=1,n<=6,n++,For[k=Length@s,k>0,k--,If[Count[Tally[IntegerDigits[s[[k]]^n]][[All,2]],n]==10,Print[{n,s[[k]]}];Break[]]]]

from itertools import permutations
a=[]
for n in range(1,7):
    for k in [int(''.join(d)) for d in permutations('9876543210', 10)]:
        if all(str(k**n).count(d) ==n for d in '0123456789'):
            a.append(k)
            break
print(a)

A154876 10-digit numbers n such that n^16 contains every digit exactly 16 times.

Original entry on oeis.org

8691229761, 8776040742, 8800099059, 8812428855, 8813522223, 8815323864, 8823675177, 8886940968, 9239038038, 9324907263, 9480130515, 9500938647, 9643844169, 9801034758, 9857840688, 9872688021, 9962545842, 9970902252

Offset: 1

Zhining Yang, Jan 16 2009

The search program was provided by wuxinren(http://bbs.emath.ac.cn/space-uid-80.html): http://bbs.emath.ac.cn/attachment.php?aid=697&k=37ef434325a887e1a6f268d69d06192a&t=1232126232

			8691229761^16=1059984945135973085116625441940958734567890938937942910046410302827750560860737374626331724228885721853160790705924439371252226476405367618058329962361885148161 means that 16th power of 8691229761 has all digit(0-9) each for 16 times exactly

Ten digit numbers

Cf. A154874 A054038 A074205 A154532 A154566 A154871 A154873

cp's OEIS Frontend

A154532 a(n) is the largest 10-digit number whose n-th power contains each digit (0-9) n times, or -1 no such number exists.

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A154871 Numbers n such that n^6 contains every digit exactly 4 times.

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A154873 Numbers n such that n^5 contains every digit exactly 3 times.

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Mathematica

A154874 Numbers k such that k^3 contains every digit exactly twice.

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Maple

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A154818 Numbers k such that k^4 contains every digit exactly twice.

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A357755 Number of solutions for a 10-digit number whose n-th power contains each digit (0-9) exactly n times.

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Python

A154875 Numbers k such that k^4 contains every digit exactly 3 times.

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Mathematica

PARI

A217535 Least number having in its decimal representation each digit n times.

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Maple

PARI

Formula

A370667 Largest pandigital number whose n-th power contains each digit (0-9) exactly n times.

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