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

1,2

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

Links

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

Example

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

Mathematica

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

Prog

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

Crossrefs

Cf. A007953, A004159, A058369.

Sequence in context: A107433 A090570 A131417 * A058369 A110939 A260042

Adjacent sequences: A268132 A268133 A268134 * A268136 A268137 A268138

Keyword

nonn,base

Author

Melvin Peralta, Jan 26 2016

Extensions

More terms from Michel Marcus, Jan 27 2016

Status

approved