A357080 Numbers k such that the sum of the digits of k multiplied by the sum of the digits of k^2 equals k.

0, 1, 80, 162, 243, 476, 486

Offset

1,3

Comments

Suppose k has m digits, then the sum of the digits of k multiplied by the sum of the digits of k^2 is bounded by 9m times 9*(2m), which equals 162m^2. On the other hand, k is greater than 10^(m-1), which grows much faster than 162m^2. It follows that k can't have more than 4 digits.

Links

Table of n, a(n) for n=1..7.

Example

The sum of the digits of 80 is 8, the sum of the digits of 80^2 = 6400 is 10. The number 80 itself is 8*10. Thus, 80 is in this sequence.

Mathematica

Select[Range[100000], # == Total[IntegerDigits[#]] Total[IntegerDigits[#^2]] &]

Prog

(PARI) isok(k) = k == sumdigits(k)*sumdigits(k^2); \\ Michel Marcus, Sep 11 2022
(Python)
def sd(n): return sum(map(int, str(n)))
def ok(n): return sd(n) * sd(n*n) == n
print([k for k in range(10**5) if ok(k)]) # Michael S. Branicky, Sep 11 2022

Crossrefs

Cf. A007953, A004159, A130181.

Sequence in context: A031496 A044331 A044712 * A044412 A044793 A359517

Adjacent sequences: A357077 A357078 A357079 * A357081 A357082 A357083

Keyword

nonn,base,fini,full

Author

Tanya Khovanova, Sep 10 2022

Status

approved