Problems & Puzzles: Puzzles

Puzzle 915. Follow-up to Puzzles 172 & 914

Here we will explore another variety of the so-called "Minimal-prime-complete-rules (MPCR)" that were the main subject of the Puzzles 172 & 914.

This variety will consist in a ruler composed by two parts:

Part I: The first KP primes set in strict ascending order from left to right, from 2 to P(KP).

Part II: a) K1B ones before or b)  K1A after the Kp primes.

In any case, the total length L of the ruler is: L=SP+K1, where SP is the sum of the primes from 2 to P(KP).

A couple of examples:

Example #1 of MPCR for KP=2

a) (K1B=2), L=5+2=7
b) 2.3.1 (K1A=1), L=5+1=6

Example #2 of MPCR for KP=4

a) (K1B=6), L=17+6=23
b) (K1A=6), L=17+6=23

Please check by yourself that these examples are real MPCR rulers, that is to say that in each ruler you can obtain any integer from 1 to L adding a pertinent set of contiguous distances in the ruler, for the K1 stated and this K1 is minimal.

I have obtained all the rulers from KP=1 to 149. See the whole results in this file. From that work I derived the following three empirical results:

R1: K1B = P(KP)-1 for all 1<= KP <=149
R2: K1A <= K1B for all 1<= KP <=149
R3: K1B-K1A = {0, 1, 2, or 3} for all 1<= KP <=149, with just one exception in the same range.

The exception mentioned above is for KP = 38 where K1B-K1A = 8

This is the statistics:

K1B-B1A Frec. %
0 67 44.97%
1 36 24.16%
2 32 21.48%
3 13 8.72%
8 1 0.67%
  149 100%

Q1. Can you verify & explain R1 & R2.
Q2. Are there more exceptions for KP>149



Contributions came from Jan vand Delden and Emmanuel Vantieghem


Jan wrote:

Q1. Both statements are true.


Extension to the left with 1’s (K1B):



Given a list of primes P[1,n]=[p[1],..,p[n]]
Define S[i,j]=sum(p[k],k=i..j) and T={S[i,j], 1<=i<=j<=n} the set of all sums of consecutive primes.

We know that S[1,n]-1 is not part of T.



Extend P[1,m] with k 1’s to the left,  EP[k,m]=[1,…,1,p[1],..,p[m]]
Since S[1,n]-1 is not part of T we need to construct it using EP[k,m] for some k and m.


The sum of the elements of EP[k,m] is k+S[1,m] and this should equal S[1,n]-1: k+S[1,m]=S[1,n]-1
k = S[m+1,n]-1


The smallest value for k is attained if m+1=n, or m=n-1, or k=S[n,n]-1=p[n]-1

We need k to be at least equal to p[n]-1.




We need to proof that with k=p[n]-1 the induced set ET[n] of partial sums is complete.

Where ET[n] is the set of partial sums belonging to EP[p[n]-1,n].


We proceed by induction on n.

If n=1 we have p[1]=2 and k=2-1=1, EP[1,2]=[1,2] and ET[2]={1,2,3} is complete.              

Induction step:  ET[n] is complete implies the resulting ET[n+1] is complete


ET[n] = {1,2,3,…,p[n]-1+S[1,p[n]]} implies ET[n+1]={1,2,3,…,p[n+1]-1+S[1,p[n+1]]}


Consider EP[k,n].

If k=p[n]-1 we have ET[n]={1,2,3,..,p[n]-1+S[1,p[n]]} by assumption and hence a part of ET[n+1].
For every increase in k in EP[k,n] the maximum value in the set above is increased by 1, this can be repeated p[n+1]-p[n] times.

If k=p[n+1] -1 we therefore know that  U={1,2,3,…,p[n]-1+p[n+1]-p[n]+S[1,p[n]]}={1,2,3,..,p[n+1]-1+S[1,p[n]]} is in ET[n+1].

We must still account for the values of ET[n+1]\U = {p[n+1]+S[1,p[n]],…,p[n+1]-1+S[1,p[n+1]]}

But cleary S[1,p[n]]+p[n+1]=S[1,p[n+1]] is in ET[n+1] (The sum of all primes in P[1,n+1]).
Which we extended to the left by using p[n+1]-1 extra 1’s, so all these values are in ET[n+1].



So we found that we need k to be at least equal to p[n]-1 and proved that the induced set of partial sums is complete for this k.




Extension to the right with 1’s (K1A):


Define F[k]=[1,..,1] (a list of k 1’s)

Extend a partial list of P[m,n] with F[k] to the right: EP[m,n,k]=[p[m],…,p[n],1,…,1]

Define ET[k,n] equal to all partial sums of EP[1,n,k].
Define S[i,j]=sum(p[k],k=i..j) as before.


First we show that ET[p[n]-1,n] is complete.


  1. The partial sums of F[k]:                       {1,2,..,p[n]-1} are part of the set ET[p[n]-1,n].
  2. Extend with p[n], EP[n,n,p[n]-1]:        {p[n],p[n]+1,…,p[n]+p[n]-1} or {S[n,n],..,S[n,n]+p[n]-1} are part of the set.
  3. Extend with p[n-1], EP[n-1,n,p[n]-1]: {p[n]+p[n-1],…,p[n]+p[n-1]+p[n]-1} or {S[n-1,n],..,S[n-1,n]+p[n]-1} are part of the set.



Please note that we have  S[n,n]<S[n-1,n]=S[n,n]+p[n-1]<=S[n,n]+p[n]-1 because all primegaps are larger than 1, so the sets given in B) and C) overlap.


Finally ET[p[n]-1,n] is equal to the union of {1..p[n]-1} and the sets {S[k,n],…,S[k,n]+p[n]-1} = {1,…S[1,n]+p[n]-1} and hence complete.

k might not be minimal, we have: k<=p[n]-1.


If one has k<p[n]-1 it still might happen that the values of {k+1,…,p[n]-1} are equal to S[i,j] for some i and j.
This might solve “filling the gap” from A to B, but then one should check this doesn’t lead to a gap from B to C.


An example is easier than giving a general rule:


Consider n=27, p=103. The last few primes in L are [97,101,103]

For this prime we have that p-1=102,p-2=101,p-3=100 are in S[i,j], so one is tempted to try k=99 (or K1B-K1A=3)

  1. {1,..,99} are in the set, {100,101,102} are in S[i,j] so in the set ET[99,27]
  2. {103,..,202} are in the set ET[99,27]
  3. {204,…,303} are in the set ET[99,27]
  4. {303,…,400} are in the set ET[99,27]

So 203 is missing from ET[99,27] unless it is in S[i,j] for some i and j, BUT IT ISN’T.
There is no gap between the numbers in the sets belonging to C) and D) anymore (or further down the line).


A more careful analysis gives that the following procedure will take care of the gaps A)-B) and B)-C):


For the primes in P[1,n] keep all partial sums S[i,j] if S[i,j]<=2*p[n] in a set T. [Considerable speedup]



delta=0;                 //The value of K1B-K1A
while  k is a member of T {
   if k<=p[n-1]-1{
      if  (p[n]+k) is not a member of T{





This will give a correct result if delta<p[n]-p[n-2]. If the last condition is not fullfilled the sets in C) and D) don’t overlap.
And one either has to keep a larger set of partial sums (slower), or check the resulting numbers between C) and D) (rarely, but very slow).




The following values of delta=K1B-K1A>8 exist for p<10^6, with the additional check (p[n]-p[n-2]):

n: 12404 p[n] 132887 delta 9 (28)

n: 14570 p[n] 158611 delta 10 (20)

n: 15296 p[n] 167437 delta 9 (14)

n: 19005 p[n] 212437 delta 11 (18)

n: 23388 p[n] 266947 delta 9 (20)

n: 37830 p[n] 451481 delta 13 (42)

n: 42899 p[n] 517901 delta 10 (28)

n: 49492 p[n] 605173 delta 9 (26)

n: 58357 p[n] 724219 delta 9 (32)

n: 58967 p[n] 732449 delta 10 (76)

n: 68231 p[n] 858427 delta 10 (54)

n: 75145 p[n] 953221 delta 10 (42)


I found the following two critical values for delta=8:

n: 34305 p[n] 405997 delta 8 (8)

n: 55578 p[n] 686977 delta 8 (8)

I didn’t perform a check on these in order to determine if delta should be decreased or not.


Emmanuel wrote:

It is possible to prove  R1.

Let's first introduce some notations :
     p(n)  is the nth prime
     S(n) = the sum of the first  n  primes.
     B(n) = ( 1, 1, ..., 1, 2, 3, 5, ..., p(n) )  where the number of ones is precisely  p(n) - 1.
     T(n) = Sum of all elements of  B(n)
             = p(n) - 1 + S(n)
We must prove that
(1) every number  k  from  1  till  T(n)  can be obtained as sum of consecutive elments out of  B(n)
     (we then shall say that k is 'reached')
(2) if we put  B'(n) = B(n) with one one dropped, there is at least one number < T(n) - 1
     that cannot be obtained as sum of consecutive elments out of  B'(n).
We prove (1) with mathematical induction.

The assertion (1) is true when  n = 1 (obvious).

Asume the assertion (1) holds for  n.
It is clear that  B(n)  is a 'substring' of  B(n+1).
Hence, without using p(n+1) or any of the first  p(n+1) - p(n)  'ones',
all numbers from  1  till  T(n)  can be obtained as sums of consecutive elements out of  B(n+1),
If we involve these 'ones', we also can reach all numbers from  T(n) + 1  till  T(n) + p(n+1) - p(n) = S(n+1) - 1.
Thus, we certainly can reach all numbers from  1  till  S(n+1) - 1  without involving  p(n+1).
Now, let us look at all substrings of  B(n+1)  that involve  p(n+1)  but none of the 'ones'.
Their sums provide  n+1  numbers <= S(n+1)
Involving the 'ones' then finally will give all numbers from  S(n+1)  till  T(n+1)
Thus, all numbers form  1  till  T(n+1)  can be reached by  B(n+1).
This proves (1).

To prove (2) we consider  B'(n) = ( 1, 1, ..., 1, 2, 3, ..., p(n) )  (with  p(n) - 2  'ones').
By the fact that (1) is proved now, every number from  1  till  S(n-1)+p(n) - 2 = S(n) - 2
is reached by substrings of  B'(n)  that do not involve  p(n).
We see immediately that no substring that involves  p(n)  will ever reach  S(n) - 1 !
That gives Q.E.D !

Eilas, that way of reasoning fails for the proof of the assertions R2 and R3.
I verified them for  N  up to 240.  This is what I got :

K1B-B1A Frec.    %
0 112
1   60
2   49
3   16  
4     2
8     1

Further examination leads to memory overflow on my PC.



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