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

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

Offset

2,5

Comments

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)

Links

Table of n, a(n) for n=2..18.

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

Formula

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

Example

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 = 195^2.
For n=8 there is one penholodigital square, 6532471_8 = 1750329 = 1323^2.
For n=10 there are 30 penholodigital squares listed in A036744.

Prog

(Python)
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

Crossrefs

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

Sequence in context: A154989 A064307 A165883 * A260950 A110905 A205447

Adjacent sequences: A370947 A370948 A370949 * A370951 A370952 A370953

Keyword

nonn,base,more

Author

Chai Wah Wu, Mar 06 2024

Extensions

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

Status

approved