A199632 -id:A199632 - cp's OEIS frontend

A050278 Pandigital numbers: numbers containing the digits 0-9. Version 1: each digit appears exactly once.

1023456789, 1023456798, 1023456879, 1023456897, 1023456978, 1023456987, 1023457689, 1023457698, 1023457869, 1023457896, 1023457968, 1023457986, 1023458679, 1023458697, 1023458769, 1023458796, 1023458967, 1023458976, 1023459678, 1023459687, 1023459768

Offset: 1

Eric W. Weisstein, Dec 11 1999

This is a finite sequence with 9*9! = 3265920 terms: a(9*9!) = 9876543210.
A171102 is the infinite version, where each digit must appear at least once.
More precisely, this is exactly the subset of the first 9*9! terms of A171102. - M. F. Hasler, Jan 05 2020
Subsequence of A134336 and of A178403; A178401(a(n)) = 1. - Reinhard Zumkeller, May 27 2010
Smallest prime factors: A178775(n) = A020639(a(n)). - Reinhard Zumkeller, Jun 11 2010
A178788(a(n)) = 1. - Reinhard Zumkeller, Jun 30 2010
All these numbers are composite because the sum of the digits, 45, is divisible by 9. - T. D. Noe, Nov 09 2011
This is the 10th row of the array T(k,n) = n-th number in which the number of distinct base-10 digits is k. A031969 is the 4th row. A220063 is the 5th row. A220076 is the 6th row. A218019 is the 7th row. A219743 is the 8th row. - Jonathan Vos Post, Dec 05 2012
From Hieronymus Fischer, Feb 13 2013: (Start)
The sum of all terms is 9!*49444444440 = 17942399998387200.
General formula for the sum of all terms of the finite sequence of the corresponding base-p pandigital numbers with p places: sum = ((p^2 - p - 1)*(p^p - 1) + p - 1)*(p-2)!/2.
General formula for the sum of all terms (interpreted as decimal permutational numbers with exactly d+1 different digits from the range 0..d < 10): sum = (d+1)!*((10d - 1)*10^d - d + 1)/18, d > 1.
(End)

Robert G. Wilson v, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Pandigital Number
Chai Wah Wu, Pandigital and penholodigital numbers, arXiv:2403.20304 [math.GM], 2024. See p. 1.

Cf. A171102, A050288, A050289.
Cf. A199630, A199631, A114260, A199632, A199633.
Cf. A031969, A050278, A220063, A220076, A218019, A219743.

Select[ FromDigits@# & /@ Permutations[ Range[0, 9]], # > 10^9 &, 20] (* Robert G. Wilson v, May 30 2010, Jan 17 2012 *)

A050278(n)={ my(b=vector(9,k,1+(n+9!-1)%(k+1)!\k!), t=b[9]-1, d=vector(9,i,i+(i>t)-1)); for(i=1,8, t=10*t+d[b[9-i]]; d=vecextract(d,Str("^"b[9-i]))); t*10+d[1]} \\ M. F. Hasler, Jan 15 2012

is_A050278(n)={ 9<#vecsort(Vecsmall(Str(n)),,8) & n<1e10 } /* assuming that n is a nonnegative integer */ /* M. F. Hasler, Jan 10 2012 */

a(n)=my(d=numtoperm(10,n+9!-1));sum(i=1,#d,(d[i]-1)*10^(#d-i)) \\ David A. Corneth, Jun 01 2014

from itertools import permutations
A050278_list = [int(''.join(d)) for d in permutations('0123456789',10) if d[0] != '0'] # Chai Wah Wu, May 25 2015

A050278 = 9*A171571. - M. F. Hasler, Jan 12 2012

A050278(n) = A171102(n) for n <= 9*9!.

Edited by N. J. A. Sloane, Sep 25 2010 to clarify that this is a finite sequence

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

Original entry on oeis.org

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

A114261 Numbers k such that the 5th power of k contains exactly 5 copies of each digit of k.

Original entry on oeis.org

961527834, 7351062489, 8105632794, 8401253976, 8731945026, 9164072385, 9238750614, 9615278340, 9847103256, 72308154699, 73510624890, 81056327940, 83170652949, 83792140506, 84012539760, 87319450260, 91602408573, 91640723850, 92387506140, 96152783400, 98471032560

Offset: 1

Giovanni Resta, Nov 18 2005

Some of the early terms of the sequence are also pandigital, i.e. they contain all the 10 digits once. This is probably accidental, but quite curious!

All terms are divisible by 9. First decimal digit of a term is 6 or larger. - Chai Wah Wu, Feb 27 2024

			E.g. 961527834 is in the sequence since its 5th power 821881685441327565743977956591832631269739424 contains five 9's, five 6's, five 1's and so on.

Chai Wah Wu, Table of n, a(n) for n = 1..26

Cf. A114258, A114259, A114260, A199632.

from itertools import count, islice
from sympy import integer_nthroot
def A114261_gen(): # generator of terms
    for l in count(1):
        a = integer_nthroot(10**(5*l-1),5)[0]
        if (a9:=a%9):
            a += 9-a9
        for b in range(a,10**l,9):
            if sorted(str(b)*5)==sorted(str(b**5)):
                yield b
A114261_list = list(islice(A114261_gen(),5)) # Chai Wah Wu, Feb 27 2024

a(8)-a(9) from Ray Chandler, Aug 23 2023

a(10)-a(21) from Chai Wah Wu, Feb 28 2024

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

A050278 Pandigital numbers: numbers containing the digits 0-9. Version 1: each digit appears exactly once.

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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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A114261 Numbers k such that the 5th power of k contains exactly 5 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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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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