Does parity of f(a) and f(a+1) are same whenever a is even? Question related to sum of digits.

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sa3b4or9i9 · Accepted Answer

Step 1Assume a is even. That means we already know the parity with i=2j+1 odd: D(a,2j+1)=2kj (even) and j=1,⋯,a2−1Now take a look for i=2j even: We don&#039;t know the concrete value of D(a,2j), but we can draw a connection between D(a,2j) and D(a+1,2j). As a is even the last digit in any even-numbered system must be even as well (otherwise we will get a contradiction mod 2)! This means the last digit can never be i−1 so if we add 1 we will only increase the last digit to a value ≤i−1 and no overflow to the next digit is generated. So all digits besides the last one stay untouched.That means we get D(a+1,2j)=D(a,2j)+1Summing all information up we can conclude:f(a)=∑j=1a2D(a,2j)+∑j=1a2−1D(a,2j+1)f(a+1)=∑j=1a2D(a+1,2j)+∑j=1a2−1D(a+1,2j+1)+D(a+1,a+1)=∑j=1a2D(a,2j)+a2⋅1+∑j=1a2−1D(a+1,2j+1)+1Considering everything mod 2 we can use thatD(a+1,2j+1)≡D(a,2j+1)+1f(a+1)≡∑j=1a2D(a,2j)+a2⋅1+∑j=1a2−1D(a,2j+1)+(a2−1)⋅1+1≡f(a)+a2+a2−1+1=f(a)+a≡f(a)mod2Here we have it thatf(a)≡f(a+1)mod2if a is even.

Does parity of f(a) and \(\displaystyle{f{{\left({a}+{1}\right)}}}\) are

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