A036744 -id:A036744 - cp's OEIS frontend

A204139 Primes p such that 9p^2 is a penholodigital square (A036744).

Original entry on oeis.org

5227, 7673, 8147, 8269, 8353, 8647, 8803, 9043, 9091

Offset: 1

Zak Seidov, Jan 11 2012

Subsequence of A071519/3.

Cf. A036744, A071519.

A071519 Numbers whose square is a zeroless pandigital number (i.e., use the digits 1 through 9 once).

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

Offset: 1

Lekraj Beedassy, Jun 20 2002

A subset of A054037.

Cf. A036744, A054038, A156977.

lim:=floor(sqrt(987654321)): for n from floor(sqrt(123456789)) 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): fi: od: # Nathaniel Johnston, Jun 22 2011

Sqrt[#]&/@Select[FromDigits/@Permutations[Range[9]],IntegerQ[Sqrt[#]]&] (* Harvey P. Dale, Sep 23 2011 *)
Select[Range[11112, 31427,3], DigitCount[#^2] == {1,1,1,1,1,1,1,1,1,0} &]  (* Zak Seidov, Jan 11 2012 *)

A071519 = select( {is_A071519(n,L=[1..9])=vecsort(digits(n^2))==L}, [1e5\9..1e5\3]) \\ M. F. Hasler, Jun 28 2023

a(n) = sqrt(A036744(n)). - Zak Seidov, Jan 11 2012

A235724 Squares which have one or more occurrences of exactly nine different digits.

Original entry on oeis.org

102495376, 102576384, 102738496, 104325796, 105637284, 139854276, 152843769, 157326849, 158306724, 158407396, 172843609, 176039824, 176305284, 178035649, 180472356, 183467025, 187635204, 198753604, 208571364, 215384976, 217356049, 218034756, 235714609

Offset: 1

Colin Barker, Jan 15 2014

The first term having a repeated digit is 1005397264.

The smallest penholodigital square is a(6) = A036744(1) = 139854276 and the largest one is a(83) = A036744(30) = 923187456 (see Penguin references). - Bernard Schott, Feb 07 2022

			102495376 is in the sequence because 102495376 = 10124^2 and 102495376 contains exactly nine different digits: 0, 1, 2, 3, 4, 5, 6, 7 and 9.

David Wells, The Penguin Dictionary of Curious and Interesting Numbers (Revised Edition), Penguin Books, 1997, entry 139854276, page 184 and entry 923187456, page 186.

Michael S. Branicky, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Colin Barker)

Cf. A054037.
Cf. A235717-A235723, A225218.
A036744 is a subsequence.

s=[]; for(n=1, 100000, if(#vecsort(eval(Vec(Str(n^2))),,8)==9, s=concat(s, n^2))); s

from itertools import count, islice
def agen(): yield from (r*r for r in count(10**4) if len(set(str(r*r)))==9)
print(list(islice(agen(), 23))) # Michael S. Branicky, May 24 2022

a(n) = A054037(n)^2.

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

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

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

A383691 Square numbers with distinct digits from 1-9 that have an initial string of two or more digits forming a square number.

Original entry on oeis.org

169, 256, 361, 3249, 16384, 18496, 36481, 81796, 237169, 729316, 2537649, 3481956, 5184729, 36517849, 81432576, 254817369, 361874529, 529874361

Offset: 1

Hilarie Orman, May 05 2025

The sequence is based on correspondence with Donald S. McDonald. He originated the idea and computed the complete list of terms.

The last two numbers in the sequence are perfect squares with square roots that are digit permutations of each other: 361874529 = 19023*19023 and 529874361 = 23019*23019.

			169 is a term, because 169 and 16 are both squares, and each digit of 169 is unique.
18496 is a term, because 18496 and 1849 are both squares, and each digit of 18496 is unique.

Cf. A036744, A036745, A352329.

dd=Select[Range[11,25000]^2,IntegerLength[#]==CountDistinct[IntegerDigits[#]]&&ContainsNone[IntegerDigits[#],{0}]&];f[n_]:=AnyTrue[Table[FromDigits[Take[IntegerDigits[n],i]],{i,2,IntegerLength[n]-1}]^(1/2),IntegerQ];Select[dd,f[#]&] (* James C. McMahon, May 07 2025 *)

cp's OEIS Frontend

A204139 Primes p such that 9p^2 is a penholodigital square (A036744).

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A071519 Numbers whose square is a zeroless pandigital number (i.e., use the digits 1 through 9 once).

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A235724 Squares which have one or more occurrences of exactly nine different digits.

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PARI

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

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

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Mathematica

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

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PARI

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

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A383691 Square numbers with distinct digits from 1-9 that have an initial string of two or more digits forming a square number.

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