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Given the number sequence

15, 18, 30, 39, 54, 69, 78, 90, 93

can you predict the next two values (or more...?)

Note that

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    $\begingroup$ I was thinking of adding the lateral-thinking tag because it may help to think outside the box, but it is a mathematical number sequence problem rather than some coded message. $\endgroup$ – tom Feb 21 '20 at 18:36
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    $\begingroup$ The problem with problems like these is that they don't have unique solutions, as evidenced by the answers here. What can you point to to declare their answers as wrong compared to yours? They fit the quesiton. $\endgroup$ – Don Thousand Feb 22 '20 at 0:51
  • $\begingroup$ @DonThousand - Ok fair point.. just seeing your point now.. There is one answer which is close - so I will accept that and explain the final points. - I hope the comments with the accepted answer explain the problem more clearly and what I was trying to hold out for in an answer. I hope you agree that the relationship between the numbers in the sequence is governed by a very simple mathematical rule. $\endgroup$ – tom Feb 22 '20 at 12:33
  • $\begingroup$ For next time, you have to be a lot more careful about designing your puzzle so that it is obvious once solved that the solution is unique. Otherwise, it's a bad puzzle. $\endgroup$ – Don Thousand Feb 22 '20 at 13:33
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    $\begingroup$ @DonThousand, thanks for the comments, point taken. $\endgroup$ – tom Feb 22 '20 at 13:35
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Values:

102, 108

Because:

The change pattern is:

6, 9, 3, 12, 9, 15, 15, 9, 12, 3, 9, 6

It is shared by the sum of the digits in each number, offset by one position.

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  • $\begingroup$ So I accept your answer, but would like to make the following points. Note that the nth member of the series is equal to the sum of the digits of the first (n+1) members. This makes it similar in a way to OEIS series oeis.org/A248050 where the nth member is the sum of the digits of the first n members of the series. Note that as pointed our in the OEIS that means there is an element of 'forward look' in the sequence and numbers in the sequence are not necessarily unique. $\endgroup$ – tom Feb 22 '20 at 12:25
  • $\begingroup$ Thus the question was posed above with 'can you predict the next two .... ', and so the answer is - yes and no - possible values for the next two numbers include 102 and 108, but could also be 102 and 117. If we move to 201 or 1002 as the next number after 93 then the next number in the sequence would have a particularly large number of digits so that the sum of them was equal to the difference between 93 and 201 or 1002. $\endgroup$ – tom Feb 22 '20 at 12:29
  • $\begingroup$ Note that, for example 15 = 1+5+1+8 from 15, 18,..... and 18 = 1+5+1+8+3+0 from 15, 18, 30...... etc. $\endgroup$ – tom Feb 22 '20 at 12:45
  • $\begingroup$ Interesting theory, but it appears to be incorrect $\endgroup$ – Barry Chapman Feb 22 '20 at 16:07
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The rule is:

1st number plus 3 give 2nd, 2nd plus 12 give 3rd, 3rd plus 9 give 4th, 4th plus 15 give 5th and then in reverse 5th plus 15, 6th plus 9... so sequence is +3,+12,+9,+15,+15,+9,+12,+3,+3,+12... Next are: 96, 108.

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  • $\begingroup$ This was my first result... Back to the drawing board $\endgroup$ – TCooper Feb 22 '20 at 0:22
  • $\begingroup$ I think you are right about the next two numbers being in the sequence. However, I believe that there will be two numbers occurring before you get to those two. $\endgroup$ – Barry Chapman Feb 22 '20 at 6:56
  • $\begingroup$ This answer is closer than I realised at first, but not correct. you already have the +1 from me and can see the expected answer with the accepted solution above. $\endgroup$ – tom Feb 22 '20 at 12:39
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I think it's

105, 114

because

the change in the sum of the digits in each number is: +3 -6 +9 -3 +6 +0 -3 +3

so we continue with

-6 +9

meaning

the sum of digits following 93 must be 12-6, or 6 and the next number that satisfies that is 105. Then the sum of the digits must be 6+9 or 15, and 114 is the next number to satisfy.

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  • $\begingroup$ interesting answer +1 - see the accepted answer above and comments for the solution $\endgroup$ – tom Feb 22 '20 at 12:36
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This was the only pattern I saw:

15,     18,     30,     39,     54,     69,     78,     90,     93       <-- starting


5 3     6 3     10 3    13 3    18 3    23 3    26 3    30 3    31 3     <-- factors (5*3 etc)


        1       4       3       5       5       3       4       1        <-- difference between non-3

Now what jumped out at me was that they were multiples of 3. I just saw the differences between the factor of the other value (not the 3), that is what yields the bottom row.

The first blank spot indicates your starting point. My inclination is that is if this is the full sequence, then it will follow the pattern at the third line:

1, 4, 3, 5, 5, 3, 4, 1,  ,  , 1, 4 ... and so forth

I am making an assumption here, and that is that the full sequence repeats, thats why there are two blank spots. It wouldnt make sense for it to play through once, and then bump one of the numbers in the sequence off only to display the rest.

So in response to your question:

the next two numbers would be 93 and 93. Followed by 96, 108 and 117.

15,     18,     30,     39,     54,     69,     78,     90,     93      93      93      96      108     117


5 3     6 3     10 3    13 3    18 3    23 3    26 3    30 3    31 3    31 3    31 3    32 3    36 3    39 3


        1       4       3       5       5       3       4       1       ''      ''      1       4       3       5       5       3       4       1

Of course, it is entirely possible that the blank does not repeat at the end, but once.

In that case, the answer is: 93, 96, 108, 117

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  • $\begingroup$ Why would it just repeat after what we know though? That seems a bit arbitrary. Is there any significance to the sequence 1,4,3,5,5,3,4,1? $\endgroup$ – Rand al'Thor Feb 22 '20 at 7:34
  • $\begingroup$ Nice answer. It strikes me as being perhaps similar to MKCafe's answer, but you have explained it very clearly. I hope that the solution with the accepted answer above makes sense - and you have the +1 from me... (which I should have, but forgot to add before) $\endgroup$ – tom Feb 22 '20 at 16:15
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The answer is

108, 117

Because

If we divide the numbers in the posted sequence by three we have the sequence shown below.

   15, 18, 30, 39, 54, 69, 78, 90, 93
   5,  6, 10, 13, 18, 23, 26, 30, 31
Using the formula $(a+b)+2b-a$ you obtain the next term in the posted sequence. Below are three examples.

First pair in the lower sequence: Set $a=5, b=6$, then $(5+6)+2*6-5=18$
Second pair in the lower sequence: Set $a=6, b=10$, then $(6+10)+2*10-6=30$
Third pair in lower sequence: Set $a=10, b=13$, then $(10+13)+2*13-10=39$

and so on...

Set $a=31, b=36$ and we obtain $(31+36)+2*36-31=108$.
Set $a=36, b=39$ and we obtain $(36+39)+2*39-36=117$.

Looking at the first 5 terms of the bottom sequence, obtained by dividing the numbers in the original sequence by 3, I observed that differences of these numbers create the permutation 1, 4, 3, 5. The next terms reveal the permutation 5, 3, 4, 1. Then I use the same permutation 5, 3, 4, 1 to obtain the terms 108 and 117. In all you have 24 permutations so there can be multiple answers.

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    $\begingroup$ hmm sorry no, and I don't follow your reasoning I'm afraid. - i've edited to hide the answer from casual views $\endgroup$ – tom Feb 21 '20 at 22:05
  • $\begingroup$ @tom. I'm sorry, I got mixed up. Now I hope it is clear. $\endgroup$ – Vassilis Parassidis Feb 21 '20 at 22:42
  • $\begingroup$ I do not think this is correct. Close. But I don't think it is correct. $\endgroup$ – Barry Chapman Feb 22 '20 at 6:58
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    $\begingroup$ note that your formula (a+b)+2b-a is equivalent to 3b +a -a = 3b $\endgroup$ – tom Feb 22 '20 at 12:43
  • $\begingroup$ @Tom.The formula has to be used as is. My explanation is based on solid mathematic reasoning. $\endgroup$ – Vassilis Parassidis Feb 22 '20 at 18:25

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