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How is this constructed? What would the next step look like?

06, 12, 14, 228, 456, 812, 1024, 3048, 6, 1, 2, 5, 1, 2

HINT:

The next step would add something at the far right of the list, as well as something after the last 6.

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2 Answers 2

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After the next step, the sequence is:

06, 12, 14, 228, 456, 812, 1024, 3048, 6096, 1, 2, 5, 1, 2, 4

The pattern is:

The sequence originally begins 0,1,1,2,4,8.
During Step $n$, the $n$th element of the sequence has all-but-the-first-digit of $2^{n+3}$ appended, and the first digit of $2^{n+3}$ is added to the end of the sequence.

Full construction:

Step 1: We have the 1st element ($0$) and the power-of-two ($2^4=16$), so the first element becomes $06$ and the sequence is now 06,1,1,2,4,8,1.
Step 2: We have the 2nd element ($1$) and the power-of-two ($2^5=32$), so the second element becomes $12$ and the sequence is now 06,12,1,2,4,8,1,3
Step 3: We have the 3rd element ($1$) and the power-of-two ($2^6=64$), so the third element becomes $14$ and the sequence is now 06,12,14,2,4,8,1,3,6
Step 4: We have the 4th element ($2$) and the power-of-two ($2^7=128$), so the fourth element becomes $228$ and the sequence is now 06,12,14,228,4,8,1,3,6,1
Step 5: We have the 5th element ($4$) and the power-of-two ($2^8=256$), so the fifth element becomes $456$ and the sequence is now 06,12,14,228,456,8,1,3,6,1,2
Step 6: We have the 6th element ($8$) and the power-of-two ($2^9=512$), so the sixth element becomes $812$ and the sequence is now 06,12,14,228,456,812,1,3,6,1,2,5
Step 7: We have the 7th element ($1$) and the power-of-two ($2^{10}=1024$), so the seventh element becomes $1024$ and the sequence is now 06,12,14,228,456,812,1024,3,6,1,2,5,1
Step 8: We have the 8th element ($3$) and the power-of-two ($2^{11}=2048$), so the eighth element becomes $3048$ and the sequence is now 06,12,14,228,456,812,1024,3048,6,1,2,5,1,2
Step 9: We have the 9th element ($6$) and the power-of-two ($2^{12}=4096$), so the ninth element becomes $6096$ and the sequence is now 06,12,14,228,456,812,1024,3048,6096,1,2,5,1,2,4

Alternate understanding:

Equivalently, you could generate the sequence 0,1,1,2,4,8,16,32,64,... by starting with 0,1,... and then appending the total sum to the end of the sequence at each step. After the start, this immediately becomes the power-of-two sequence.
Taking that sequence, make one sequence using just the first digits of each entry, and one sequence using all-but-the-first-digit:
First digits: 0,1,1,2,4,8,1 ,3 ,6 ,1 ,2 ,5 ,1 ,2 ,4 ,...
Non-Firsts: _,_,_,_,_,_, 6, 2, 4, 28, 56, 12, 024, 048, 096,...
Slide the Non-First sequence over to the left, so that the first non-empty element (the actual start of the sequence) aligns with the start of the First digit sequence. Then combine the two sequences by gluing the numbers together First before Non-First.
First digits: 0 ,1 ,1 ,2 ,4 ,8 ,1 ,3 ,6 ,1 ,2 ,5 ,1 ,2 ,4 ,...
Non-Firsts: 6, 2, 4, 28, 56, 12, 024, 048, 096,...
Combined: 06,12,14,228,456,812,1024,3048,6096,1 ,2 ,5 ,1 ,2 ,4 ,...
By only generating the original 0,1... sequence up to a certain point, the last six First digits will be left without a Non-First element to add on, and so will be left as single-digit numbers alone. The sequence given in the question does exactly this, by stopping the 0,1... sequence at 2048; the answer given does exactly this, by stopping the 0,1... sequence at 4096, one step further.

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Partial

06, 12, 14, 228, 456, 812, 1024, 3048, 6, 1, 2, 5, 1, 2

The first group of numbers from 06 to 2048 are powers of 2 starting from 16 to 2048 with the first digit being incorrect. The second group of numbers is what should be the first digit, starting from 14 (should be 64) to 2048 (should be 2048). Are you allowed to add numbers to the sequence?

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  • $\begingroup$ So, what is the next number in this sequence? $\endgroup$ Jun 9, 2018 at 16:01
  • $\begingroup$ It is only partial, and I can solve it if the question I asked is answered. $\endgroup$ Jun 9, 2018 at 16:02
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    $\begingroup$ Awwww.... But is it close? It seems too coincidental to be a coincidence $\endgroup$ Jun 10, 2018 at 2:27
  • $\begingroup$ @ZaniXu I have added a hint to the question now. $\endgroup$
    – O0123
    Jun 10, 2018 at 12:35
  • $\begingroup$ Ok, so there is supposed to be something after the 6 too? $\endgroup$ Jun 10, 2018 at 12:38

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