A357756 a(n) is the least k > 0 such that A007953(n*k) equals A007953((n*k)^2), where A007953 is the sum of the digits.

1, 1, 5, 3, 25, 2, 3, 27, 62, 1, 1, 5, 15, 27, 128, 3, 31, 17, 1, 1, 5, 9, 9, 2, 75, 4, 18, 7, 64, 5, 3, 16, 56, 3, 85, 17, 5, 27, 5, 9, 25, 9, 45, 13, 27, 1, 1, 27, 66, 54, 2, 9, 9, 18, 22, 1, 32, 15, 25, 135, 3, 18, 8, 3, 28, 9, 3, 43, 47, 72, 27, 8, 25, 126, 27

Offset: 0

Author

Thomas Scheuerle, Oct 12 2022

Keywords

• nonn
• base

Comments

A task in the German competition "Bundeswettbewerb Mathematik 2021" was to prove that for each positive integer n there exists a k such that A007953(n*k) = A007953((n*k)^2).

One of the proposed proofs uses the argument that numbers of the form m = (10^x-1)*(10^y) will have the desired property A007953(m) = A007953(m^2). Thus we need to prove that we can find for all n a k, x and y such that n*k = (10^x-1)*(10^y). Let n be of the form b*2^c*5^d with b odd and not divisible by 5, then we know that y = max(c, d). From Euler's totient theorem we know that 10^x-1 will be divisible by e if x = A000010(e) where A000010 is Euler's totient function. See the formula section for the corresponding resulting k.

a(n) will never be divisible by 10.

If n is divisible by 3 but not by 9, then a(n) is divisible by 3. - Robert Israel, Oct 13 2022

Links

• Johann Beurich, Wie kann man das beweisen? (Bundeswettbewerb Mathematik 2021), YouTube video, 2021 (in German).
• Bundeswettbewerb Mathematik 2021, Die Aufgaben_und_Loesungen der 2. Runde 2021 (in German). The proof and solutions.

Crossrefs

Cf. A000010, A007953, A051628, A058369, A060284, A132740, A178505.

Programs

Maple

f:= proc(n) local k;
   for k from 1 do if sd(n*k) = sd((n*k)^2) then return k fi od
end proc:
map(f, [$1..100]); # Robert Israel, Oct 13 2022

PARI

a(n) = {my(k = 1); while(sumdigits(n*k)!=sumdigits((n*k)^2),k++);k}

Python

def sd(n): return sum(map(int, str(n)))
def a(n):
    k = 1
    while not sd(n*k) == sd((n*k)**2): k += 1
    return k
print([a(n) for n in range(75)]) # Michael S. Branicky, Oct 13 2022

Formula

a(A058369(n)) = 1.

a(a(n)) <= n.

a(n) <= A132740(n)*A060284(A132740(n))*10^A051628(n)/n.

or a(n) <= (10^A000010(A132740(n))-1)*10^A051628(n)/n.