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

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

Offset: 2

Author

Adam J.T. Partridge, May 20 2015

Keywords

• base
• nonn
• more

Comments

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

Examples

			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.
		

Links

• 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.

Crossrefs

Cf. A036745, A054038, A071519.

Programs

PARI

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))]

Python

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

Extensions

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