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%I #57 Nov 17 2022 14:12:03
%S 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,
%T 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,
%U 1,32,15,25,135,3,18,8,3,28,9,3,43,47,72,27,8,25,126,27
%N 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.
%C 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).
%C 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.
%C a(n) will never be divisible by 10.
%C If n is divisible by 3 but not by 9, then a(n) is divisible by 3. - _Robert Israel_, Oct 13 2022
%H Johann Beurich, <a href="https://www.youtube.com/watch?v=mfJax_fab-c">Wie kann man das beweisen? (Bundeswettbewerb Mathematik 2021)</a>, YouTube video, 2021 (in German).
%H Bundeswettbewerb Mathematik 2021, <a href="https://www.mathe-wettbewerbe.de/fileadmin/Mathe-Wettbewerbe/Bundeswettbewerb_Mathematik/Dokumente/Aufgaben_und_Loesungen_BWM/loes_21_2_e.pdf">Die Aufgaben_und_Loesungen der 2. Runde 2021</a> (in German). The proof and solutions.
%F a(A058369(n)) = 1.
%F a(a(n)) <= n.
%F a(n) <= A132740(n)*A060284(A132740(n))*10^A051628(n)/n.
%F or a(n) <= (10^A000010(A132740(n))-1)*10^A051628(n)/n.
%p f:= proc(n) local k;
%p for k from 1 do if sd(n*k) = sd((n*k)^2) then return k fi od
%p end proc:
%p map(f, [$1..100]); # _Robert Israel_, Oct 13 2022
%o (PARI) a(n) = {my(k = 1); while(sumdigits(n*k)!=sumdigits((n*k)^2),k++);k}
%o (Python)
%o def sd(n): return sum(map(int, str(n)))
%o def a(n):
%o k = 1
%o while not sd(n*k) == sd((n*k)**2): k += 1
%o return k
%o print([a(n) for n in range(75)]) # _Michael S. Branicky_, Oct 13 2022
%Y Cf. A000010, A007953, A051628, A058369, A060284, A132740, A178505.
%K nonn,base
%O 0,3
%A _Thomas Scheuerle_, Oct 12 2022
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