Problems & Puzzles: Puzzles

 

Puzzle 821. Prime numbers and complementary sequences

Let's starts with the sequence of prime numbers:

i 1 2 3 4 5 6 7 ...
pi 2 3 5 7 11 13 17 ...

Now, lets define qi as the quantity of integers less than i in pi:

i 1 2 3 4 5 6 7 ...
pi 2 3 5 7 11 13 17 ...
qi 0 0 1 2 2 3 3 ...

Now, let add i to pi and to qi, respectively:

i 1 2 3 4 5 6 7 ...
pi 2 3 5 7 11 13 17 ...
qi 0 0 1 2 2 3 3 ...
pi+i 3 5 8 11 16 19 24 ...
qi+i 1 2 4 6 7 9 10 ...

Q. Demonstrate that the sequences {pi+1} and {qi+i} are "complementary", that is to say that all and every integer greater than zero is present in one and only one of these two sequences.


Contributions came from John Arioni and Jan van Delden

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John wrote:

 Let A1, A2, A3, ......Ai .. be any numeric sequence such that:

                       A1 < A2 < A3< .......<Ai < ...............
 

The sequence of the prime numbers is one of these.

The following is a general scheme to understand what happens:

   i --->    1    2    3 ..... A1  A1+1  A1+2 ......... A2   A2+1  A2+2 ......... A3     A3+1 .

 Ai  ---> A1 A2 A3 ............................................... ..............................................................

 q i  ---> 0    0    0 ...... 0         1        1 .............. 1         2           2 .........     2         3

qi+i ---> 1   2    3            A1  A1+2   A1+3 ....... A2+1   A2+3 ......           A3+2    A3+4 not present in qi+1 --->  A1+1 A2+2  A3+3

The sequence qi starts with a 0 and continues with zeroes as long as i <= A1, so that qi+i = 0+i produces the sequence 1, 2, 3..... A1.

When the index i passes the value A1 (from A1 to A1+1), qi has an increment of 1 and keeps the value 1 as long as i <=A2 , so that qi+i = 1+i produces the sequence A1+2, A1+3, A1+4 ......A2 of consecutive integers.

So far qi+i has produced the sequence 1, 2, 3, ... A1, A1+2, A1+3, A1+4, ....A2, i.e. all the integers from 1 to A2, but the value A1+1.

Every time the index i passes one of the values present in the sequence Ai, qi has an increment of 1 so that qi+i has an increment of 2 instead of an increment of 1, then the value Ai+i is regularly bypassed, while all of the remaining integers are present in the sequence qi+i.

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Jan wrote:

Let N = {i, i integer >0}
Let S be an ascending ordered (denumerable) subset of N (for instance the primes), with elements S(i).
 
Define the function M() on N: N->{0,1}
 
M(i) = 1 if i-1 in S
M(i) = 0 if i-1 not in S
[Note since 0 is not in S we have M(1)=0]
 
Define the monotic non decreasing counting function  C() on N:
 
C(i) = sum(M(j), j=1..i) [In the question this is qi(i)]
[We have C(0)=0]
 
We have C(i)-C(i-1)=M(i), so a jump indicates that i-1 is a (the next) member in S.
 
This function counts the total number of j<i present in S, so we have:
C(S(i))=i-1 and C(S(i)+1)=i
[This is the essential part..]
 
Define:
 
T = { S(i)+i | i in N}
U = { C(i)+i | i in N }
 
To proof:
 
For every k in N we either have k in T or k in U.
 
A) 1 is in U. Because M(1)=0, C(1)=M(1)=0, C(1)+1=1.
 
B) Proof that for all k in U we have that k+1 is either in T in U (but not in both)
 
If k is in U, there exist an unique i such that C(i)+i=k, since the inverse exists.
If C(i+1)=C(i), i.e. M(i+1)=0, we have C(i+1)+i+1=C(i)+i+1=k+1 is in U.
If C(i+1)=C(i)+1, i.e. M(i+1)=1, we have C(i+1)+i+1=C(i)+i+2=k+2 is in U.
Since M(i+1)=1 we know that i is in S, so we have i=S(j) for some j and
k+1=C(i)+i+1=C(S(j))+S(j)+1=j-1+S(j)+1=j+S(j), therefore k+1 is in T.
 
C) Proof that for all k in T we have that k+1 is in U.
 
If k is in T, we know that there is a j for which S(j)+j=k.
We know that if j increases S(j) increases as well, so S(j+1)+j+1>=S(j)+1+j+1=k+2,
the sequence S(j)+j is an increasing sequence with jumps of size at least 2, so k+1 is not in T.
We have i=S(j) for some i. Consider m=i+1. Then C(m)+m=C(i+1)+i+1=C(S(j)+1)+i+1=j+S(j)+1=k+1.
Therefore k+1 is in U.
 
So by induction we know that all k in N are present and every value of k is present in either T or U. q.e.d.
 
An example where S is finite:
i   1 2 3 4 5  6
S   1 2 3 4
M   0 1 1 1 1  0 
C   0 1 2 3 4  4 
S+i 2 4 6 8
C+i 1 3 5 7 9 10
 
Shows that we can also define the operations when sets are (finite or) empty,
if we impose the condition that addition is only applied when meaningfull.
 
i  1  2 3  4 5  6
S  5  3 2  7 1  6
M  0  1 1  1 0  1
C  0  1 2  3 3  4
S+i 6 5 5 11 6 12
C+i 1 3 5  7 8 10
 
Shows that we do need and ascending ordered set.

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