A199630 -id:A199630 - 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

A114258 Numbers k such that k^2 contains exactly 2 copies of each digit of k.

Original entry on oeis.org

72576, 406512, 415278, 494462, 603297, 725760, 3279015, 4065120, 4152780, 4651328, 4915278, 4927203, 4944620, 4972826, 4974032, 4985523, 4989323, 5002245, 5016125, 6032970, 6214358, 6415002, 6524235, 7257600, 9883667

Offset: 1

Giovanni Resta, Nov 18 2005

From Chai Wah Wu, Feb 27 2024: (Start)
If k is a term, then k == 0 (mod 9) or k == 2 (mod 9) (see A370676).
First decimal digit of each term is 3 or larger. (End)

			72576 is in the sequence since its square 5267275776 contains four 7's, two 2's, two 5's and two 6's.

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (terms 1..645 from Seiichi Manyama)

Cf. A114259, A114260, A114261, A199630.

Cf. A061656, A061657, A061658, A061659, A061660, A061661, A061662, A061663, A061664.

from math import isqrt
from itertools import count, islice
def A114258_gen(): # generator of terms
    for l in count(1):
        a = isqrt(10**((l<<1)-1))
        if (a9:=a%9):
            a -= a9
        for b in range(a,10**l,9):
            for c in (0,2):
                k = b+c
                if sorted(str(k)*2)==sorted(str(k**2)):
                    yield k
A114258_list = list(islice(A114258_gen(),20)) # Chai Wah Wu, Feb 27 2024

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

A199632 Numbers having each digit once and whose 5th power has each digit five times.

Original entry on oeis.org

7351062489, 8105632794, 8401253976, 8731945026, 9164072385, 9238750614, 9615278340, 9847103256

Offset: 1

T. D. Noe, Nov 09 2011

There are 8 numbers total. Subsequence of A114261.

			7351062489 ^5 = 21465972705539303240727164839587886180361092651449.

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

Cf. A050278 (pandigital numbers), A199630, A199631, A114260, A114261, A199633.

t = Select[Permutations[Range[0, 9]], #[[1]] > 0 &]; t2 = Select[t, Union[DigitCount[FromDigits[#]^5]] == {5} &]; 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

A365144 Numbers having each digit once and whose 4th power has each digit four times.

Original entry on oeis.org

5702631489, 7264103985, 7602314895, 7824061395, 8105793624, 8174035962, 8304269175, 8904623175, 8923670541, 9451360827, 9785261403, 9804753612, 9846032571

Offset: 1

T. D. Noe, Nov 09 2011

Currently same terms as A114260, but that sequence has more terms to follow. - Ray Chandler, Aug 23 2023

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

Cf. A050278 (pandigital numbers), A199630, A199631, A199633. Subsequence of A114260.

t = Select[Permutations[Range[0, 9]], #[[1]] > 0 &]; t2 = Select[t, Union[DigitCount[FromDigits[#]^4]] == {4} &]; FromDigits /@ t2 (* T. D. Noe, Nov 08 2011 *)

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]]

A358705 Zeroless pandigital numbers whose square has each digit 1 to 9 twice.

Original entry on oeis.org

345918672, 351987624, 359841267, 394675182, 429715863, 439516278, 487256193, 527394816, 527498163, 528714396, 572493816, 592681437, 729564183, 746318529, 749258163, 754932681, 759142683, 759823641, 762491835, 783942561, 784196235, 845691372, 891357624, 914863275, 915786423, 923165487, 928163754, 976825431

Offset: 1

Zhining Yang, Nov 27 2022

			345918672 is a term since its square 119659727638243584 contains all digits 1..9 twice each.

Cf. A050289, A199630.

R:= NULL:
for t in combinat:-permute([$1..9]) do
  x:= add(t[i]*10^(i-1),i=1..9);
  if sort(convert(x^2,base,10)) = [seq(i$2,i=1..9)] then
    R:= R, x
  fi
od:
sort([R]); # Robert Israel, Nov 27 2022

from itertools import permutations as per
a=[]
for n in [int(''.join(d)) for d in per('123456789', 9)]:
    if all(str(n**2).count(d) ==2 for d in '123456789'):
        a.append(n)
print(a)

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 *)

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

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A114258 Numbers k such that k^2 contains exactly 2 copies of each digit of k.

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

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A199632 Numbers having each digit once and whose 5th power has each digit five times.

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

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A365144 Numbers having each digit once and whose 4th power has each digit four times.

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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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