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

1,2

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)

Links

Table of n, a(n) for n=1..48.

Example

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.

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

Prog

(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)
(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

Crossrefs

Cf. A052040, A034838.

Sequence in context: A127845 A306899 A096932 * A212976 A352621 A176681

Adjacent sequences: A376000 A376001 A376002 * A376004 A376005 A376006

Keyword

nonn,base

Author

Sam N. Harrison, Sep 28 2024

Status

approved