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

Author

Tanya Khovanova, Sep 10 2022

Keywords

• nonn
• base
• fini
• full

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.

Examples

			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.
		

Crossrefs

Cf. A007953, A004159, A130181.

Programs

Mathematica

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

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