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%I #18 Jan 17 2017 05:53:14
%S 1,9,10,18,19,45,46,55,90,99,100,145,149,180,189,190,198,199,289,351,
%T 361,369,379,388,450,451,459,460,468,495,496,549,550,558,559,568,585,
%U 595,639,729,739,775,838,855,900,954,955,990,999,1000,1049,1098,1099,1179,1188,1189,1198
%N Numbers n such that the digit sum of n^2 is a divisor of the digit sum of n.
%C Because A058369 (with offset 1) is a subsequence, this sequence is infinite.
%C Conjecture: The relative complement of A058369 with respect to this sequence is infinite. That is, there are infinitely many n such that the digit sum of n^2 is a proper divisor of the digit sum of n.
%C If the digit sum of n^2 is a proper divisor of the digit sum of n, then this property holds for 10*n as well, i.e. the digit sum of n = 149*10^k has as a proper divisor the digit sum of n^2 for all k > 0. Are there infinitely many n that are not a multiple of 10 such that the digit sum of n^2 is a proper divisor of the digit sum of n? The first few such numbers are: 149, 549, 1049, 14499, 19499, 55679, 59499, 64499, 73499, 118499, 144999, 145949, 179249, 244949, 244998, 334679, 347855, 473499, 548735, 549549, 549639, 556965, 837855, ... - _Chai Wah Wu_, Mar 16 2016
%H Indranil Ghosh, <a href="/A268135/b268135.txt">Table of n, a(n) for n = 1..10001</a>
%e Digit sum of 149^2 = 7. Digit sum of 149 = 14. Since 7 is a divisor of 14, 149 is in the sequence.
%t Select[Range[200], Mod[Total[IntegerDigits[#]], Total[IntegerDigits[#^2]]] == 0 &]
%o (PARI) isok(n) = (sumdigits(n) % sumdigits(n^2)) == 0; \\ _Michel Marcus_, Jan 27 2016
%Y Cf. A007953, A004159, A058369.
%K nonn,base
%O 1,2
%A _Melvin Peralta_, Jan 26 2016
%E More terms from _Michel Marcus_, Jan 27 2016
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