The Dartboard Sequence The Dartboard Sequence The arrangement of the numbers around the circumference of a standard dart board is as shown below 20 1 18 4 13 6 10 15 2 17 3 19 7 16 8 11 14 9 12 5 Oddly enough, no one seems to know for sure how this particular arrangement was selected. It evidently dates back at least 100 years. Some say the pattern was devised by a carpenter named Brian Gamlin in 1896, while others attribute it to someone named Thomas William Buckle in 1913, but both of these attributions are relatively recent, and neither can be traced back to a contemporary source. Also, although it's clear that the numbers are ordered to mix the large and small together, and possibly to separate numerically close values as far as possible (e.g., 20 is far from 19), no one seems to know of any simple criterion that uniquely singles out this particular arrangement as the best possible in any quantitative sense.

It may be just an accident of history that this particular arrangement has been adopted as the standard dart board format. It's interesting to consider various possible criteria for choosing a circular arrangement of the first n positive integers. In order to get as "flat" a distribution as possible, we might try to minimize the sum of the squares of each k consecutive terms. For example, setting k = 3, the standard dard board sequence gives (20+1+18)^2 + (1+18+4)^2 + (18+4+13)^2 + ... + (5+20+1)^2 = 20478 Apparently the standard board layout described above is called the "London" dart board, and there is another, less common, version called the "Manchester" dart board, which has the sequence 20 1 16 6 17 8 12 9 14 5 19 2 15 3 18 7 11 10 13 4 for which the sum of squares of each set of three consecutive numbers is 20454, just slightly less than the London arrangement.

In contrast, if we were to arrange the numbers by just inter-weaving the largest and smallest numbers like this 20 1 19 2 18 3 17 4 16 5 15 6 14 7 13 8 12 9 11 10 the resulting sum of squares of each 3 consecutive elements is 20510, so the standard dart boards are, in this sense, more flat distributions. Needless to say, all of these arrangements are much more flat than the natural monotonic sequence 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 which has a sum of 24350. By the way, note that if the sum of the squares of every sum of three consecutive numbers for a given arrangement is S, then we can form another arrangement with the same sum simply by taking the "21-complement", i.e., subtracting each number from 21.

For example, the complement of the standard London arrangement is 1 20 3 17 8 15 11 6 19 4 18 2 14 5 13 10 7 12 9 16 which has the same sum (20478) as the London arrangement. This works because if we begin with an arrangement a,b,c,d,... having the sum S = (a+b+c)^2 + (b+c+d)^2 + (c+d+e)^2 + ... and replace each of the numbers a,b,c,... with 21-a, 21-b, 21-c,... respectively, the sum S' of this complementary arrangement is S' = [(21-a)+(21-b)+(21-c)]^2 + [(21-b)+(21-c)+(21-d)]^2 + ... = [63-(a+b+c)]^2 + [63-(b+c+d)]^2 + ...

= S + 20(63)^2 - 2(63)[(a+b+c)+(b+c+d)+...] Each of the numbers from 1 to 20 appears three times in the summation inside the square brackets in the last term, so that summation equals 630, and hence S' = S. (The same identity applies to the N+1 complement for sums of squares of every sum of k consecutive terms of a circular arrangement of the first N integers.) How would we go about finding the circular arrangement of the integers 1 to 20 that gives the smallest sum of squares of every sum of three consecutive numbers? One possible approach would be to begin with the monotonic arrangement and then check each possible transposition of two numbers to see which one gives the lowest result. Then make that change and repeat the process, at each stage always choosing the transposition that gives the steepest reduction in the sum.

This "greedy algorithm" produces arrangements with the following sums (of squares of each 3 consecutive terms around the cycle): 24350 21650 20678 20454 20230 20110 19990 19970 19950 19946 19938 19936 19930 19926 19918 Once it reaches the arrangement with the sum 19918, no further transposition of two numbers gives any reduction in the sum. Of course, this doesn't imply that 19918 is the minimum possible sum, it simply means that it is a "local" minimum. We might try to make our search algorithm more robust by considering all possible permutations of THREE numbers at each stage.

(This includes permutations of two, since some of the permutations of three numbers leave one of the numbers fixed.) Applying the greedy algorithm to permutations of any three numbers gives dartboard arrangements with the sums 24350 21542 20362 20098 19978 19954 19942 19930 Once we reach 19930, no further permutation of three numbers gives any reduction in the sum. Interestingly, this doesn't even produce as low a result as the simple transpositions, and it illustrates the fact that a local minimum need not be a global minimum. By applying permutations of three elements, the algorithm is too greedy and enters a region of the configuration space that cannot be extended by such permutations, whereas the transpositions follow a less-steep path that leads them ultimately to a lower level.

Expanding our algorithm to examine all permutations of FOUR numbers, we get a sequence of dartboard arrangements with the following sums: 24350 20678 20190 19974 19932 19918 19910 19908 19902 19900 19896 19894 Thus we arrive at the lowest sum we've seen so far, but of course this is still just a local minimum, with no guarantee that it is the lowest possible sum.

Expanding our algorithm to take the best of all the permutations of FIVE number at each stage, we get the sequence of dartboard arrangements 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 6 2 19 4 5 16 7 8 9 10 11 12 13 14 15 3 17 18 1 20 1 2 19 9 5 18 7 8 16 10 11 12 4 14 15 3 17 13 1 20 10 2 19 9 5 18 7 8 16 6 11 15 4 14 13 3 17 12 1 20 These arrangements have the sums 24350, 20406, 1999

2, and 19874 respectively.

By applying various combinations of these algorithms to various initial arrangements, we can often arrive at the ultimate sum 19874, but never to any lower sum. However, there appear to be three distinct arrangements with this sum (up to rotations and reflections). Each of them has 1 adjacent to 20, so to compare the arrangements directly we will rotate and reflect them if necessary so that they begin with 20 and 1.

With this convention, the three minimal sequences, labeled (a), (b), and (c), are (a) 20 1 11 19 2 12 16 3 14 13 5 15 10 6 17 7 8 18 4 9 (b) 20 1 11 18 2 13 15 4 14 12 5 16 9 7 17 6 8 19 3 10 (c) 20 1 12 17 3 13 14 4 15 11 6 16 8 7 18 5 9 19 2 10 The differences between the (a) and (b) sequences, and between the (c) and (b) sequences, are shown below: 0 0 0 1 0 -1 1 -1 0 1 0 -1 1 -1 0 1 0 -1 1 -1 0 0

 1 -1 1 0 -1 0 1 -1 1 0 -1 0 1 -1 1 0 -1 0 Interestingly, if we reverse the order of the lower differences and then rotate two places to the right, the result is exactly the negative of the upper differences.

This is because the (a) and (c) arrangements are the 21-complements of each other (as defined above). The (b) arrangement is "self-dual", i.e., it is its own complement. We also note that (a) and (b) differ by the transpositions (3,4) (6,7) (9,10) (12,13) (15,16) (18,19) whereas (b) and (c) differ by the transpositions (3,2) (6,5) (9,8) (12,11) (15,14) (18,17) Thus the three minimal sequences differ from each other by permutations of six numbers, and no permutations of just five or fewer numbers can transform one of these to the others using the greedy algorithm, if we require the sum to drop or remain constant on each permutation. But if we allow permutations of six numbers it becomes possible to oscillate between these three arrangements in steps with constant sums. This is an interesting example of "symmetry breaking".

At lower "energies" (permutations of fewer terms) every sequence of arrangements progresses to one of several different possible stable limiting arrangements, but at higher "energies" (permutations of more terms) these asymptotic arrangements can transform into each other, so the sequence can oscillate between them. (Of course, if we allow permutations of all 20 terms at once, then any arrangement can be transformed to any other in a single step.) Despite the extensive numerical evidence, and the apparently unique symmetry of the (a), (b), and (c) arrangements, one could still question whether our search algorithm based on permutations of five elements is guaranteed to find the global minimum. To prove that the three arrangements (a),(b),(c) presented above are indeed the absolute minimal solutions, note that the sum of the sums of three consecutive elements must be 630, which is three times the sum of the integers from 1 to 20. If we didn't require integer values, the minimal solution would be given by uniformly distributing this, so each sum of three consecutive terms would be 31.5, but since we require integer values, this is ruled out.

We could consider arrangements such that every sum of three consecutive terms is either 31 or 32, but it's easy to see that this cannot lead to an acceptable solution. Notice that the two consecutive 3-sums for the four elements n1,n2,n3,n4 are n1+n2+n3 and n2+n3+n4, so if the two 3-sums are equal, it follows that n4=n1, and hence this is not an acceptable solution (the 20 elements are distinct). Similarly we can show that two 3-sums can't alternate more than twice. Hence the flattest possible arrangements that are not ruled out by these simple considerations must have more than two distinct values for the 3-sums Indeed the solutions with 19874 consist of the 3-sum values 30, 31, 32, and 33 with valences 6,4,4,6 respectively.

These 3-sums for the (a), (b) and (c) arrangements are as shown below (a) 32 31 32 33 30 31 33 30 32 33 30 31 33 30 32 33 30 31 33 30 (b) 32 30 31 33 30 32 33 30 31 33 30 32 33 30 31 33 30 32 33 31 (c) 33 30 32 33 30 31 33 30 32 33 30 31 33 30 32 33 30 31 32 31 By examining each sequence of the values 30, 31, 32, and 33, checking to see which ones correspond to 3-sums of the integers 1 to 20,

 we find that indeed the only viable sequences are those corresponding to the arrangements (a), (b), and (c).

Thus these are the circular arrangements of the integers 1 through 20 such that the sum of squares of every 3 consecutive terms has the smallest possible value, namely 19874. (If we evaluate the sum of squares of every three consecutive elements of these 3-sum sequences we find that they yield 178614, 178618, and 178614 respectively.) The (a), (b), and (c) sequences each consist of three interleaved arithmetic progressions. If we designate the position of each number by the integers modulo 20, then the positions of the values are as shown in the table below.

positions modulo 20 values (a) (b) (c) 3k+1 -3k-3 6k 6k k = 0 to 6 3k+2 6k 6k+3 -3k-3 k = 0 to 6 3k+3 6k+3 -3k-3 6k+3 k = 0 to 5 By the way, to find the arrangement that maximizes (rather than minimizes) the sum, it's fairly intuitive that we would cluster the largest numbers together as tightly as possible. This leads to the arrangement 20 19 17 15 13 11 9 7 5 3 1 2 4 6 8 10 12 14 16 18 which has the sum 25406.

Indeed this is the highest sum I've found using the greedy algorithm with permutations of 2, 3, 4, and 5 elements (selecting the highest rather than the lowest at each stage), although it's interesting that there are many initial arrangements from which this algorithm does not lead to this global maximum. In general if H(n,k) and L(n,k) are the highest and lowest sums of squares of every k consecutive elements in a circular arrangement of the first n positive integers, are the values of H(n,k) and L(n,k) well known and/or easily computed? Another possible way of "optimizing" the arrangement of the numbers 1 through 20 on a dart board would be to minimize the sum of the squares of every sum of TWO (rather than three) consecutive numbers. In general, I think the minimum sum of squares of every sum of two consecutive numbers in a cyclical arrangement of the integers 1 through N is S_min(N) = N^3 + 2N^2 + 2N - j where j is 1 if N is odd, and j is 2 if N is even. For the particular case N=20 this formula gives a minimum sum of 8838.

For even N the minimum arrangement has the odd and even numbers restricted to separate halfs of the cycle, as illustrated below for N=20 1 3 5 7 9 10 8 6 4 2 19 17 15 13 11 12 14 16 18 20 For odd N the minimum arrangement is very simple, as shown below for N=19. 1 3 5 7 9 11 13 15 17 19 18 16 14 12 10 8 6 4 2 This raises some interesting questions. Given any circular arrangement of the integers 0 through n-1, let S denote the sum of squares of every sum of two contiguous numbers, and let v(n) denote the number of distinct values of S for all n! possible arrangements.

Following is a table of the number of distinct values of v(n) n v(n) ----- --------- 1 1 2 1 3 1 4 3 5 8 6 21 7 43 8 69 9 102 10 145 Hugh Montgomery, A. M. Odlyzko, and Bjorn Poonen developed a very nice approach to this problem, showing that the general term with n>6 is given by / (n^3 - 16n + 27)/6 if n is odd v(n) = ( \ (n^3 - 16n + 30)/6 if n is even A whole family of interesting sequences can be produced by generalizing the definition as follows: Given any circular arrangement of the integers 0 through n-1, let S denote the sum of the qth powers of every sum of k contiguous numbers. Then let v(q,k,n) denote the number of distinct values of S for all possible arrangements. With this nomenclature, the previous sequence is denoted as v(2,2,n).

Of course, we have v(1,k,n) = 1 for all k and n, because the sum of the 1st powers is independent of the arrangement. We also have v(q,1,n) = 1 because the sum of any fixed power of the individual numbers is also independent of the arrangement. Also, for fixed values of q and n, the function v is PERIODIC in k. Another generalization is to add some constant integer j to each of the numbers 0 to n-1. Thus, the general function has four indices, v(q,k,j,n). Notice that v is independent of j for q<3, but for larger values of q, j becomes significant. Can v(q,k,j,n) be expressed in closed form as a function of the indices? Which other integer sequences are contained in this family? Which continuous functions (e.g., sin(x), cos(x), exp(x), etc) can be approximated by sequences of this form? Maths Sequence Puzzles Join For Free >> Tips on solving sequence puzzles logically Sequence puzzles are commonly used in tests and quizzes. Sometimes they may be related to words and knowledge.

This article looks purely at maths related puzzles where you need to find the mathematical pattern between members of the sequence, and thereby work out what the next term in the puzzle is. As this works through, a couple of simple but effective techniques are explored for helping to discover the pattern between different elements of the sequence. Here's a very simple sequence: 1,2,3,4,5 Instantly you know that '6' comes next. But lets look at something very similar: 154,162,170,178,186 It may take a little longer to spot, but here we see that the next number is '194'. So how do we solve these puzzles quickly? Well, this is a key technique for analysing sequences: work out the difference between different terms of the sequence, thus here the differences are +8, +8, +8, +8 and therefore we instantly find out what the sequence is. When there is a constant addition or subtraction, we know the next term is n + 8, where 'n' is here the last item in the sequence we know the value of. So it is always useful to check the difference between terms.

Let's do this again: 3, 9, 27, 81 Here the difference path looks like this: +6, +18, +54 (which in turn is +12, +36). What we notice here is that the difference between terms is not constant, but is instead increasing. This means we are going to be performing some sort of multiplication, so it won't be n+ or n- a figure, but rather xn, where x is a multiplying factor. To find the multiplication factor, we now divide the terms by each other to see that 54 / 18 and 18 / 3 is 3, so the multiplication factor is 3 each time, giving us that the next term is 81 x 3 or 243. So seeing what is going on with the differences can help to solve these sums. Remember that if you get alternating negative and positive terms in a sequence, you should consider that there is a negative multiplication factor being applied, e.g. 2, -4, 8, -16, 32 is a sequence being multiplied by -2n each time.

How about sequences that are not steadily increasing by addition, subtraction, multiplication or division? For instance look at this one: 3, 6, 18, 72, 360 This time the differences show us this: 3, 12, 54, 288. Well, what we do know is there is a large increase each time, so there must be multiplication involved. However, it's not steady, as it increases. By dividing the terms we soon see that we are multiplying by 2 then 3 then 4 then 5... problem solved! If you like this sort of game, then try these sequence puzzles Maths Sequence Puzzles Page Views: 91670 | Page Ranking: 1 Popular Tags: maths sequences, patterns, solving maths puzzles Last Updated: Nov 1st 2007 Questions Dan has Answered: HI Dan, which number logically ends this sequence? 100 500 1 50 1000 5 ?

Many thanks So the sequence is: 100 500 1 50 1000 5 Multiply by 5, divide by 500 Multiply by 50, multiply by 20, divide by 200. No idea! What is the next term in the following 1) 6,8,5,10,3,14,1 2) B,0,C,2,D,0,E,3,F,3,G 30 2,3,6,1,8,6,8,4,8,4,8,3,2,3,2,3 1) 6,8,5,10,3,14,1 +2, -3, + 5, -7, + 11, - 13 So looks like this alternates adding then subtracting primary numbers. Next primary is 17, so therefore add 17 to get the next term which is 18.

2) B,0,C,2,D,0,E,3,F,3,G 30 Alternating letters of the alphabets: B, C, D etc so the next term is H. Triangle Sequences. 1 Triangle = 3 Match sticks 2 Triangles = 5 Match Sticks 3 Triangles = 7 Match Sticks. 1509 Triangles = How many Match Sticks 1 Triangle = 3 Match sticks 2 Triangles = 5 Match Sticks 3 Triangles = 7 Match Sticks. 1509 Triangles = How many Match Sticks Ok, so the pattern is 2n + 1. Therefore this (1509 x2) + 1. This is 3019. What is the next number in this sequence: 1,18,4,13,6,10 So the sequence is: 1,18,4,13,6,10 There are many possible answers to this sequence so no absolute answer. One option is that this is two sequences interlinked, so the relevant one here is 1, 4, 6 ?

which is add three, add two, then add one so the answer is 7. There are many other possible answers. Hi, if a sequence is 3, 3, 5, 4, 4, 3, 5 how was this worked out and what is the next number? 3, 3, 5, 4, 4, 3, 5 is the sequence. Add 0, add 2, minus one, add zero, minus one, add 2. No obvious sequence here so not sure. Complete the missing numbers - - - 112 - 162 - - 237 112 162 237 Add 50, add 75, so next add 100 to get 337. Whqt is the sequence? 1,2,4,7,11,16 1,2,4,7,11,16 Add one, two, three, four, five so add six to get 22. 1 8 27 64 1,8,27,64. This is the cube of each number, e.g.

1x1x1, 2x2x2 etc so here we have 5x5x5 which is 125 as the next term. Hi dan what no ends this sequence 1 8 27 64 125 Find the next 3 numbers in the following sequence: 2,3,5,7,11,13,17 2,3,5,7,11,13,17 This looks like a list of prime numbers, so next comes 19 then 23. I dont understand how to work out n-3 from 1,5,6,11,17,24,32,41,51,62 please could you help from zak 6,11,17,24,32,41,51,62 Here you are adding 5 then 6 then 7 then 8 then 9,10,11 so next 12 to get 74. Whose famous sequence starts1123581321 Yes this is called the fibbonaci sequence.

3, 6, 18, 72, 360 what is the next number Multiply by 2, then 3, then 4 and so on. So here we have 360 x 6 so it is 2160. Hi could you please tell me the next two numbers in this sequence 0,2,3,6,7,1,_,_, So the sequence is 0,2,3,6,7,1,_,_, What comes next is 9 then 4 then 5 then 8. It is the sequence of numbers written as words in english in reverse alphabetical order - quite a nasty sequence to work out as it's not a mathematical relationship. What are the next two numbers in this sequence? 2, 5, 11 2, 5, 11... there are only three terms here so there are many possible sequences. However, the most obvious is to double and add one each time. This makes 11x2 + 2 the next term: 24, followed by 50. In sum then, this is 2,5,11,24,50.

To find a missing number in a pattern, you may have to try more than one ____________. ... possible combination. This is basically saying a trial and error method could work; however there is more than one possible way to finish this sentence - in fact several answers, so your teacher must be looking for a sensible answer not a specific one! What is next in the sequence: 1,4,?,?,?,36,? 1,4,?,?,?,36,? This is the square numbers, e.g. 1x1, 2x2. So we have 1,4,9,16,25,36,49 I have to give the next three answers to this sequence 2 3 5 7 11 13 17 So we have... 2 3 5 7 11 13 17 This is the sequence of prime numbers so we have 19 then 23 next.

Write a number in the brackets so they are in order -8,( ),-5,( )-2( ) There are lots of choices - it could be -7 or -6 in the first bracket, then -4 or -3 in the second bracket, then any number from -1 up to positive infinity in the third bracket What are the next terms and the formula for the sequence : 3;6;7;-2 ? Can't see anything obvious here. Hi.please can you tell me what comes next in the following 3 thanks 0 2 3 6 7 1 - - 20 1 18 4 13 - - 8 5 4 9 1 - - 0 2 3 6 7 1 9 then 4 20 1 18 4 13 6 then 10 8 5 4 9 1 7 then 6 What is the sequence 15 19 _ _ _ 65 84 Not sure on this one. What is the next number in the sequence 2,4,3,9,8,-?

Sequence appears to be 'square it, subtract one, square it, subtract one' therefore next number is 64. Hi can you help with the next 2 numbers in the sequence 20 1 18 4 13 Several possible sequences here depending whether you look for one or two interweaving sequences etc. One possible answer is 6,10. Hi Dan, Could you help me work out this sequence please 2,6,8,14,22,36 Add 22 to get 58 - basically add the previous term each time, e.g. to get 8 add 2 to 6, to get 14 add 6 to 8 and s on. Wthat is the rule for this sequence. 0, 1, 4, 11 0, 1, 4, 11: Add 1, add 3, add 7. So the sequence could be 'double the difference and add one' meaning the next term would be 11 + 15 = 26. 3354435 what is the next two numbers thanks This is one long number...

not sure how you can work out what comes next :s What are the 50th, 63rd and 100th numbers in the sequence below? 4, 5, 8, 8, 9, 9, 12, 13, 13, 13, 17, 18, ... How do you do this?

4, 4, 5, 8, 8, 9, 9, 12, 13, 13, 13, 17, 18, 21, 22, 22, 23, 26, 26, 27, 26, 30, 31, 34, 34, 35, 35, 38, 39, 39, 36, 40, 41, 44, 44, 45, 45, 48, 49, 49, 45 Hello dan, can you tell me a formula for this sequence 2 4 4 6 6 8 8 10 8 8 6 6 4 4 2, and how you found it? Looks like 'add two, add nothing, add two, add nothing etc' then start taking away from 10.

What is the connecting rule here (S) 1 2 3 4 5 (t) 1 3 7 15 31 Double and add one each time, so for the next term double and add one to get 63. What are the following pattern to 2,3,5,7,11,13? please tell me no later then 9 This looks like the sequence of odd numbers. HI Dan, Have a few sequences I need help with, need to find next 2 numbers in each sequence. 7-0-1-2-7-1-2-3 20-23-23-23-25-25-26-27-27 87-69-75-61-63-53-51 36-42-35-28-45-21-14-7 Thanks Pete 7-0-1-2-7-1-2-3 Looks like it might be 7 next... not sure on the others though!

Help 1 11 21 1211 111221 312211 13112221 1113213211 Next? 1113213211 FIND RHE MISSING NUMBER 1 3 4 13 ? FIND THE MISSING NUMBER 1 3 4 13 ? Several options. One possible sequence is to add on the square of the previous number; so 4 is 3 + 1x1, 13 is 4 + 3x3 so next would be 13 + 4x4 = 29 What numbber comes next and how do you know 1,4,9,1,6,2,5,3,6,4,9,6,4 8,1 comes next. It is the squares of each number in turn with commas inbetween. e.g. 1 squared, 2 squared, 3 squared, 4 squared etc is 1,4,9,16 but with commas added between every digit becomes 1,4,9,1,6...

1.....11.....21.....1,211.....111,221.....312,211 what would be the next # and explain 13112221 comes next. This is one of those rather silly sequences that is not mathematical but descriptive. Start with one, then describe it - there is one one, so that's 11. Next describes this, there are two ones, or 21, and so on. How do you finish this sequence 0,2,3,6,7,1 9,4,5,8 I have the sequence 9, a, 4, b. If b=8/3, how do you get that answer and what is a? 9, a, 4, 8/3 The answer is 6. The sequence is dividing by 1.5 for each term. Which best completes the sequence 3,4,6,9,13 18 comes next. This is add one, then add two, then add three etc, so we add five to get 18. Dan, help.... trying to do my son's homework and stuck on this one...

2, 2, 3, 4, 6, 9, 14, ___, ___, ___ help 2, 2, 3, 4, 6, 9, 14, The next number is 22; it looks like we sum the differences between the previous two terms to find the next item in the sequence: 0,1,1,2,3,5 - so since 5+3 is 8 then the next number is 14 + 8 = 22. Please solve What are the next terms in the sequence 25,36,47? 25,36,47... Looks like this is adding 11 each time, so next up will be 58. 1,1,2,3,5,8,13,21 1,1,2,3,5,8,13,21 Add the previous two terms each time, so next is 21 + 13 = 34 Whats next in the sequence? 3, 8, 17, 26, 34,.... See previous answer!

How do you do 1,1,2,3,5,8,13,21...... pretty pretty please with a cherry on top answer my question Add the previous two numbers; next is 34. What is the next number in the number pattern 2 10 12 17 18 19 200, these numbers start with 'D' in French What is the rule in this sequence 1 , 4 , 14 Add three, add ten, so looks like adding seven more each time, so now add 17, so 31. Dan : I have a number sequence for you to solve? I am wondering if it is a valid sequence or not? Here it is as follows? 1,4,10,19,31,__. Possible answers are:a(36), b(48),c (46), or no apperent pattern is answer d? Dale Add 3, 6, 9, 12 so next add fifteen making the answer c - 46.

What is the next number in this sequence 2, 18, 1, 9, 14, 16, 15, 23, 5, 6 What is the nth term of this sequence? 4,7,12,19,28 Add 3, 5, 7, 9 so next add 11 to get 39. So n = n-2 + (n-2 - n-1) + 2 4 3 6 3 2 7 2 1 8 1 .. What is the next term in this sequence? 0, 9 64,16,4,1,1/4 which comes next Divide by four each time, next in this maths sequence is therefore 1/16 Hi Dan, Can you explain how you get 6 & 10 as the next two numbers in the sequence 20, 1, 18, 4, 13? Many thanks. 20, 1, 18, 4, 13, 6, 10. No idea sorry!

The sequence 3,3,5,4,4,3,5 is explained thus: the nth terms of the sequence, 1,2,3,4,5,6,7 are spelt one two three four five six seven the sequence is thus the number of letters in each word. Good answer. What is the next number in tis sequence: 1 2 4 3 6 8 7 14 16 ? Take away one, double it, add two in turn, so 15 next. 1, 2, 6, 42, 1806 Hi Dan can you find the next number in the above sequence please? 3263442 All you do is multiply by the previous number plus 1, eg 2 is (1+1)x1 6 is (2+1)x2 42 is (6+1)x6 2n+6=0 Re-arrange for n is it? Well 2n = -6 Therefore n = -3 Hi Dan I\'m stumped! Can you complete this sequence please?

...,7.5,3,1.2,...,0.192,0.0768 It is 0.48 - just multply by 2.5 at each stage of the sequence. Childs Homework im stuck the sequence is CYGTNLIT The answer is 'S': can you get the next letter in this Sequence. How do you find a formula for finding the nth term in a square number sequence? The formula is just n x n For instance in the sequence of squares: 1, 4, 9, 16 this is 1 x 1, 2 x 2, 3 x 3, 4 x 4 - so therefore the nth term is just n x n Dan, we know the numerical answer to a sequencial problem but we do not know how to write it out. (1,6,12,19). we know the answer is 27, but how does he state it in words. Add 5, then 6, then 7, then 8.

So the nth term is (n-1) + (i+3) where n is the given term in the sequence and i is the position of the term. So to get six for instance we get the n - 1 term which is 1. Then we add to this the position of six in the sequence (position two) and add three to that, so 1 + 5 = 6 What are the next three numbers in the pattern 1 11 21 1211 111221 312211 13112221? 1113213211 comes next 0,2,5,10,17,28,41,58,__,__,__ find the pattern and the next three numbers in the sequence Add 2,3,5,7,9... Hi. i am stuck on a question. what type of numbers are 2 4 8 16 32 and 64. thanks These are even numbers - all multiples of 2! Find the missing term in the sequence? Q.1.

7 9 5 11 4 15 12 7 13 8 11 ? --------------------------------------- Q.2. 1 3 4 8 15 ? --------------------------------------- Q.3. 58 35 ? 11 6 3 --------------------------------------- Q2: various options, could be sum the previous three numbers to get the next number, if so 15+8+4 = 27. What is the missing number (and explanation) for this sequence? 12 15 19 21 22 22 24 26 26 ?? 29 Don't know What is the rest of the pattern 1 5 13 29? Double it and add three, eg 2 x 29 + 3 = 61 What comes next in these 2 sequences please. 4 5 4 2 5 3 ? 20 20 8 10 15 ? thanks sc Not sure on this one... looking at all these sequence questions on this page is like a full time job!!!

;) Hey Dan, I just took a test with random sequence questions, and I think I remembered the numbers from one that really stumped me: 123,615,1002,___ I\'m stumped! And, for some reason, I *need* to know the answer. Any help? With only three terms there are many possible sequences that fit. The differences between these terms are: 497, 387 so one sequence could be add 497 then 110 less each time so next you would add 277 to get 1279. But with so few terms there are many possible answers so they should really have asked you a sequence with more terms!

What is the formula for this sequence: 0, 100, 300, 700, 1400, 2500, 4100, 6300, 9200, 12900 Add 200, 400, 700, 1100, 1600, 2200, 2900 What is the missing number in this sequence 1, 8, 27, ?, 125, 216. and how do we find it? This is the sequence of cubes, 4 x 4 x 4 = 64 1, 2, 5, 10, 20, next two numbers and why please Looks like 50,100 as the coins in legal tender in the UK in pence. What are the next 2 numbers in the sequence 12,5,20,1,18,14 Not sure. 1, 2, 6, 42, 1806 ?

Answer is 3263442. The sequence is the number x the number + 1, therefore next is (1806 x 1807). Number Sequences - Square, Cube and Fibonacci A Geometric Sequence is made by multiplying by the same value each time. Example: 1, 3, 9, 27, 81, 243, ... This sequence has a factor of 3 between each number. The pattern is continued by multiplying by 3 each time, like this: What we multiply by each time is called the "common ratio". In the previous example the common ratio was 3: We can start with any number: Example: Common Ratio of 3, But Starting at 2 2, 6, 18, 54, 162, 486, ... This sequence also has a common ratio of 3, but it starts with 2. Example: 1, 2, 4, 8, 16, 32, 64, 128, 256, ... This sequence starts at 1 and has a common ratio of 2.

The pattern is continued by multiplying by 2 each time, like this: The common ratio can be less than 1: Example: 10, 5, 2.5, 1.25, 0.625, 0.3125, ... This sequence starts at 10 and has a common ratio of 0.5 (a half). The pattern is continued by multiplying by 0.5 each time. But the common ratio can't be 0, as we get a sequence like 1, 0, 0, 0, 0, 0, 0, ... What is the next number in the sequence 3, 10, 29, 66, 127? - Quora Quora Written Aug 14, 2014 First of all let me clarify that the next term can be a lot of terms, because theoretically there exists more than one function which can connect these points. graphically speaking, these 5 points can be joined by infinitely mane curves, each of which corresponds to a different function. I am assuming a polynomial function for finding that particular function.

Since I have only 5 points, I can assume that the polynomial has degree four( since the number of unknown coefficients in that case is 5, which is precisely why we can solve the simultaneous equation from the given data).

So, let, [math]f(x)=ax^4+bx^3+cx^2+dx+e[/math] Therefore, [math]f(1)=3=a+b+c+d+e[/math] [math]f(2)=10=16a+8b+4c+2d+e[/math] [math]f(3)=29=81a+27b+9c+3d+e[/math] [math]f(4)=66=256a+64b+16c+4d+e[/math] [math]f(5)=127=625a+125b+25c+5d+e[/math] Now, after doing a lot of mathematics, which will do nothing but lengthen my answer, we find that, [math]a=c=d=0, b=1 [/math] and [math]e=2[/math] Thus [math]f(x)=x^3+2[/math] Hence, [math]f(6)=6^3+2=218[/math] Thus 218 is one of

 the infinitely other numbers in the next position of the series.

Question: In darts, what is the next in the sequence 1, 18, 4, 13, 6?
Answer: 10