A376003 - OEIS

%I #61 Oct 05 2024 22:52:58

%S 1,6,12,36,54,108,156,168,192,204,288,306,408,432,486,696,804,1104,

%T 1146,1188,1488,1512,1632,1764,1806,1932,2232,2904,3114,3408,3456,

%U 3528,4014,4104,4392,4596,4608,4704,4788,4872,4932,4944,5208,5304,5868,6012,6696,6792

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

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

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

%C From _Andrew Howroyd_, Sep 28 2024: (Start)

%C 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.

%C 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)

%e k = 12 is a term since k^2 = 144 has digits 1 and 4 and both are factors of k.

%e k = 2 is not a term since k^2 = 4 has a digit 4 which is not a factor of k.

%p q:= n-> andmap(x-> x>0 and irem(n, x)=0, convert(n^2, base, 10)):

%p select(q, [$1..10000])[];  # _Alois P. Heinz_, Sep 28 2024

%o (Python)

%o def is_valid_k(k):

%o     k_squared = k ** 2

%o     for digit in str(k_squared):

%o         d = int(digit)

%o         if d == 0 or k % d != 0:

%o             return False

%o     return True

%o def find_valid_k(max_k):

%o     valid_k = []

%o     for k in range(1, max_k + 1):

%o         if is_valid_k(k):

%o             valid_k.append(k)

%o     return valid_k

%o max_k = 10000

%o result = find_valid_k(max_k)

%o print(result)

%o (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

%Y Cf. A052040, A034838.

%K nonn,base

%O 1,2

%A _Sam N. Harrison_, Sep 28 2024