Bit shifting repeating pattern

Question

I'm looking for a way to bit shifting this pattern:

6 24 96

All these numbers are multiples of 2 so I was thinking there would be a way to shift them

I want to shift them so I get the pattern but it keeps repeating possibly in a loop

6,24,96,6,24,96...

Programming language isn't important, the concept is

Answer 1

Multiplying a number by 2ⁿ is equivalent to shifting it left by n bits. Multiplying by 4 can be done by shifting left 2 bits.

06₁₀ = 00000110₂
24₁₀ = 00011000₂
96₁₀ = 01100000₂

If you want to loop this pattern, you could rotate left rather than simply shifting left. To rotate left 2 bits you would shift left 2 bits and then bitwise OR the leftmost two bits back onto the right side of the number. If you were working with 8-bit numbers this would be written in C-like syntax as:

(n << 2) | (n >> 6)

Interestingly, if you rotate 96 left two bits you won't get 384 since that's bigger than an 8-bit byte can hold. Instead you get 129, because one of the 1 bits ends up rotating back to the right side.

006₁₀ = 00000110₂
024₁₀ = 00011000₂
096₁₀ = 01100000₂
129₁₀ = 10000001₂

If you then rotate 129 one more time you end up back at the beginning at 6.

Here's an interactive Python session demonstrating this. Note that {0:3} formats n as a decimal and {0:08b} as a zero-padded binary number.

>>> n = 6
>>> n = ((n << 2) | (n >> 6)) & 0xFF; print '{0:3} {0:08b}'.format(n)
 24 00011000
>>> n = ((n << 2) | (n >> 6)) & 0xFF; print '{0:3} {0:08b}'.format(n)
 96 01100000
>>> n = ((n << 2) | (n >> 6)) & 0xFF; print '{0:3} {0:08b}'.format(n)
129 10000001
>>> n = ((n << 2) | (n >> 6)) & 0xFF; print '{0:3} {0:08b}'.format(n)
  6 00000110
>>> n = ((n << 2) | (n >> 6)) & 0xFF; print '{0:3} {0:08b}'.format(n)
 24 00011000
>>> n = ((n << 2) | (n >> 6)) & 0xFF; print '{0:3} {0:08b}'.format(n)
 96 01100000
>>> n = ((n << 2) | (n >> 6)) & 0xFF; print '{0:3} {0:08b}'.format(n)
129 10000001
>>> n = ((n << 2) | (n >> 6)) & 0xFF; print '{0:3} {0:08b}'.format(n)
  6 00000110