Can someone tell me why this happens?

I observed this awkward repetition while making a strategy for the game "odd-eve". Can someone tell me why this happens?

Let a1{ a }_{ 1 } be the sum of digits of xx in base 10.

Let an{ a }_{ n} be the sum of digits of an1{ a }_{ n-1} in base 10, for integers n>1n>1.

Let g(x)=limnang\left( x \right) =\lim _{ n\rightarrow \infty }{ { a }_{ n } }

I observed that:-

g(2)=2g\left( 2 \right) =2 g(4)=4g\left( 4 \right) =4 g(8)=8g\left( 8 \right) =8 g(16)=7g\left( 16\right) =7 g(32)=5g\left( 32 \right) =5 g(64)=1g\left( 64 \right) =1

then

g(128)=2g\left( 128 \right) =2 g(256)=4g\left( 256\right) =4 g(512)=8g\left( 512 \right) =8 g(1024)=7g\left( 1024 \right) =7 g(2048)=5g\left( 2048 \right) =5 g(4096)=1g\left( 4096 \right) =1

and this pattern is repeating again and again.

g(8192)=2g\left( 8192 \right) =2 g(16384)=4g\left( 16384 \right) =4

and so on.........

So, The pattern for powers of 2 is 2,4,8,7,5,1,2,4,8,7,5,12,4,8,7,5,1,2,4,8,7,5,1

and

The pattern for powers of 3 is 3,9,9,9,9,9,9,9,9,9,9,93,9,9,9,9,9,9,9,9,9,9,9

The pattern for powers of 4 is 4,7,1,4,7,1,4,7,1,4,7,14,7,1,4,7,1,4,7,1,4,7,1

The pattern for powers of 5 is 5,7,8,4,2,1,5,7,8,4,2,..5,7,8,4,2,1,5,7,8,4,2,..


Why does it work?

Does it work for powers of all integers?

Does it work in all bases?

Note by Archit Boobna
4 years, 5 months ago

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1 vote

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Hint: mod9 \bmod 9 , divisibility rule.

Pi Han Goh - 4 years, 5 months ago

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Thanks so much!

I got it. The sum of digits mod 9 is same as the number mod 9.

And 2^n mod 9 is repeating every few terms.

Archit Boobna - 4 years, 5 months ago

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One line. That beautiful number 99! +1

Raghav Vaidyanathan - 4 years, 5 months ago

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