A078255 -id:A078255 - cp's OEIS frontend

A052046 Squares whose digits occur with the same frequency.

0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 169, 196, 256, 289, 324, 361, 529, 576, 625, 729, 784, 841, 961, 1024, 1089, 1296, 1369, 1764, 1849, 1936, 2304, 2401, 2601, 2704, 2809, 2916, 3025, 3249, 3481, 3721, 4096, 4356, 4761, 5041, 5184, 5329, 5476, 6084, 6241

Offset: 1

Patrick De Geest, Dec 15 1999

Contains all members of A078255 that are < 10^10. 7744 is the first member that is not a member of A078255. - David Wasserman, Jun 27 2006

Robert Israel, Table of n, a(n) for n = 1..10000
P. De Geest, Numbers whose digits occur with same frequency

Cf. A045540, A052047, A052048, A052049, A052050, A052051, A052052.

Cf. A078255.

filter:= proc(n) local L,M;
  L:= convert(n,base,10);
  M:= {seq(numboccur(i,L),i=0..9)} minus {0};
  nops(M) = 1
end proc:
select(filter, [seq(i^2,i=0..200)]); # Robert Israel, Jan 08 2018

t={}; Do[If[Length[DeleteDuplicates[Transpose[Tally[IntegerDigits[n^2]]][[2]]]] == 1, AppendTo[t,n^2]], {n,0,80}]; t (* Jayanta Basu, May 10 2013 *)
sfQ[n_]:=Length[Union[Select[DigitCount[n],#!=0&]]]==1; Select[ Range[ 0,80]^2,sfQ] (* Harvey P. Dale, May 05 2019 *)

Offset corrected by Michel Marcus, Aug 12 2015

A154566 Smallest 10-digit number whose n-th power contains each digit (0-9) n times, or -1 if no such number exists.

Original entry on oeis.org

1023456789, 3164252736, 4642110594, 5623720662, 6312942339, 6813614229, 7197035958, 7513755246, 7747685775, 7961085846, 8120306331, 8275283289, 8393900487, 8626922994, 8594070624, 8691229761, 8800389678, 8807854905, 9873773268, 8951993472, 9473643936, 9585032094

Offset: 1

Zhining Yang, Jan 12 2009, Jan 13 2009

A number with 10*n digits could contain all ten digits(0-9) 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 some 10-digit numbers. So there are about (10n)!*(10^10-10^(10-1/n)))/((n!)^10 * 10^((10*n)-10^(10*n-1)) numbers which satisfy the requirements.
Fortunately, I found a larger number than those shown here, for n=26, a(n)=9160395852. Since (10n)!*(10^10-10^(10-1/n))/((n!)^10 * 10^((10*n)-10^(10*n-1)) = 0.31691419..., this is a lucky event!
The sequence is -1 beyond a certain point because when n > 23025850928 we have 9999999999^n < 10^(10*n-1), i.e., it is impossible to obtain a power with 10*n digits. From a(23) to a(600) the only terms which are not -1 are a(24)=9793730157, a(26)=9160395852, a(35)=9959167017, and a(38)=9501874278. - Giovanni Resta, Jan 17 2020

			For n=18, a(n)=8807854905. That means 8807854905^18 has all digits 0-9 each 18 times and 8807854905 is the smallest 10-digit number which has this property.

Zhining Yang, Smallest Ten Digit Powers
Zhining Yang, Largest Ten Digit Powers

Cf. A010784, A078255, A154532.

def flag(p, n):
    for i in range(10):
        if not p.count(str(i)) == n:
            return False
    return True
def a(n):
    for i in range(3*int(10**(10-1/n)/3), 10**10, 3):
        if flag(str(i**n), n):
            return i
for i in range(1, 41):
    print(a(i), end=", ") # Zhining Yang, Oct 05 2022

Function Flag(ByVal s As String, ByVal num As Long) As Long
Dim b&(9), t&, i&
Flag = 1
If Len(s) <> 10 * num Then
    Flag = 0
    Exit Function
End If
For i = 1 To Len(s)
    t = Val(Mid(s, i, 1))
    b(t) = b(t) + 1
    If b(t) > num Then
        Flag = 0
        Exit Function
    End If
Next
End Function
'
Function Mypower(ByVal num As Currency, ByVal power As Long) As String
Dim b(), temp, i&, j&
ReDim b(1 To 2 * power)
ReDim s(1 To 2 * power)
b(2 * power - 1) = Val(Left(num, 5))
b(2 * power) = Val(Right(num, 5))
For i = 2 To power
    temp = 0
    For j = 2 * power To 1 Step -1
        temp = b(j) * num + temp
        b(j) = Format(Val(Right(temp, 5)), "00000")
        temp = Int(temp / 10 ^ 5)
    Next
Next
Mypower = Join(b, "")
End Function
'
Function a(ByVal n As Long)
Dim j As Currency, s As String, num&
For j = 3 * Int(1 + 10 ^ (10 - 1 / n) / 3) To 9999999999# Step 3
    DoEvents
    If Flag(Mypower(j, n), n) = 1 Then
        a = j
        Exit Function
    End If
Next
End Function ' Zhining Yang, Oct 11 2022

Edited by N. J. A. Sloane, Jan 13 2009
Edited by Charles R Greathouse IV, Nov 01 2009
Further edits by M. F. Hasler, Oct 05 2012
a(19)-a(22) from Giovanni Resta, Jan 17 2020
Added escape clause to definition. - N. J. A. Sloane, Nov 22 2022

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.

Original entry on oeis.org

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

A119509 Positive numbers whose square contains no digit more than once.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 13, 14, 16, 17, 18, 19, 23, 24, 25, 27, 28, 29, 31, 32, 33, 36, 37, 42, 43, 44, 48, 49, 51, 52, 53, 54, 55, 57, 59, 61, 64, 66, 69, 71, 72, 73, 74, 78, 79, 82, 84, 86, 87, 89, 93, 95, 96, 98, 99, 113, 116, 117, 118, 124, 126, 128, 133

Offset: 1

Tanya Khovanova, Jul 26 2006

There are exactly 610 terms. a(610) = 99066 and 99066^2 = 9814072356. - Rick L. Shepherd, Jul 27 2006

If we count 0, there is one more term, for a total of 611. - T. D. Noe, Jun 21 2013

Rick L. Shepherd, Table of n, a(n) for n = 1..610 (full sequence)

Subsequence of A045540 = numbers whose squares contain an equal number of each digit that they contain. The first number that belongs to A045540 and doesn't belong to this sequence is number 88.

Cf. A078255, A036745, A075309, A162950.

[n: n in [1..10^5] | #Set(d) eq #d where d is Intseq(n^2)];  // Bruno Berselli, Aug 02 2011

lim:=floor(sqrt(9876543210)): A119509:={}: for n from 1 to lim do pandig:=true: d:=convert(n^2,base,10): for k from 0 to 9 do if(numboccur(k, d)>1)then pandig:=false: break: fi: od: if(pandig)then A119509 := A119509 union {n}: fi: od: op(sort(convert(A119509,list))); # Nathaniel Johnston, Jun 23 2011

Select[Range[1000000], Length[IntegerDigits[ # ^2]] == Length[Union[IntegerDigits[ # ^2]]] &] (* Tanya Khovanova, May 29 2007 *)
Select[Range[10^5], Max[DigitCount[#^2]] <= 1 &] (* T. D. Noe, Aug 02 2011 *)

is_A119509(n)=#(n=digits(n^2))==#Set(n) \\ M. F. Hasler, Sep 08 2017

def ok(n): s = str(n**2); return n > 0 and len(set(s)) == len(s)
afull = [k for k in range(10**5) if ok(k)] # Michael S. Branicky, Nov 27 2022

More terms from Rick L. Shepherd, Jul 27 2006

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)

A259187 Primes p such that both p and p^2 are distinct-digit numbers.

Original entry on oeis.org

2, 3, 5, 7, 13, 17, 19, 23, 29, 31, 37, 43, 53, 59, 61, 71, 73, 79, 89, 137, 179, 193, 269, 281, 367, 397, 463, 487, 509, 571, 593, 647, 709, 829, 839, 1307, 1873, 2069, 2731, 2801, 3041, 4157, 4967, 4987, 6043, 7549, 7621, 8623, 21397

Offset: 1

Zak Seidov, Jun 20 2015

Corresponding squares are 4, 9, 25, 49, 169, 289, 361, 529, 841, 961, 1369, 1849, 2809, 3481, 3721, 5041, 5329, 6241, 7921, 18769, 32041, 37249, 72361, 78961, 134689, 157609, 214369, 237169, 259081, 326041, 351649, 418609, 502681, 687241, 703921, 1708249, 3508129, 4280761, 7458361, 7845601, 9247681, 17280649, 24671089, 24870169, 36517849, 56987401, 58079641, 74356129, 457831609 (subsequence of A078255).

Subsequence of A029743 and of A119509. Cf. A078255.

Select[Prime[Range[2500]],Max[DigitCount[#]]<2&&Max[DigitCount[#^2]]<2&] (* Harvey P. Dale, May 25 2020 *)

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)

A366940 a(n) is the number of positive squares with n digits, all distinct.

Original entry on oeis.org

3, 6, 13, 36, 66, 96, 123, 97, 83, 87, 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

Tanya Khovanova, Oct 29 2023

a(n) = 0, for n > 10.

			a(1)=3 because all three 1-digit squares, 1, 4, and 9, have trivially distinct digits.
a(2)=6 because all six 2-digit squares, 16, 25, 36, 49, 64, and 81, have distinct digits.
158407396 = 12586^2: has 9 distinct digits. Thus, this number contributes to a(9). On the other hand, 158382225 = 12585^2 has repeated digits. Thus, it doesn't contribute.

Index entries for linear recurrences with constant coefficients, signature (1).

Cf. A000290, A045540, A049415, A073532, A078255.

Table[Length[Select[Range[100000], Length[Union[IntegerDigits[#^2]]] == k &&  Length[IntegerDigits[#^2]] == k &]], {k, 10}]

from math import isqrt
from itertools import permutations
def sqr(n): return isqrt(n)**2 == n
def a(n):
    if n > 10: return 0
    return sum(1 for p in permutations("0123456789", n) if p[0] != '0' and sqr(int("".join(p))))
print([a(n) for n in range(1, 31)]) # Michael S. Branicky, Oct 29 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

A052046 Squares whose digits occur with the same frequency.

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A154566 Smallest 10-digit number whose n-th power contains each digit (0-9) n times, or -1 if no such number exists.

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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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A119509 Positive numbers whose square contains no digit more than once.

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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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A259187 Primes p such that both p and p^2 are distinct-digit numbers.

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

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A366940 a(n) is the number of positive squares with n digits, all distinct.

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