A351650 Integers m such that digsum(m) divides digsum(m^2) where digsum = sum of digits = A007953.

1, 2, 3, 9, 10, 11, 12, 13, 18, 19, 20, 21, 22, 24, 27, 30, 31, 33, 36, 42, 45, 46, 54, 55, 63, 72, 74, 81, 90, 92, 99, 100, 101, 102, 103, 108, 110, 111, 112, 113, 117, 120, 121, 122, 123, 126, 128, 130, 132, 135, 144, 145, 153, 162, 171, 180, 189, 190, 191, 198

Offset

1,2

Comments

This is a generalization of a problem proposed by French site Diophante in link.

The smallest term k such that the corresponding quotient = n is A280012(n).

The quotient is 1 iff m is in A058369 \ {0}.

If k is in A061909, then digsum(k^2) = digsum(k)^2.

If k is a term, 10*k is also a term.

There are infinitely many m such that both m and m+1 are in the sequence, for example subsequence A002283 \ {0}.

Corresponding quotients are in A351651.

Links

Table of n, a(n) for n=1..60.

Diophante, A1730 - Des chiffres à sommer pour un entier (in French).

Formula

A004159(a(n)) = A007953(a(n)) * A351651(n).

Example

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.

Mathematica

Select[Range[200], Divisible[Total[IntegerDigits[#^2]], Total[IntegerDigits[#]]] &] (* Amiram Eldar, Feb 16 2022 *)

Prog

(PARI) is(n)=sumdigits(n^2)%sumdigits(n) == 0 \\ David A. Corneth, Feb 16 2022
(Python)
def sd(n): return sum(map(int, str(n)))
def ok(n): return sd(n**2)%sd(n) == 0
print([m for m in range(1, 200) if ok(m)]) # Michael S. Branicky, Feb 16 2022

Crossrefs

Cf. A004159, A007953, A351651.

Subsequences: A002283, A011557, A052268, A058369, A061909, A093136, A093138, A254066, A280012.

Sequence in context: A007316 A163905 A353653 * A235499 A242590 A135204

Adjacent sequences: A351647 A351648 A351649 * A351651 A351652 A351653

Keyword

nonn,base

Author

Bernard Schott, Feb 16 2022

Extensions

More terms from David A. Corneth, Feb 16 2022

Status

approved