A376003 Positive integers k such that each digit of k^2 is a factor of k.

1, 6, 12, 36, 54, 108, 156, 168, 192, 204, 288, 306, 408, 432, 486, 696, 804, 1104, 1146, 1188, 1488, 1512, 1632, 1764, 1806, 1932, 2232, 2904, 3114, 3408, 3456, 3528, 4014, 4104, 4392, 4596, 4608, 4704, 4788, 4872, 4932, 4944, 5208, 5304, 5868, 6012, 6696, 6792

Offset: 1

Author

Sam N. Harrison, Sep 28 2024

Keywords

• nonn
• base

Comments

0 is never a factor so k^2 must be zeroless and this sequence is a subset of A052040.

The first term > 1 that is not divisible by 6 is 47768.

From Andrew Howroyd, Sep 28 2024: (Start)

Except for the first term, all terms are even since all squares with at least 2 digits contain an even digit. This implies k^2 cannot contain the digit 5.

All numbers of the form (100*1000^k-1)/3+3 are terms. These are the numbers 36, 33336, 33333336, 33333333336, etc. This shows that the sequence is infinite. (End)

Examples

			k = 12 is a term since k^2 = 144 has digits 1 and 4 and both are factors of k.
k = 2 is not a term since k^2 = 4 has a digit 4 which is not a factor of k.
		

Crossrefs

Cf. A052040, A034838.

Programs

Maple

q:= n-> andmap(x-> x>0 and irem(n, x)=0, convert(n^2, base, 10)):
select(q, [$1..10000])[];  # Alois P. Heinz, Sep 28 2024

PARI

isok(k) = my(d=Set(digits(k^2))); if(!vecmin(d), return(0)); for (i=1, #d, if (k % d[i], return(0))); return(1); \\ Michel Marcus, Sep 28 2024

Python

def is_valid_k(k):
    k_squared = k ** 2
    for digit in str(k_squared):
        d = int(digit)
        if d == 0 or k % d != 0:
            return False
    return True
def find_valid_k(max_k):
    valid_k = []
    for k in range(1, max_k + 1):
        if is_valid_k(k):
            valid_k.append(k)
    return valid_k
max_k = 10000
result = find_valid_k(max_k)
print(result)