u/Pixel-Jones3117

Useful fact: Near powers of 3 are 8X

Because of the 3x role in Collatz rise steps, it's useful to look at the relationship between powers of 3 and powers of 2.

Specifically, even powers of 3 (e) and odd powers of 3 (o) are always near multiples of 8 (and higher powers of 2):

3^e-1 = 8x = x*2^3 (and x*2^4, x*2^5, x*2^6, etc. for higher even powers of 3)

3^o-3 = 8x = x*2^3 (and x*2^4, x*2^5, x*2^6, etc. for higher odd powers of 3)

(Examples and proof below)

I think this may be useful when thinking about how upper terms of Collatz values change after N rise steps.

For instance, we know that a Collatz sequence starting at 2^H*C+L (where L is odd) rises to 3^H*C+Cz(L) after H steps (where Cz(L) is Collatz steps applied to lower term L).

So, any sequence that rises an even number of steps will have an upper term 3^E which by the fact given above will be (8x+1)*C. Sometimes that will be 16x, 32x, 128x.

That's more clearly interesting in binary. If you start with a number:
CCCC00000LLLL
It will rise to:
XXXCCCMMMM where MMM is Collatz expansion of L
That means the lower three binary digits (maybe more) of C are untouched after H rises.
So, even though it seems like multiple 3x+1 steps completely scramble the upper bits of the starting number, some bits are preserved.

3^(o)-3 3^(o)-3 Binary Multiple of 2^(N)
3^(3)-3 24 ...0000000000011000 8x
3^(5)-3 240 ...0000000011110000 16x
3^(7)-3 2184 ...0000100010001000 8x
3^(9)-3 19680 ...0100110011100000 32x
3^(11)-3 177144 ...0011001111111000 8x
... ... ... ...
3^(17)-3 129140160 ...0000010111000000 64x
3^(33)-3 5.55906E+15 ...0011101110000000 128x
3^(e)-1 3^(e)-1 Binary Multiple of 2^(N)
3^(2)-1 8 ...0000000000001000 8x
3^(4)-1 80 ...0000000001010000 16x
3^(6)-1 728 ...0000001011011000 8x
3^(8)-1 6560 ...0001100110100000 32x
3^(10)-1 59048 ...0110011010101000 8x
... ... ... ...
3^(16-1) 43046720 ...0101011101000000 64x
3^(32-1) 1.85302E+15 ...0011111010000000 128x

Its easy enough to prove that these near powers of 3 are always 8x.

If 3^o-3 = 8y for any odd o (which is true for 3^3 - 3 = 24)
then: 9*(8y+3)-3 = 72y + 27 - 3 = 72y + 24 = 8*(9y+3)
so: 3^(o+2)-3 is also a multiple of 8 and by induction all 3^o-3 = 8x

If 3^e-1 = 8y for any even e (which is true for 3^2 - 1 = 8)
then: 9*(8y+1)-1 = 72y + 9 - 1 = 72y + 8 = 8*(9y+1)
so: 3^(e+2)-1 is also a multiple of 8 and by induction all 3^e-1 = 8x

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u/Pixel-Jones3117 — 9 days ago