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Next Number in the Sequence. 
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#1
Feb1006, 12:19 AM

P: 75

You know the routine, what is the next number in this sequence.
4, 8, 9, 10, 12, 16, 18, 20, 24, 25, 27, 28, ... I think that I would rate this as difficult. 


#2
Feb1006, 03:00 PM

Sci Advisor
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P: 2,586

Okay, well I primefactored first the list of numbers that weren't in the sequence, then those that were in:
1, 2, 3, 5, 6, 7, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 1, 2, 3, 5, 2x3, 7, 11, 13, 2x7, 3x5, 17, 19, 3x7, 2x11, 23, 2x13 4, 8, 9, 10, 12, 16, 18, 20, 24, 25, 27, 28 2x2, 2x2x2, 3x3, 2x5, 2x2x3, 2x2x2x2, 2x3x3, 2x2x5, 2x2x2x3, 5x5, 3x3x3, 2x2x7 The list of numbers in the sequence all appear to have prime factorizations with a repeated factor, and the list of nonelements does not EXCEPT the above characterization fails for 10, which appears to be the only anomaly. What's so special about 10? Bah! There's probably nothing special about 10, and even though the above solution is "off by 1" and seems tempting to think it's close, chances are its not, so I shouldn't get hung up on it. I'll have to keep thinking. 


#3
Feb1006, 03:10 PM

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Sci Advisor
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P: 16,091

I'm annoyed 25 comes before 27 in the original sequence.



#4
Feb1006, 03:37 PM

P: 160

Next Number in the Sequence.
If all we want is "the next number" and not an infinite sequence then we can find rules that apply to this number only. Example: when a number plus 4 (the starting point) is an integer multiple of 8 then add 4. After 4: 8. After 12: 16. After 20: 24. So after 28: 32. There must be others like it.



#5
Feb1006, 04:07 PM

P: 75

AKG: That is the first direction I headed with this as well, and really couldn't find any patterns right off having to do with primes. I have a good feeling that primes are somehow involved in the solution.
Also, this is a nontrivial problem. There is something interesting actually going on with this sequence of numbers. If you would like more terms of the sequence, I'd be happy to give them to you. I stopped where I did rather arbitrarily. 


#6
Feb1006, 04:22 PM

P: 160




#7
Feb1306, 12:32 AM

P: 75

AKG, I feel like a total ***. 10 is not in the sequence. I noticed that today while going over this again, and immediately thought of what you said, and wondered if 10 was the number. Sure enough. I should have taken note of that and checked 10. Flagrant mistake on my part.
I would imagine that you have solved the problem. My work with this sequence is still far from over, though. 


#8
Feb2106, 04:31 AM

P: 3

According to me the sequence is a combination of 4 sequences triggered by four input numbers,in this case these four no.'s are 4,8,9,10.
these sequnces follow there own rules unaffected by others. All have only one thing in common that there first two numbers show anamolous behaviour setting the basic rule for the rest. For the first input4 ,the sequence is 4 , 4*(41)*1 , 4*(41)*2 ,4*(41)*3 ,4*(41)*4 ,4*(41)*5 so on Giving the sequence as 4,12,24,36,48,60 .... For the second parameter 8 ,the sequence is 8 , 8*2 + Z , 8*3 + Z , 8*4 +Z so on such that Z = (n1) \ 2 ( where n is the no. of term) and Z belong to integers .If Z is not an integer it is taken as 0 .This means that n is defined only for odd integers. .e.g. for n= 2 the term is given by T = 8*2 + (21)\2 .Taking Z=0 as z is not integeral for n=2 we get T =16. This rule gives the sequence as follows  8 , 16 , 25 , 32 , 42 ..... For the third parameter 9 the rule is the simplest 9*1 ,9*2 , 9*3 ,9*4 ,9*5 so on to give the sequence as 9,18,27,36,45,54 ............. For the fourth parameter 10 , The rule is 10 , 10*2  X , 10*2  X , 10* 3 X so on such that X = [n  { (n 1) \2 } ] where n is the no. of term and the part in curly brackets i.e. A={ (n 1) \2 } is the smallest integer greater than or equal to A itself. e.g. for n = 2 the term is calculated as T=10*2  [2  { (2 1)\2 } ] =20  0 = 20 to give the sequence as follows 10, 20 , 28 , 38 ......... The four sequeces obtained are  4, 12, 24, 36, 48 .... 8 , 16 , 25 , 32 , 42 ..... 9, 18, 27, 36, 45....... 10, 20 , 28 , 38, 47 ......... Combing the correspong terms in the above written sequeces we get the final sequence as  4,8,9,10,12,16,18,20,24,25,27,28,36,32,36,38,48,42,45,47,...... Note that the the sequence doesnot follow any increasing decreasing p[attern as it is a combination of several independent sequences. I hope my thinking is right . Do notify me or my rights and wrongs . Akash 


#9
Feb2106, 09:15 AM

P: 75

Wow. That is just a bit too complicated. It also doesn't work. Sorry, but 10 is not in the sequence as I stated in my last post, but it's understandable that you didn't want to read down since you were solving the puzzle. With 10 in the sequence, it's something radically different, so your solution is correct in that regard.
The sequence as I intended it (no 10) is the sequence of "n"s for the integers mod n such that that ring has a nonzero nilpotent. a nilpotent is an element of a ring such that a^2=a. If you check this out, you will see that in the integers mod 4, 2^2=4=0. In the integers mod 6, the only nilpotent is zero. To characterize the "n"s (and the sequence above) you prime factor. If the integer in question has a prime factor raised to the 2nd power or higher, it will have a nonzero nilpotent(s). 


#10
Feb1608, 08:27 PM

P: 6

i got 40.
my work looks like this; good luck figuring it out, and let me know what you think. it's probably wrong anyways. 4, 8, 9, 10, 12, 16, 18, 20, 24, 25, 27, 28, ... 40 4 1 1 2 / 4 2 2 4 / 1 2 1 12/ 8 12 16 basically, if you take the differences of the numbers given, you get my second row of numbers (4,1,1,2,4,2,2,4...) divide them into sets of four (i have done this here using slashes,) and find the sum of the differences for each set. for each set, the sum increases by 4 when going from left to right. so what i found, is that the sum of a set of 4 differences from the original numbers is 4 more than the previous set of differences' sum. am i right or totally off? 


#11
Feb1608, 11:02 PM

P: 15,319




#12
Feb1608, 11:13 PM

P: 15,319

4x1 (jump 1) 4x2 (jump 3) 4x3 (jump 1) 4x4 (jump 2) 4x5 (jump 1) 4x6 (jump 3) 4x7...
...(jump 1) 4x8 = 32 I doubt this answer though, it's trivial. 


#13
Apr408, 08:42 AM

P: 1,417

I also think 32
And if that's correct, is the following number 34? 


#14
Apr408, 09:49 AM

P: 1

Not squarefree numbers



#15
May309, 04:41 AM

P: 1

The next others after 32 are: 36,37,38,40,44,46,48,52,53,55,56,60,… This sequence can be grouped into three subgroups or subsets. A: [32,36,37,38,40], B: [40,44,46,48,52], C:[52,53,55,56,60] The sequential order of increase differs as the sequence enters the next subset. To my understanding the following are the rules of increase of each subgroup or subset: Subset A:{4,1,1,2}; B:{4,2,2,4}; C:{1,2,1,4} The beginning number of each subset is being picket from the last sequence of each subset. The order is cyclic of subset A>B>C>A>B>C. e.g. A :{ a, a+4, a+5, a+6, a+8}; B :{ [a+8], [a+8] +4, [a+8] +6, [a+8] +8, [a+8] +12}; C:{ [a+8]+12,([a+8]+12)+1,( [a+8]+12)+3,( [a+8]+12)+4,([a+8]+12)+8} OR A:{x+4,x+5,x+6,x+8}; B:{x+12,x+14,x+16,x+20}; C:{x+21,x+23,x+24,x+28} We now need to find a generic formular for this 


#16
May2309, 12:45 AM

P: 1

4, 8, 9, 10, 12, 16, 18, 20, 24, 25, 27, 28, 30, 32, 36.
and it ends there. the problem is thinking you start with 4, you don't. you start with 16, and focus on 20. 16+4(the first number)=20 then it follows 16+8=24 16+9=25 then, after 3 you jump to 18 and skip the first 2 numbers in the sequence: 18+9=27 18+10=28 18+12=(30) then you jump to 20 and jump 2 more: 20+12=32 and then, the final way to prove you're right, the check, the last calculation before the pattern begins to collapse upon itself: 20+16=36 which is 6 squared, and follows the square rule in the pattern. to put it simply: start with 16, add the first 3 sequences, 20, 24, 25 then go to 18, skip the first 2 sequences, and do the next 3. then jump to 20, skip the next 2 sequences, (or the first 4), and do the next 2, as that's all you need to check it and it eventually collapses in on itself, to where you get 32 and 36, which ultimately proves the pattern. rough stuff. 


#17
Jun209, 09:12 PM

P: 1

I think the next number in the sequence may be 30.
It seems that the numbers in the sequence are all created as a result of prime factors. For example: 4= 2,2 8=2,2,2 9=3,3 10=2,5 12=2,2,3 16=2,2,2,2 18=2,3,3 20=2,2,5 24=2,2,2,3 25=5,5 27=3,3,3 28=2,2,7 If this is correct then 2,3,5 = 30 I could be wrong of course. 


#18
Nov2511, 11:47 PM

P: 3

what is the next numbers of 3,12,39,40,45,49



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