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%I #35 Feb 19 2022 04:52:31
%S 1,2,3,9,10,11,12,13,18,19,20,21,22,24,27,30,31,33,36,42,45,46,54,55,
%T 63,72,74,81,90,92,99,100,101,102,103,108,110,111,112,113,117,120,121,
%U 122,123,126,128,130,132,135,144,145,153,162,171,180,189,190,191,198
%N Integers m such that digsum(m) divides digsum(m^2) where digsum = sum of digits = A007953.
%C This is a generalization of a problem proposed by French site Diophante in link.
%C The smallest term k such that the corresponding quotient = n is A280012(n).
%C The quotient is 1 iff m is in A058369 \ {0}.
%C If k is in A061909, then digsum(k^2) = digsum(k)^2.
%C If k is a term, 10*k is also a term.
%C There are infinitely many m such that both m and m+1 are in the sequence, for example subsequence A002283 \ {0}.
%C Corresponding quotients are in A351651.
%H Diophante, <a href="http://www.diophante.fr/problemes-par-themes/arithmetique-et-algebre/a1-pot-pourri/4786-a1730-des-chiffres-a-sommer-pour-un-entier">A1730 - Des chiffres à sommer pour un entier</a> (in French).
%F A004159(a(n)) = A007953(a(n)) * A351651(n).
%e digit sum of 42 = 4+2 = 6; then 42^2 = 1764, digit sum of 1764 = 1+7+6+4 = 18; as 6 divides 18, 42 is a term.
%t Select[Range[200], Divisible[Total[IntegerDigits[#^2]], Total[IntegerDigits[#]]] &] (* _Amiram Eldar_, Feb 16 2022 *)
%o (PARI) is(n)=sumdigits(n^2)%sumdigits(n) == 0 \\ _David A. Corneth_, Feb 16 2022
%o (Python)
%o def sd(n): return sum(map(int, str(n)))
%o def ok(n): return sd(n**2)%sd(n) == 0
%o print([m for m in range(1, 200) if ok(m)]) # _Michael S. Branicky_, Feb 16 2022
%Y Cf. A004159, A007953, A351651.
%Y Subsequences: A002283, A011557, A052268, A058369, A061909, A093136, A093138, A254066, A280012.
%K nonn,base
%O 1,2
%A _Bernard Schott_, Feb 16 2022
%E More terms from _David A. Corneth_, Feb 16 2022
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