A365144 -id:A365144 - cp's OEIS frontend

A199630 Numbers having each digit once and whose square has each digit twice.

3175462089, 3175804269, 3204957816, 3206549178, 3210754689, 3254196708, 3260974851, 3275409816, 3284591706, 3290581476, 3406829517, 3410856297, 3459186720, 3469857012, 3475806912, 3501249678, 3512067849, 3519876240, 3549716208, 3564980172, 3587902614

Offset: 1

T. D. Noe, Nov 09 2011

			3175462089^2 = 10083559478676243921.

T. D. Noe, Table of n, a(n) for n = 1..534 (all terms).
Patrick De Geest, The nine digits (page 4) with some ten digit (pandigital) exceptions
Author?, All terms

Cf. A050278 (pandigital numbers), A199631, A365144, A199632, A199633. Subsequence of A114258.

t = Select[Permutations[Range[0, 9]], #[[1]] > 0 &]; t2 = Select[t, Union[DigitCount[FromDigits[#]^2]] == {2} &]; FromDigits /@ t2

A199631 Numbers having each digit once and whose cube has each digit three times.

Original entry on oeis.org

4680215379, 4752360918, 4765380219, 4915280637, 5063248197, 5164738920, 5382417906, 5426370189, 5429013678, 5628130974, 5679321048, 5697841320, 5762831940, 5783610492, 5786430129, 5903467821, 6019285734, 6053147982, 6095721483, 6143720958, 6158723094

Offset: 1

T. D. Noe, Nov 09 2011

			4680215379^3 = 102517384602327906545167884939.

T. D. Noe, Table of n, a(n) for n = 1..74 (all terms).
Patrick De Geest, The nine digits (page 4) with some ten digit (pandigital) exceptions
Author?, All terms

Cf. A050278 (pandigital numbers), A199630, A365144, A199632, A199633. Subsequence of A114259.

t = Select[Permutations[Range[0, 9]], #[[1]] > 0 &]; t2 = Select[t, Union[DigitCount[FromDigits[#]^3]] == {3} &]; FromDigits /@ t2

A199633 Numbers having each digit once and whose 6th power has each digit six times.

Original entry on oeis.org

7025869314, 7143258096, 7931584062, 8094273561, 8920416357, 9247560381

Offset: 1

T. D. Noe, Nov 09 2011

There are 6 numbers total. There are no higher powers with this property.

			7025869314 ^6 = 120281934463386157260042215510596389732740014997586987548736.

Patrick De Geest, The nine digits (page 4) with some ten digit (pandigital) exceptions
Author?, All terms

Cf. A050278 (pandigital numbers), A199630, A199631, A365144, A199632.

t = Select[Permutations[Range[0, 9]], #[[1]] > 0 &]; t2 = Select[t, Union[DigitCount[FromDigits[#]^6]] == {6} &]; FromDigits /@ t2

A114260 Numbers k such that the 4th power of k contains exactly 4 copies of each digit of k.

Original entry on oeis.org

5702631489, 7264103985, 7602314895, 7824061395, 8105793624, 8174035962, 8304269175, 8904623175, 8923670541, 9451360827, 9785261403, 9804753612, 9846032571, 57026314890, 59730829461, 60947591328, 64017823995, 65190218436, 67024081935, 70645192839, 72641039850, 74991208356

Offset: 1

Giovanni Resta, Nov 18 2005

First 13 terms of the sequence are also pandigital, i.e., they contain all the 10 digits at least once. This is probably accidental, but quite curious.
If a(n) is in the sequence, then 10*a(n) is also in the sequence. So 10^k*a(n) is also in the sequence for positive integers k. Thus this sequence differs from A365144. - Ray Chandler, Aug 23 2023
From David A. Corneth, Aug 30 2023: (Start)
Not all terms are pandigital, for example 65190218436, 75932056341 and 83581076421 are not.
For any k ends in, say, 425742 (which has six digits) the last six digits of k^4 are fixed. So if k ends in 425742 then k^4 ends in 425742^2 mod 10^6 = 318096 and so it must have a 3, 1, 8, 0, 9 and 6 none of which are contained in 425742.
Therefore if k ends in 425742 then it must have at least 12 digits. In a search for terms <= 10^11 this eases the search, in a search for terms <= 10^12 this leaves only 6! * 5 / 6 = 600 cases to check instead of 10^6.
An additional idea one might use is that the number of digits of k^4 must be a multiple of 4. I.e. 10^(4*m + 3) <= k^4 <= 10^(4*m + 4) so 10^m * 10^0.75 < m < 10^(m + 1) (all strict inequalities as 10^0.75 (cf. A210522) is irrational) which tells us a bunch about the leading digits of k.
Checking pandigital numbers separately might ease the search. That way if the union of some k and last q digits of k^4 is all decimal digits one could end the search there. This goes for example for 100426. If a term ends in 100426 then it has all decimal digits. (End)
All terms are divisible by 9. First decimal digit of a term is 5 or larger. - Chai Wah Wu, Feb 27 2024

			5702631489 is a term since its 4th power 1057550783692741389295697108242363408641 contains four 5's, four 7's, four 0's and so on.

Chai Wah Wu, Table of n, a(n) for n = 1..69 (terms 1..53 from David A. Corneth)
Carlos Rivera, Puzzle 971. The cube of N such that..., The Prime Puzzles & Problems Connection.

Cf. A114258, A114259, A114261, A210522, A365144.

from itertools import count, islice
from sympy import integer_nthroot
def A114260_gen(): # generator of terms
    for l in count(1):
        a = integer_nthroot(10**(4*l-1),4)[0]
        if (a9:=a%9):
            a += 9-a9
        for b in range(a,10**l,9):
            if sorted(str(b)*4)==sorted(str(b**4)):
                yield b
A114260_list = list(islice(A114260_gen(), 5)) # Chai Wah Wu, Feb 27 2024

a(14) from Ray Chandler, Aug 24 2023

More terms from David A. Corneth, Aug 30 2023

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)

A199634 Number of pandigital numbers raised to the n-th power is a number in which each digit appears n times.

Original entry on oeis.org

3265920, 534, 74, 13, 8, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

T. D. Noe, Nov 09 2011

Note that a(1) is the number of pandigital numbers, 10! - 9! = 9*9!. For n > 1, it is the number of numbers in A199630, A199631, A365144, A199632, and A199633.

The Mathematica code takes many hours to run. The program stops after doing power 186 because the largest pandigital number 9876543210 raised to any greater power does not produce enough digits.

Patrick De Geest, The nine digits (page 4) with some ten digit (pandigital) exceptions
Author?, All terms

Cf. A050278 (pandigital numbers), A199630, A199631, A365144, A199632, A199633.

t = {}; perm = Select[Permutations[Range[0, 9]], #[[1]] > 0 &]; len = Length[perm]; Print[{1, len}]; AppendTo[t, len]; pwr = 1; i = 1; While[pwr++; i < len, While[IntegerLength[FromDigits[perm[[i]]]^pwr] < 10*pwr, i++]; cnt = 0; Do[If[Union[DigitCount[FromDigits[perm[[j]]]^pwr]] == {pwr}, cnt++], {j, i, len}]; Print[{pwr, cnt}]; AppendTo[t, cnt]]

A363160 Smallest positive integer m with all digits distinct such that m^n contains each digit of m exactly n times, or -1 if no such m exists.

Original entry on oeis.org

1, 406512, 516473892, 5702631489, 961527834, 7025869314, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1

Offset: 1

Jean-Marc Rebert, Sep 07 2023

For 7 <= n <= 185, I tried all possibilities with at most 10 distinct digits and I found no solution.
9876543210^186 has only 1859 < 186 * 10 = 1860 digits, so a(n) = -1 for n = 186.
So 9876543210^n has fewer than 10*n digits for n >= 186, so a(n) = -1 for n >= 186.

			a(1) = 1, because 1^1 = 1 has each digit of 1, 1 time, and no lesser number > 0 satisfies this.
a(2) = 406512, because 406512 has distinct digits, 406512^2 = 165252006144 has each digit of 406512, 2 times, and no lesser number satisfies this.
n a(n)        a(n)^n
1 1           1
2 406512      165252006144
3 516473892   137766973511455269432948288
4 5702631489  1057550783692741389295697108242363408641
5 961527834   821881685441327565743977956591832631269739424
6 7025869314  120281934463386157260042215510596389732740014997586987548736

Cf. A114258, A114259, A114260, A114261, A199630, A199631, A199632, A199633, A365144, A114260.

hasDistinctDigitsQ[m_Integer?NonNegative]:=Length@IntegerDigits@m==Length@DeleteDuplicates@IntegerDigits@m;validNumberQ[n_Integer?NonNegative,m_Integer?NonNegative]:=AllTrue[Tally@IntegerDigits@m,Function[{digitFreq},MemberQ[Tally@IntegerDigits[m^n],{digitFreq[[1]],n*digitFreq[[2]]}]]];a[n_Integer?Positive,ex_Integer?Positive]:=Module[{m=1},Monitor[While[True,If[hasDistinctDigitsQ[m]&&validNumberQ[n,m],Return[m]];m++;If[m>10^(ex*n),Return[-1]];];m,m]];Table[a[n,7],{n,1,7}] (* Robert P. P. McKone, Sep 09 2023 *)

A371469 Least pandigital number whose n-th power contains each digit (0-9) exactly n times.

Original entry on oeis.org

1023456789, 3175462089, 4680215379, 5702631489, 7351062489, 7025869314

Offset: 1

Zhining Yang, Apr 01 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.

For the largest pandigital number whose n-th power contains each digit (0-9) exactly n times, see A370667.

			a(4) = 5702631489 because it is the least 10-digit number that contains each digit (0-9) exactly once and its 4th power 1057550783692741389295697108242363408641 contains each digit (0-9) exactly 4 times.

Cf. A050278, A199630, A199631, A199632, A199633, A078255, A154532, A154566, A357755, A365144, A370667.

s = FromDigits /@ Permutations[Range[0, 9]]; For[n = 1, n < 7, n++,
 For[k = 1, k <= Length@s, k++,
  If[Count[Tally[IntegerDigits[s[[k]]^n]][[All, 2]], n] == 10,
   Print[{n, s[[k]]}]; Break[]]]]

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

cp's OEIS Frontend

A199630 Numbers having each digit once and whose square has each digit twice.

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A199631 Numbers having each digit once and whose cube has each digit three times.

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A199633 Numbers having each digit once and whose 6th power has each digit six times.

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A114260 Numbers k such that the 4th power of k contains exactly 4 copies of each digit of k.

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

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A199634 Number of pandigital numbers raised to the n-th power is a number in which each digit appears n times.

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Mathematica

A363160 Smallest positive integer m with all digits distinct such that m^n contains each digit of m exactly n times, or -1 if no such m exists.

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Mathematica

A371469 Least pandigital number whose n-th power contains each digit (0-9) exactly n times.

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Python