A268135 Numbers n such that the digit sum of n^2 is a divisor of the digit sum of n.

1, 9, 10, 18, 19, 45, 46, 55, 90, 99, 100, 145, 149, 180, 189, 190, 198, 199, 289, 351, 361, 369, 379, 388, 450, 451, 459, 460, 468, 495, 496, 549, 550, 558, 559, 568, 585, 595, 639, 729, 739, 775, 838, 855, 900, 954, 955, 990, 999, 1000, 1049, 1098, 1099, 1179, 1188, 1189, 1198

Offset: 1

Author

Melvin Peralta, Jan 26 2016

Keywords

• nonn
• base

Comments

Because A058369 (with offset 1) is a subsequence, this sequence is infinite.

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.

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

Examples

			Digit sum of 149^2 = 7. Digit sum of 149 = 14. Since 7 is a divisor of 14, 149 is in the sequence.
		

Links

• Indranil Ghosh, Table of n, a(n) for n = 1..10001

Crossrefs

Cf. A007953, A004159, A058369.

Programs

Mathematica

Select[Range[200], Mod[Total[IntegerDigits[#]], Total[IntegerDigits[#^2]]] == 0 &]

PARI

isok(n) = (sumdigits(n) % sumdigits(n^2)) == 0; \\ Michel Marcus, Jan 27 2016

Extensions

More terms from Michel Marcus, Jan 27 2016