Sicherman on steroids

Probability Level 5

Exploring this idea for twelve-sided dice, in how many different ways can one construct an (unordered) pair of 12-sided dice which

  • are not normal, 12-sided dice (i.e. not simply numbered 1 through 12),
  • bear only positive integers, and
  • when summed have the same probability distribution as the sum of a pair of normal, 12-sided dice?
0 1 4 7

Arjen Vreugdenhil by Arjen Vreugdenhil , 44, USA

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3 solutions

Mark Hennings
Jun 6, 2017

We want to factorise the probability generating function 1 144 ( t + t 2 + t 3 + ⋯ + t 11 + t 12 ) 2 = 1 144 t 2 ( t + 1 ) 2 ( t 2 + t + 1 ) 2 ( t 2 + 1 ) 2 ( t 2 − t + 1 ) 2 ( t 4 − t 2 + 1 ) 2 \tfrac{1}{144}(t + t^2 + t^3 + \cdots + t^{11} + t^{12})^2 \; = \; \tfrac{1}{144}t^2(t+1)^2(t^2+t+1)^2(t^2+1)^2(t^2-t+1)^2(t^4 - t^2+1)^2 of the random variable of the sum of two normal 12 12 -sided dice as the product P ( t ) Q ( t ) P(t)Q(t) of two polynomials where

  • neither P ( t ) P(t) nor Q ( t ) Q(t) have a constant coefficient,
  • both 12 P ( t ) 12P(t) and 12 Q ( t ) 12Q(t) have nonnegative integer coefficients,
  • P ( 1 ) = Q ( 1 ) = 1 P(1) = Q(1) = 1 .

There are a number of possible candidates:

  • P ( t ) = P a , b ( t ) = 1 12 t ( t 2 + t + 1 ) ( t + 1 ) 2 ( t 2 − t + 1 ) a ( t 4 − t 2 + 1 ) b P(t) = P_{a,b}(t) = \tfrac{1}{12}t(t^2+t+1)(t+1)^2(t^2-t+1)^a(t^4-t^2+1)^b and Q ( t ) = Q a , b ( t ) = 1 12 t ( t 2 + t + 1 ) ( t 2 + 1 ) ( t 2 − t + 1 ) 2 − a ( t 4 − t 2 + 1 ) 2 − b Q(t) = Q_{a,b}(t) = \tfrac{1}{12}t(t^2+t+1)(t^2+1)(t^2-t+1)^{2-a}(t^4-t^2+1)^{2-b} for some integers 0 ≤ a , b ≤ 2 0 \le a,b \le 2 . All of these possibilities satisfy the condition P ( 1 ) = Q ( 1 ) = 1 P(1) = Q(1) = 1 , but not all of these options give nonnegative coefficients. The cases that do give nonnegative coefficients are ( a , b ) = ( 1 , 0 ) , ( 1 , 1 ) , ( 2 , 0 ) , ( 2 , 1 ) (a,b) = (1,0), (1,1), (2,0), (2,1) . This gives us 4 4 possible dice pairs.

  • P ( t ) = R a , b ( t ) = 1 12 t ( t 2 + t + 1 ) ( t + 1 ) ( t 2 + 1 ) ( t 2 − t + 1 ) a ( t 4 − t 2 + 1 ) b P(t) = R_{a,b}(t) = \tfrac{1}{12}t(t^2+t+1)(t+1)(t^2+1)(t^2-t+1)^a(t^4-t^2+1)^b and Q ( t ) = R 2 − a , 2 − b ( t ) Q(t) = R_{2-a,2-b}(t) for some integers 0 ≤ a , b ≤ 2 0 \le a,b \le 2 . The ( a , b ) = ( 0 , 2 ) (a,b) = (0,2) case yields some negative coefficients. The cases ( 0 , 0 ) (0,0) and ( 2 , 2 ) (2,2) give the same pair of dice, as do the cases ( 0 , 1 ) (0,1) and ( 2 , 1 ) (2,1) and also the cases ( 1 , 0 ) (1,0) and ( 1 , 2 ) (1,2) . The case ( 1 , 1 ) (1,1) gives the normal pair of dice. Thus there are 3 3 possible new pairs of dice.

That makes 7 \boxed{7} pairs of dice in total.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Write D n D_n for a uniform distribution on { 0 , 1 , … , n − 1 } \{0,1,\dots,n-1\} . Then we are looking for different ways to rewrite 2 + D 12 + D 12 2 + D_{12} + D_{12} . Now we can split D 12 D_{12} into at most three parts, featuring D 3 D_3 once and D 2 D_2 twice: D 12 = D 3 + 6 D 2 + 3 D 2 = 2 D 3 + 6 D 2 + D 2 = 4 D 3 + 2 D 2 + D 2 . D_{12} = D_3 + 6D_2 + 3D_2 = 2D_3 + 6D_2 + D_2 = 4D_3 + 2D_2 + D_2.

New twelve-sided dice are created by swapping out D 2 D_2 terms. We explore the six possible decompositions of 2 + D 12 + D 12 2 + D_{12} + D_{12} :

  • ( 1 + D 3 + 6 D 2 + 3 D 2 ) + ( 1 + D 3 + 6 D 2 + 3 D 2 ) (1 + D_3 + 6D_2 + 3D_2) + (1 + D_3 + 6D_2 + 3D_2) . There is one possible swap: D A = 1 + D 3 + 6 D 2 + 6 D 2 = [ 1 , 2 , 3 , 7 , 7 , 8 , 8 , 9 , 9 , 13 , 14 , 15 ] ; D B = 1 + D 3 + 3 D 2 + 3 D 2 = [ 1 , 2 , 3 , 4 , 4 , 5 , 5 , 6 , 6 , 7 , 8 , 9 ] . D_A = 1 + D_3 + 6D_2 + 6D_2 = [1,2,3,7,7,8,8,9,9,13,14,15]; \\ D_B = 1 + D_3 + 3D_2 + 3D_2 = [1,2,3,4,4,5,5,6,6,7,8,9].

  • ( 1 + 2 D 3 + 6 D 2 + D 2 ) + ( 1 + 2 D 3 + 6 D 2 + D 2 ) (1 + 2D_3 + 6D_2 + D_2) + (1 + 2D_3 + 6D_2 + D_2) . Again, one possible swap: D A = 1 + 2 D 3 + 6 D 2 + 6 D 2 = [ 1 , 3 , 5 , 7 , 7 , 9 , 9 , 11 , 11 , 13 , 15 , 17 ] ; D B = 1 + 2 D 3 + D 2 + D 2 = [ 1 , 2 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , 6 , 7 ] . D_A = 1 + 2D_3 + 6D_2 + 6D_2 = [1,3,5,7,7,9,9,11,11,13,15,17]; \\ D_B = 1 + 2D_3 + D_2 + D_2 = [1,2,2,3,3,4,4,5,5,6,6,7].

  • ( 1 + 4 D 3 + 2 D 2 + D 2 ) + ( 1 + 4 D 3 + 2 D 2 + D 2 ) (1 + 4D_3 + 2D_2 + D_2) + (1 + 4D_3 + 2D_2 + D_2) . Another single swap here: D A = 1 + 4 D 3 + 2 D 2 + 2 D 2 = [ 1 , 3 , 3 , 5 , 5 , 7 , 7 , 9 , 9 , 11 , 11 , 13 ] ; D B = 1 + 4 D 3 + D 2 + D 2 = [ 1 , 2 , 2 , 3 , 5 , 6 , 6 , 7 , 9 , 10 , 10 , 11 ] . D_A = 1 + 4D_3 + 2D_2 + 2D_2 = [1,3,3,5,5,7,7,9,9,11,11,13]; \\ D_B = 1 + 4D_3 + D_2 + D_2 = [1,2,2,3,5,6,6,7,9,10,10,11].

  • ( 1 + D 3 + 6 D 2 + 3 D 2 ) + ( 1 + 2 D 3 + 6 D 2 + D 2 ) (1 + D_3 + 6D_2 + 3D_2) + (1 + 2D_3 + 6D_2 + D_2) . Again, one possible swap: D A = 1 + D 3 + 6 D 2 + D 2 = [ 1 , 2 , 2 , 3 , 3 , 4 , 7 , 8 , 8 , 9 , 9 , 10 ] ; D B = 1 + 2 D 3 + 6 D 2 + 3 D 2 = [ 1 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 14 ] . D_A = 1 + D_3 + 6D_2 + D_2 = [1,2,2,3,3,4,7,8,8,9,9,10]; \\ D_B = 1 + 2D_3 + 6D_2 + 3D_2 = [1,3,4,5,6,7,8,9,10,11,12,14].

  • ( 1 + 2 D 3 + 6 D 2 + D 2 ) + ( 1 + 4 D 3 + 2 D 2 + D 2 ) (1 + 2D_3 + 6D_2 + D_2) + (1 + 4D_3 + 2D_2 + D_2) . One swap: D A = 1 + 2 D 3 + 2 D 2 + D 2 = [ 1 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , 6 , 7 , 8 ] ; D B = 1 + 4 D 3 + 6 D 2 + D 2 = [ 1 , 2 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 15 , 16 ] . D_A = 1 + 2D_3 + 2D_2 + D_2 = [1,2,3,3,4,4,5,5,6,6,7,8]; \\ D_B = 1 + 4D_3 + 6D_2 + D_2 = [1,2,5,6,7,8,9,10,11,12,15,16].

  • ( 1 + 4 D 3 + 2 D 2 + D 2 ) + ( 1 + D 3 + 6 D 2 + 3 D 2 ) (1 + 4D_3 + 2D_2 + D_2) + (1 + D_3 + 6D_2 + 3D_2) . This is the most interesting case, because there are four different D 2 D_2 terms involved, allowing for six possible pairings. Thus there are five possible swaps:

D A = 1 + 4 D 3 + 2 D 2 + 3 D 2 = [ 1 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 14 ] ; D B = 1 + D 3 + 6 D 2 + D 2 = [ 1 , 2 , 2 , 3 , 3 , 4 , 7 , 8 , 8 , 9 , 9 , 10 ] . D_A = 1 + 4D_3 + 2D_2 + 3D_2 = [1,3,4,5,6,7,8,9,10,11,12,14]; \\ D_B = 1 + D_3 + 6D_2 + D_2 = [1,2,2,3,3,4,7,8,8,9,9,10]. But we must discard this because it is equivalent to one we found before! We continue:

D A = 1 + 4 D 3 + 2 D 2 + 6 D 2 = [ 1 , 3 , 5 , 7 , 7 , 9 , 9 , 11 , 11 , 13 , 15 , 17 ] ; D B = 1 + D 3 + 3 D 2 + D 2 = [ 1 , 2 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , 6 , 7 ] . D_A = 1 + 4D_3 + 2D_2 + 6D_2 = [1,3,5,7,7,9,9,11,11,13,15,17]; \\ D_B = 1 + D_3 + 3D_2 + D_2 = [1,2,2,3,3,4,4,5,5,6,6,7]. Again, this only doubles a result we found earlier.

D A = 1 + 4 D 3 + 3 D 2 + D 2 = [ 1 , 2 , 4 , 5 , 5 , 6 , 8 , 9 , 9 , 10 , 12 , 13 ] ; D B = 1 + D 3 + 6 D 2 + 2 D 2 = [ 1 , 2 , 3 , 3 , 4 , 5 , 7 , 8 , 9 , 9 , 10 , 11 ] . D_A = 1 + 4D_3 + 3D_2 + D_2 = [1,2,4,5,5,6,8,9,9,10,12,13]; \\ D_B = 1 + D_3 + 6D_2 + 2D_2 = [1,2,3,3,4,5,7,8,9,9,10,11].

D A = 1 + 4 D 3 + 6 D 2 + D 2 = [ 1 , 2 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 15 , 16 ] ; D B = 1 + D 3 + 2 D 2 + 3 D 2 = [ 1 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , 6 , 7 , 8 ] . D_A = 1 + 4D_3 + 6D_2 + D_2 = [1,2,5,6,7,8,9,10,11,12,15,16]; \\ D_B = 1 + D_3 + 2D_2 + 3D_2 = [1,2,3,3,4,4,5,5,6,6,7,8]. This, too, is a result we found earlier.

D A = 1 + 4 D 3 + 3 D 2 + 6 D 2 = [ 1 , 4 , 5 , 7 , 8 , 9 , 10 , 11 , 12 , 14 , 15 , 18 ] ; D B = 1 + D 3 + 2 D 2 + D 2 = [ 1 , 2 , 2 , 3 , 3 , 3 , 4 , 4 , 4 , 5 , 5 , 6 ] . D_A = 1 + 4D_3 + 3D_2 + 6D_2 = [1,4,5,7,8,9,10,11,12,14,15,18]; \\ D_B = 1 + D_3 + 2D_2 + D_2 = [1,2,2,3,3,3,4,4,4,5,5,6].

In total, then 7 \boxed{7} different ways to construct the new dice.

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The "different ways" would have to be clarified further. If I swap the pair around, does that constitute a different way?

Calvin Lin Staff - 4 years ago

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I believe that question is sufficiently answered by the case of the Sicherman dice, which are unique, up to swapping the pair around.

Arjen Vreugdenhil - 4 years ago

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Would you mind if it stated "a (unordered) pair of 12-sided dice"?

Calvin Lin Staff - 4 years ago

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@Calvin Lin – I don't mind at all!

Arjen Vreugdenhil - 4 years ago

Comparing this solution and that of @Mark Hennings , an interesting question is how different the solutions really are. A partial translation, at least, is possible. Define d n ( x ) = x n − 1 + ⋯ + x 2 + x + 1. d_n(x) = x^{n-1} + \cdots + x^2 + x + 1. Then my " D D " notation above can be translated to generating function polynomials as follows:

  • 1 ↦ t 1 \mapsto t ,

  • + ↦ × +\ \mapsto\ \times ,

  • D n ↦ d n ( t ) D_n \mapsto d_n(t) ,

  • a D n ↦ d n ( t a ) aD_n \mapsto d_n(t^a) .

Thus D A = 1 + 4 D 3 + 3 D 2 + 6 D 2 ↦ t ( t 8 + t 4 + 1 ) ( t 3 + 1 ) ( t 6 + 1 ) . D_A = 1 + 4D_3 + 3D_2 + 6D_2 \mapsto t(t^8 + t^4 + 1)(t^3 + 1)(t^6 + 1). Factorize these factors further, and we find Mark's solution. In this case, for instance, t 3 + 1 = ( t 2 − t + 1 ) ( t + 1 ) , t 6 + 1 = ( t 4 − t 2 + 1 ) ( t 2 + 1 ) , t 8 + t 4 + 1 = ( t 2 + t + 1 ) ( t 2 − t + 1 ) ( t 4 − t 2 + 1 ) . t^3 + 1 = (t^2 - t + 1)(t + 1),\ \ t^6 + 1 = (t^4 - t^2 + 1)(t^2 + 1), \\ t^8 + t^4 + 1 = (t^2 + t + 1)(t^2 - t + 1)(t^4 - t^2 + 1). This further factorization introduces negative coefficients and therefore non-solutions. On the other hand, it results a unique factorization, while my D D notation has non-unique decomposition and therefore results in generating identical solutions. D 12 = { 4 D 3 + 2 D 2 + D 2 6 D 2 + 2 D 3 + D 2 6 D 2 + 3 D 2 + D 3 ↦ t 11 + ⋯ + t + 1 = { ( t 8 + t 4 + 1 ) ( t 2 + 1 ) ( t + 1 ) ( t 6 + 1 ) ( t 4 + t 2 + 1 ) ( t + 1 ) ( t 6 + 1 ) ( t 3 + 1 ) ( t 2 + t + 1 ) = ( t 4 − t 2 + 1 ) ( t 2 + t + 1 ) ( t 2 − t + 1 ) ( t 2 + 1 ) ( t + 1 ) . D_{12} = \begin{cases} 4D_3 + 2D_2 + D2 \\ 6D_2 + 2D_3 + D_2 \\ 6D_2 + 3D_2 + D_3 \end{cases} \mapsto t^{11} + \dots + t + 1 = \begin{cases} (t^8 + t^4 + 1)(t^2 + 1)(t + 1) \\ (t^6 + 1)(t^4 + t^2 + 1)(t + 1) \\ (t^6 + 1)(t^3 + 1)(t^2 + t + 1) \end{cases} = (t^4 - t^2 + 1)(t^2 + t + 1)(t^2 - t + 1)(t^2 + 1)(t + 1). Reconciling the two methods could be accomplished by introducing the pseudo-die D 3 ⋆ = { 2 , 1 ⋆ , 0 } D^\star_3 = \{2,1^\star,0\} with the rules D 3 + D 3 ∗ = { 4 , 3 , 3 ⋆ , 2 , 2 , 2 ⋆ , 1 , 1 ⋆ , 0 } = { 4 , 2 , 0 } = 2 D 3 , D 2 + D 3 ∗ = { 3 , 2 , 2 ⋆ , 1 , 1 ⋆ , 0 } = { 3 , 0 } = 3 D 2 . D_3 + D^\ast_3 = \{4,3,3^\star,2,2,2^\star,1,1^\star,0\} = \{4,2,0\} = 2D_3, \\ D_2 + D^\ast_3 = \{3,2,2^\star,1,1^\star,0\} = \{3,0\} = 3D_2. where the ∗ \ast operator indicates removal from the set. Now we get D 12 = 2 D 3 ⋆ + D 3 + D 3 ⋆ + 2 D 2 + D 2 , D_{12} = 2D^\star_3 + D_3 + D^\star_3 + 2D_2 + D_2, and the former ambiguity is resolved: D 12 = { 4 D 3 + 2 D 2 + D 2 = ( 2 D 3 ⋆ + D 3 + D 3 ⋆ ) + 2 D 2 + D 2 6 D 2 + 2 D 3 + D 2 = ( 2 D 3 ⋆ + 2 D 2 ) + ( D 3 ⋆ + D 3 ) + D 2 6 D 2 + 3 D 2 + D 3 = ( 2 D 3 ⋆ + 2 D 2 ) + ( D 3 ⋆ + D 2 ) + D 3 . D_{12} = \begin{cases} 4D_3 + 2D_2 + D2 = (2D^\star_3 + D_3 + D^\star_3) + 2D_2 + D_2 \\ 6D_2 + 2D_3 + D_2 = (2D^\star_3 + 2D_2) + (D^\star_3 + D_3) + D_2 \\ 6D_2 + 3D_2 + D_3 = (2D^\star_3 + 2D_2) + (D^\star_3 + D_2) + D_3.\end{cases} The problem becomes a redistribution of the ten D D -terms in D 12 + D 12 = D A + D B D_{12} + D_{12} = D_A + D_B in two groups in such a way, that

  • each group has one non-starred D 3 D_3 term, and two D 2 D_2 terms,

  • each D 3 ⋆ D^\star_3 remains combined with D 2 D_2 or D 3 D_3 , and each 2 D 3 ⋆ 2D^\star_3 remains combined with 2 D 2 2D_2 or 2 D 3 = D 3 + D 3 ⋆ 2D_3 = D_3 + D^\star_3 .

The eight solutions (or fifteen possible dice) are easily reproduced by considering where the four starred terms go:

  • D 3 + 2 D 3 ⋆ + 2 D 3 ⋆ + D 3 ⋆ + D 3 ⋆ + 2 D 2 + D 2 D_3 + 2D^\star_3 + 2D^\star_3 + D^\star_3 + D^\star_3 + 2D_2 + D_2 ; die with maximum value = 18.

  • D 3 + 2 D 3 ⋆ + 2 D 3 ⋆ + D 3 ⋆ + 2 D 2 + 2 D 2 D_3 + 2D^\star_3 + 2D^\star_3 + D^\star_3 + 2D_2 + 2D_2 ; maximum value = 17.

  • D 3 + 2 D 3 ⋆ + 2 D 3 ⋆ + D 3 ⋆ + 2 D 2 + D 2 D_3 + 2D^\star_3 + 2D^\star_3 + D^\star_3 + 2D_2 + D_2 ; maximum value = 16.

  • D 3 + 2 D 3 ⋆ + 2 D 3 ⋆ + 2 D 2 + 2 D 2 D_3 + 2D^\star_3 + 2D^\star_3 + 2D_2 + 2D_2 ; maximum value = 15.

  • D 3 + 2 D 3 ⋆ + D 3 ⋆ + D 3 ⋆ + 2 D 2 + D 2 D_3 + 2D^\star_3 + D^\star_3 + D^\star_3 + 2D_2 + D_2 ; maximum value = 14.

  • D 3 + 2 D 3 ⋆ + D 3 ⋆ + D 3 ⋆ + D 2 + D 2 D_3 + 2D^\star_3 + D^\star_3 + D^\star_3 + D_2 + D_2 ; maximum value = 13.

  • D 3 + 2 D 3 ⋆ + D 3 ⋆ + 2 D 2 + 2 D 2 D_3 + 2D^\star_3 + D^\star_3 + 2D_2 + 2D_2 ; maximum value = 13.

  • D 3 + 2 D 3 ⋆ + D 3 ⋆ + 2 D 2 + D 2 D_3 + 2D^\star_3 + D^\star_3 + 2D_2 + D_2 ; maximum value = 12 (the "normal" die).

  • D 3 + 2 D 3 ⋆ + D 3 ⋆ + D 2 + D 2 D_3 + 2D^\star_3 + D^\star_3 + D_2 + D_2 ; maximum value = 11.

  • D 3 + 2 D 3 ⋆ + 2 D 2 + 2 D 2 D_3 + 2D^\star_3 + 2D_2 + 2D_2 ; maximum value = 11.

  • D 3 + 2 D 3 ⋆ + 2 D 2 + D 2 D_3 + 2D^\star_3 + 2D_2 + D_2 ; maximum value = 10.

  • D 3 + D 3 ⋆ + D ⋆ 3 + 2 D 2 + D 2 D_3 + D^\star_3 + D^\star 3 + 2D_2 + D_2 ; but the other die would be wrong

  • D 3 + D 3 ⋆ + D ⋆ 3 + D 2 + D 2 D_3 + D^\star_3 + D^\star 3 + D_2 + D_2 ; maximum value = 9.

  • D 3 + D 3 ⋆ + 2 D 2 + 2 D 2 D_3 + D^\star_3 + 2D_2 + 2D_2 ; but the other die would be wrong

  • D 3 + D 3 ⋆ + 2 D 2 + D 2 D_3 + D^\star_3 + 2D_2 + D_2 ; maximum value = 8.

  • D 3 + D 3 ⋆ + D 2 + D 2 D_3 + D^\star_3 + D_2 + D_2 ; maximum value = 7.

  • D 3 + 2 D 2 + 2 D 2 D_3 + 2D_2 + 2D_2 ; but the other die would be wrong.

  • D 3 + 2 D 2 + D 2 D_3 + 2D_2 + D_2 ; maximum value = 6.

  • D 3 + D 2 + D 2 D_3 + D_2 + D_2 ; but the other die would be wrong.

Arjen Vreugdenhil - 4 years ago
K T
Nov 9, 2019

Found these, using a Python script: A=[1,4,5,7,8,9,10,11,12,14,15,18] ; B=[1,2,2,3,3,3,4,4,4,5,5,6] A=[1,3,5,7,7,9,9,11,11,13,15,17] ; B=[1,2,2,3,3,4,4,5,5,6,6,7] A=[1,3,4,5,6,7,8,9,10,11,12,14] ; B=[1,2,2,3,3,4,7,8,8,9,9,10] A=[1,3,3,5,5,7,7,9,9,11,11,13] ; B=[1,2,2,3,5,6,6,7,9,10,10,11] A=[1,2,5,6,7,8,9,10,11,12,15,16] ; B=[1,2,3,3,4,4,5,5,6,6,7,8] A=[1,2,4,5,5,6,8,9,9,10,12,13] ; B=[1,2,3,3,4,5,7,8,9,9,10,11] A=[1,2,3,7,7,8,8,9,9,13,14,15] ; B=[1,2,3,4,4,5,5,6,6,7,8,9]

For dice with other number of sides 1-15: 4: A=[1,3,3,5] ; B=[1,2,2,3] 6: A=[1,3,4,5,6,8] ; B=[1,2,2,3,3,4] 8: A=[1,3,5,5,7,7,9,11] ; B=[1,2,2,3,3,4,4,5], A=[1,3,3,5,5,7,7,9] ; B=[1,2,2,3,5,6,6,7], A=[1,2,5,5,6,6,9,10] ; B=[1,2,3,3,4,4,5,6] 9: A=[1,4,4,7,7,7,10,10,13] ; B=[1,2,2,3,3,3,4,4,5] 10: A=[1,3,5,6,7,8,9,10,12,14] ; B=[1,2,2,3,3,4,4,5,5,6] 14: A=[1,3,5,7,8,9,10,11,12,13,14,16,18,20] ; B=[1,2,2,3,3,4,4,5,5,6,6,7,7,8] 15: A=[1,4,6,7,9,10,11,12,13,14,15,17,18,20,23] ; B=[1,2,2,3,3,3,4,4,4,5,5,5,6,6,7]

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