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A351650
Integers m such that digsum(m) divides digsum(m^2) where digsum = sum of digits = A007953.
3
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
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.
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
KEYWORD
nonn,base
AUTHOR
Bernard Schott, Feb 16 2022
EXTENSIONS
More terms from David A. Corneth, Feb 16 2022
STATUS
approved

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Last modified September 20 17:28 EDT 2024. Contains 376075 sequences. (Running on oeis4.)