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%I #14 Sep 26 2022 20:35:20
%S 0,1,80,162,243,476,486
%N Numbers k such that the sum of the digits of k multiplied by the sum of the digits of k^2 equals k.
%C 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.
%e 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.
%t Select[Range[100000], # == Total[IntegerDigits[#]] Total[IntegerDigits[#^2]] &]
%o (PARI) isok(k) = k == sumdigits(k)*sumdigits(k^2); \\ _Michel Marcus_, Sep 11 2022
%o (Python)
%o def sd(n): return sum(map(int, str(n)))
%o def ok(n): return sd(n) * sd(n*n) == n
%o print([k for k in range(10**5) if ok(k)]) # _Michael S. Branicky_, Sep 11 2022
%Y Cf. A007953, A004159, A130181.
%K nonn,base,fini,full
%O 1,3
%A _Tanya Khovanova_, Sep 10 2022
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