A071519 -id:A071519 - cp's OEIS frontend

A036744 Penholodigital squares: squares containing each of the digits 1..9 exactly once.

139854276, 152843769, 157326849, 215384976, 245893761, 254817369, 326597184, 361874529, 375468129, 382945761, 385297641, 412739856, 523814769, 529874361, 537219684, 549386721, 587432169, 589324176, 597362481, 615387249, 627953481, 653927184, 672935481, 697435281, 714653289, 735982641, 743816529, 842973156, 847159236, 923187456

Offset: 1

David W. Wilson

Improved Mathematica formula provided. Because the range involved is only from Ceiling[Sqrt[123456789]]=11112 and Floor[Sqrt[987654321]]=31427, it only requires analyzing 20,315 numbers, versus 362,880 permutations of nine digits (as in the current formula). - Harvey P. Dale, Apr 17 2002
Since the sum of the digits is 45, the squares are all divisible by 3, so the given Mathematica formula could be sped up by a factor of 3, checking only multiples of 3 rather than all squares. - Joshua Zucker, Nov 28 2005
Eight-digit analog gives 5 squares: 13527684, 34857216, 65318724, 73256481, 81432576. - Zak Seidov, Mar 01 2011

Cf. A036745, A071519.

lim:=floor(sqrt(987654321)): for n from 11112 by 3 to lim do d:=[op(convert(n^2, base, 10))]: pandig:=true: for k from 1 to 9 do if(numboccur(k, d)<>1)then pandig:=false: break: fi: od: if(pandig)then printf("%d, ", n^2): fi: od: # Nathaniel Johnston, Jun 22 2011

Select[Range[11112, 31427]^2, Union[Drop[DigitCount[ # ], -1]] == {1} &]

A036744 = [n^2 | n <- A071519] \\ or less efficient & more explicit:
A036744 = [n^2 | n <- [1e5\9..1e5\3], vecsort(digits(n^2)) == [1..9]] \\ M. F. Hasler, Jun 28 2023

a(n) = A071519(n)^2.

More terms from Harvey P. Dale, Sep 26 2001

Keyword base added by Reinhard Zumkeller, May 16 2010

A258103 Number of pandigital squares (containing each digit exactly once) in base n.

Original entry on oeis.org

0, 0, 1, 0, 1, 3, 4, 26, 87, 47, 87, 0, 547, 1303, 3402, 0, 24192, 187562

Offset: 2

Adam J.T. Partridge, May 20 2015

For n = 18, the smallest and largest pandigital squares are 2200667320658951859841 and 39207739576969100808801. For n = 19, they are 104753558229986901966129 and 1972312183619434816475625. For n = 20, they are 5272187100814113874556176 and 104566626183621314286288961. - Chai Wah Wu, May 20 2015
When n is even, (n-1) is a factor of the pandigital squares.  When n is odd, (n-1)/2 is a factor with the remaining factors being odd.  Therefore, when n is odd and (n-1)/2 has an odd number of 2s as prime factors there are no pandigital squares in base n (e.g. 5, 13, 17 and 21). - Adam J.T. Partridge, May 21 2015
If n is odd and (n-1)/2 has an odd 2-adic valuation, then there are no squares in base n using all the digits from 1 to n-1 once, or all the digits from 0 to n-2 once or all the digits from 1 to n-2 once. This can be proved using the same argument as in the linked blogposts. - Chai Wah Wu, Feb 25 2024

			For n=4 there is one pandigital square, 3201_4 = 225 = 15^2.
For n=6 there is one pandigital square, 452013_6 = 38025 = 195^2.
For n=10 there are 87 pandigital squares (A036745).
There are no pandigital squares in bases 2, 3, 5 or 13.
Hexadecimal has 3402 pandigital squares, the largest is FED5B39A42706C81.

A. J. T. Partridge, Why there are no pandigital squares in base 13
Chai Wah Wu, Square pandigital numbers
Chai Wah Wu, Pandigital and penholodigital numbers, arXiv:2403.20304 [math.GM], 2024. See p. 2.

Cf. A036745, A054038, A071519.

a(n) = if(n%2==1 && valuation(n-1,2)%2==0, 0, my(lim=sqrtint(n^n - (n^n-n)/(n-1)^2), count=0); for(m=sqrtint((n^n-n)/(n-1)^2 + n^(n-2)*(n-1) - 1), lim, if(#Set(digits(m^2,n))==n, count++)); count) \\ Jianing Song, Feb 23 2024. Note that the searching range for m is [sqrt(A049363(n)), sqrt(A062813(n))]

from gmpy2 import isqrt, mpz, digits
def A258103(n): # requires 2 <= n <= 62
    c, sm, sq = 0, mpz(''.join([digits(i, n) for i in range(n-1, -1, -1)]), n), mpz(''.join(['1', '0']+[digits(i, n) for i in range(2, n)]), n)
    m = isqrt(sq)
    sq = m*m
    m = 2*m+1
    while sq <= sm:
        if len(set(digits(sq, n))) == n:
            c += 1
        sq += m
        m += 2
    return c # Chai Wah Wu, May 20 2015

a(17)-a(19) from Giovanni Resta, May 20 2015

A363905 Numbers whose square and cube taken together contain each decimal digit.

Original entry on oeis.org

69, 128, 203, 302, 327, 366, 398, 467, 542, 591, 593, 598, 633, 643, 669, 690, 747, 759, 903, 923, 943, 1016, 1018, 1027, 1028, 1043, 1086, 1112, 1182, 1194, 1199, 1233, 1278, 1280, 1282, 1328, 1336, 1364, 1396, 1419, 1459, 1463, 1467, 1472, 1475

Offset: 1

M. F. Hasler, Jun 27 2023

The first term, a(1) = 69, is the only number for which the square and the cube together contain each decimal digit 0 to 9 exactly once.

a(820) = 6534 is the only number of which the square and cube taken together contain each digit 0 to 9 exactly twice.

			69^2 = 4761, 69^3 = 328509, which together contain each digit 0-9 exactly once.

Robert G. Wilson v, Table of n, a(n) for n = 1..10000
Harold Suarez, Interesting..., Number Theory group on LinkedIn, June 2023.

Cf. A036744, A054038, A071519 and A156977 for "pandigital" squares.

Cf. A119735: Numbers n such that every digit occurs at least once in n^3.

fQ[n_] := Union[ Join[ IntegerDigits[n^2], IntegerDigits[n^3]]] == {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}; Select[Range@1500, fQ] (* Robert G. Wilson v, Jun 27 2023 *)

is(k)=#setunion(Set(digits(k^2)),Set(digits(k^3)))>9
select(is,[1..9999])

from itertools import count, islice
def A363905_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n:len(set(str(n**2))|set(str(n**3)))==10,count(max(startvalue,1)))
A363905_list = list(islice(A363905_gen(),20)) # Chai Wah Wu, Jun 27 2023

A294661 Numbers whose square contains all of the digits 1 through 9.

Original entry on oeis.org

11826, 12363, 12543, 14676, 15681, 15963, 18072, 19023, 19377, 19569, 19629, 20316, 22887, 23019, 23178, 23439, 24237, 24276, 24441, 24807, 25059, 25572, 25941, 26409, 26733, 27129, 27273, 29034, 29106, 30384, 32043, 32286, 33144, 34273, 35172, 35337, 35713, 35756, 35757, 35772, 35846, 35853

Offset: 1

M. F. Hasler, Nov 08 2017

The sequence has asymptotic density 1: it contains "almost all" numbers.

			11826^2 = 139854276 contains all digits from 1 to 9 exactly once.
The same is true for all terms up to 30384 whose square is 923187456. These terms are also listed in A071519, they form a subsequence of A054037.
The next 3 terms, 32043 (32043^2 = 1026753849), 32286 (32286^2 = 1042385796) and 33144 (33144^2 = 1098524736) contain all of the digits '0' through '9' exactly once: They are the first terms of A054038.
The next term, 34273 with 34273^2 = 1174638529, does not have this property, but the next two are again of that type (35172^2 = 1237069584 and 35337^2 = 1248703569).

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A054037, A071519 (finite subsequence of the first 30 terms), A054038.

Select[Range[#, # + 3*10^4] &@ 11111, AllTrue[Most@ DigitCount[#^2], # > 0 &] &] (* Michael De Vlieger, Nov 08 2017 *)

is_A294661(n)=#select(t->t,Set(digits(n^2)))>8
N=100;for(k=10^4,oo,is_A294661(k)||next;print1(k",");N--||break)

A363927 Numbers N such that in the concatenation of N^2 and N^3, each of the 10 decimal digits appears equally often.

Original entry on oeis.org

69, 6534, 497375, 539019, 543447, 586476, 589629, 601575, 646479, 858609, 895688, 959097, 46839081, 47469378, 47693199, 47760623, 47841576, 48038964, 48527792, 48733506, 48886836, 48965892, 49229103, 49397283, 49594832, 49670616, 50013116, 50247423, 50359157

Offset: 1

Charles R Greathouse IV and M. F. Hasler, Jun 28 2023

a(3) = 497375 and a(11) = 895688 are the only terms < 10^6 that are not divisible by 3.
Each term has an even number of decimal digits, k, and a corresponding value between 10^(k-1)*100^(1/3) and 10^k. - Michael S. Branicky, Jun 29 2023
Indeed, the number of digits of concat(N^2, N^3) is floor(2*L + 1) + floor(3*L + 1) where L = log_10(N). This is a multiple of 10 iff L mod 2 is in the interval [5/3, 2), which means that N is in the above range for some even k. - M. F. Hasler, Jul 02 2023

Michael S. Branicky, Table of n, a(n) for n = 1..10000

Cf. A363905, A363909: concat(n^2, n^3) has each digit at least once / twice.
Cf. A171102: pandigital numbers.
Cf. A036744, A054038, A071519 and A156977 for "pandigital squares".
Cf. A119735: n^3 is pandigital.

fQ[n_] := Length@ Union[ Count[ Sort[ Join[ IntegerDigits[n^2], IntegerDigits[n^3]]], #] & /@ Range[0, 9]] == 1; Select[ Range@ 52000000, fQ] (* Robert G. Wilson v, Jul 01 2023 *)

is(n)={my(v=concat(digits(n^2),digits(n^3)), c=#v); c%10==0 && vecsort(v)==[0..c-1]\(c\10)}
for(n=1,1e6, is(n)&& print1(n","))

a(13) and beyond from Michael S. Branicky, Jun 28 2023

A261213 Odd numbers n such that n^2 = m + (m+1), where both m and m+1 have no repeated digits.

Original entry on oeis.org

1, 3, 5, 7, 9, 11, 13, 23, 27, 29, 31, 35, 37, 39, 41, 43, 57, 63, 69, 77, 81, 87, 89, 95, 109, 113, 121, 125, 127, 129, 137, 163, 193, 219, 239, 271, 273, 279, 281, 285, 305, 311, 315, 331, 339, 353, 357, 377, 381, 395, 403, 409, 435, 441, 443, 597

Offset: 1

Pieter Post, Aug 12 2015

This sequence is finite and a(146) = 40797 is the last term. 40797^2 = 1664395209 and 1664395209 = 832197604 + 832197605. These last two numbers both have no repeating digits.

			5 is in the sequence, because 5^2 = 25. 25 = 12 + 13. 12 and 13 both have no repeating digits.

Pieter Post, Table of n, a(n) for n = 1..146

Cf. A036745, A071519, A156977.

nr[n_] := 1 == Max@ DigitCount@ n; Select[ Range[1, 10^5, 2], nr[x= Floor[#^2 / 2]] && nr[x + 1] &] (* Giovanni Resta, Aug 12 2015 *)

A345875 Numbers whose fourth powers are zeroless pandigital.

Original entry on oeis.org

608, 809, 897, 924, 1166, 1241, 1458, 1459, 1506, 1547, 1718, 1729, 1832, 1932, 1977, 1982, 2112, 2162, 2179, 2188, 2211, 2279, 2283, 2291, 2296, 2336, 2337, 2408, 2427, 2541, 2592, 2594, 2613, 2634, 2684, 2689, 2704, 2764, 2776, 2779, 2854, 2941, 2984, 2988, 3009

Offset: 1

Tanya Khovanova, Jun 27 2021

Zeroless pandigital means that it contains all the digits 1 through 9, but doesn't contain a zero.

			608^4 = 136651472896. Thus, 608 belongs to this sequence.

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A071519 (for squares), A124628 (for cubes).

Subsequence of A121321 (4th power is pandigital).

q:= n-> is({convert(n^4, base, 10)[]}={$1..9}):
select(q, [$1..3000])[];  # Alois P. Heinz, Jun 29 2021

Select[Range[8000], Union[IntegerDigits[#^4]] == {1, 2, 3, 4, 5, 6, 7, 8, 9} &]

isok(k) = my(d=digits(k^4)); vecmin(d) && (#Set(d) == 9); \\ Michel Marcus, Jun 30 2021

def ok(n): return set(str(n**4)) == set("123456789")
print(list(filter(ok, range(3000)))) # Michael S. Branicky, Jun 27 2021

A363909 Numbers whose square and cube taken together contain each decimal digit at least twice.

Original entry on oeis.org

6534, 11027, 11994, 21906, 22178, 22195, 23317, 24567, 27019, 27963, 28354, 29099, 29309, 29339, 29375, 29558, 29621, 30184, 30552, 30584, 31578, 31727, 32447, 32633, 32793, 32912, 32923, 33087, 33257, 33527, 34284, 35717, 36943, 36958, 37697, 38463

Offset: 1

M. F. Hasler, Jun 27 2023

The first term, a(1) = 6534 is the only number of which the square and cube taken together contain each digit 0 to 9 exactly twice.

Presumably a(n) ~ A363905(n) ~ n. - Charles R Greathouse IV, Jul 03 2023

			6534^2 = 42693156, 6534^3 = 278957081304, which together contain each digit 0-9 exactly twice.

Harold Suarez, Interesting..., Number Theory group on LinkedIn, (capture of the original post), June 2023

Cf. A363905: square and cube together contain each digit at least once.
Cf. A036744, A054038, A071519 and A156977 for "pandigital" squares.
Cf. A119735: Numbers n such that every digit occurs at least once in n^3.

is(n)=#Set(n=concat(digits(n^2),digits(n^3)))>9&&(n=vecsort(n))[#n-1]==9&&!n[2]&&!for(i=3,#n-2,n[i]>n[i-1]&&n[i]

A370950 Number of penholodigital squares (containing each nonzero digit exactly once) in base n.

Original entry on oeis.org

1, 0, 0, 0, 2, 1, 1, 10, 30, 20, 23, 0, 160, 419, 740, 0, 5116, 47677

Offset: 2

Chai Wah Wu, Mar 06 2024

From Chai Wah Wu, Mar 10 2024: (Start)
Theorem: Let m^2 be a pandigital square (containing each digit exactly once) or a penholodigital square (containing each nonzero digit exactly once) in base n. If n-1 is an odd squarefree number (A056911), then m is divisible by n-1. If n-1 is an even squarefree number (A039956), then m-(n-1)/2 is divisible by n-1.
Proof: Since n^k == 1 (mod n-1), and the sum of digits of m^2 is n*(n-1)/2, this implies that m^2 == n*(n-1)/2 (mod n-1). If n-1 is odd and squarefree, then n is even and m^2 == 0 (mod n-1). Since n-1 = Product_i p_i for distinct odd primes p_i, and m^2 == 0 (mod p_i) if and only if m == 0 (mod p_i), this implies that m == 0 (mod n-1) by the Chinese Remainder Theorem.
Similarly, if n-1 is even and squarefree, then m^2 == n(n-1)/2 == (n-1)/2 (mod n-1) and (n-1)/2 = Product_i p_i for distinct odd primes p_i, i.e., m^2 == 0 (mod p_i) and m^2 == 1 (mod 2). This implies that m == 0 (mod p_i) and m == 1 (mod 2). Again by the Chinese Remainder Theorem, m == (n-1)/2 (mod n-1).
(End)

			For n=2 there is one penholodigital square, 1_2 = 1 = 1^2.
For n=6 there are two penholodigital squares, 15324_6 = 2500 = 50^2 and 53241_6 = 7225 = 85^2.
For n=7 there is one penholodigital square, 623514_7 = 106929 = 327^2.
For n=8 there is one penholodigital square, 6532471_8 = 1750329 = 1323^2.
For n=10 there are 30 penholodigital squares listed in A036744.

Chai Wah Wu, Pandigital and penholodigital numbers, arXiv:2403.20304 [math.GM], 2024. See p. 2.

Cf. A039956, A056911, A036744, A071519, A258103.

from gmpy2 import mpz, digits, isqrt
def A370950(n): # requires 2 <= n <= 62
    if n&1 and (~(m:=n-1>>1) & m-1).bit_length()&1:
        return 0
    t = ''.join(digits(d,n) for d in range(1,n))
    k = mpz(''.join(digits(d,n) for d in range(n-1,0,-1)),n)
    k2 = mpz(t,n)
    c = 0
    for i in range(isqrt(k2),isqrt(k)+1):
        if i%n:
            j = i**2
            s = ''.join(sorted(digits(j,n)))
            if s == t:
                c += 1
    return c

If n is odd and n-1 has an even 2-adic valuation, then a(n) = 0 (see A258103).

a(18) from Chai Wah Wu, Mar 07 2024

a(19) from Chai Wah Wu, Mar 15 2025

A162859 Numbers n such that 9n^2 is a zeroless pandigital number.

Original entry on oeis.org

3942, 4121, 4181, 4892, 5227, 5321, 6024, 6341, 6459, 6523, 6543, 6772, 7629, 7673, 7726, 7813, 8079, 8092, 8147, 8269, 8353, 8524, 8647, 8803, 8911, 9043, 9091, 9678, 9702, 10128

Offset: 1

Zak Seidov, Jul 14 2009

Cf. A071519 Numbers whose square is a zeroless pandigital number.

With[{c=Range[9]},Select[Range[11000],Sort[IntegerDigits[9 #^2]]== c&]] (* Harvey P. Dale, Jun 26 2011 *)

A071519(n)/3.

Edited by Charles R Greathouse IV, Aug 02 2010

cp's OEIS Frontend

A036744 Penholodigital squares: squares containing each of the digits 1..9 exactly once.

Views

Author

Keywords

Comments

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A258103 Number of pandigital squares (containing each digit exactly once) in base n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Python

Extensions

A363905 Numbers whose square and cube taken together contain each decimal digit.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Python

A294661 Numbers whose square contains all of the digits 1 through 9.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

A363927 Numbers N such that in the concatenation of N^2 and N^3, each of the 10 decimal digits appears equally often.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Extensions

A261213 Odd numbers n such that n^2 = m + (m+1), where both m and m+1 have no repeated digits.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A345875 Numbers whose fourth powers are zeroless pandigital.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica