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A357756
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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.
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0
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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
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OFFSET
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0,3
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COMMENTS
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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
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LINKS
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FORMULA
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a(a(n)) <= n.
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MAPLE
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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:
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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