Problems & Puzzles: Puzzles

Puzzle 232. Primes and Cubic polynomials

Sebastian Martín Ruiz and myself started studying a few days ago the cubic polynomials f(x) =a.x^3+b.x^2+c.x+d, looking for those producing only primes in the x range [0, k]. As it has been usual, we studied this property simultaneously in f(x) and in abs(f(x)).

In the Ribenboim well know book we found two polynomials reported by Goetgheluck, and our first purpose was to find out if their productivity could be improved.

The Goetgheluck's polynomials and our own better results can be seen in the Table below.

Regarding the primes produced by each polynomial we report the total primes (T), their subdivision into negative (N) and positive (P) ones, and also  the quantity of distinct (D) primes found in the total primes. 

As you can see CR found (unexpectedly soon) a polynomial that produced more total and distinct, moreover only positive, primes than the produced by Goetgheluck.

A few days later SMR found another polynomial clearly superior to the Goetgheluck's one and to the CR's one except that the SMR's one contains some -let's say this way - some 'negative primes' while the CR's one doesn't.

A fast comparison between the CR & SMR polynomials and the champion quadratic well known polynomials (by Euler and Ruby) shows that the Euler's one is superior to the CR's one by far; while the SMR's polynomial is better than the Ruby's one in total primes (47 vs. 45) but inferior in distinct primes (43 vs. 45).

Question:

1. Find better cubic polynomials than the CR & SMR ones

Solution:

Adam Stinchcombe and Faride Firoozbakht sent some interesting polynomials related to the only question of this puzzle. While Sebastián Martín Ruiz found another interesting quadratic polynomial that matches to the Euler's one in certain aspects.

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Adam Stinchcombe:

A slight improvement over CR in the distinct column is 20*x^3-683*x^2+7553*x-26881 produces only distinct primes numbering D=26, but some negatives, N=11 and P=15...

I also get

 

poly                                  n               p                 d

x^3-43*x^2+572*x-2143      7              23                26

x^3-43*x^2+654*x-3701     19             12                31

x^3-48*x^2+629*x-1213      3              27                26

x^3-44*x^2+563*x-1811     11             19                26

x^3-44*x^2+601*x-2341      7              21                24

 

and a couple more that are equally lackluster, no where near the SMR one.  Hopefully, I can improve on these over the next couple of weeks.

 

(Note: I have replied to him that the very core of my polynomial is that N=0)

***

Faride Firoozbakht:

a) f(x)=x^3-65x^2+908x-617

a b c d T N P D Comment

1 -65 908 -617 31 12 19 30 one prime is repeated and only two times (T-D=1)

Question.1: Find cubic polynomials such that T > 30 and D = T.
Question.2: Find another cubic polynomial such that T > 30 and D = T-1.

In each of the two polynomials (of CR & SMR) two primes is repeated and each of them is repeated three times (T-D=2(3-1)). In the first polynomial of Goetgheluck one prime is repeated three times (T-D=2).In the second polynomial of Goetgheluck two primes is repeated (each of them is repeated two times (T-D=2(2-1)).

...

b) Note that in cubic polynomial that is found by you ,if we replace x by x-6 we obtain the polynomial x^3-53x+836-3251 which is prime for x=0,...,34 (for x=0,...,5 the terms are negative) .For this polynomial we have :

a b c d T N P D

1 -53 836 -3251 35 6 29 31

c) I found another cubic polynomials which for them 29 < T < 45 and T-D=4. 

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Sebastián Martín Ruiz found a quadratic polynomial that has 40 distinct and positive primes matching the Euler's one in productivity but using non-unitary coefficients. He also observes that both polynomials share in a very peculiar manner 27 of the 40 primes. Here goes in short his discoveries:

a) The polynomials:

P(n)=9n²-231n+1523 (SMR)
Q(n)=n²+n+41 (Euler)

b) The properties

b.1) Both generate 40 primes distinct and positive (for n=0 to 39).

b.2) P(n) & Q(n) share 27 primes according to this equations:

P(n)=Q(38-3n) for n=0,1,...,12
P(n)=Q(3n-39) for n=13,14,...,26

This adds two interesting questions to this puzzle:

2. Can you find another quadratic polynomial with the same or better productivity of primes distinct and all positive, than the Euler and SMR ones?

3. Regarding the quadratic SMR's polynomial, can you explain or interpret the equations shown in b.2)?

***

Challenged by the Polynomial found by SMR, C. Rivera scanned a solid box of the coefficients for the quadratic polynomial f(x) = a.x² - b.x + c, trying to find other polynomials as primes-productive (or better) than the Euler's one. This is what he found:

Constraints of the search

• a, b & c are integers
• a>0
• f(x)>0 is prime for 0<=x<=k, x= integer
• f(x) ≠ f(y) for x ≠ y (which means that for each polynomial the primes must be positive and distinct)

The before constraints imply the following simplifying conditions:

• c=prime
• b<2.isqrt(a.c), by calculus
• a & b have the same parity.

My search/code scanned the following solid region:

• 1<=a<=50
• 3<=c<=99991
• b<2.isqrt(a.c)

Results

For all the following six square polynomials each produces 40 positive and distinct primes for 0<=x<=39

Polynomial	Author
e(x) = x2 + x + 41	Euler (?)
es(x) = x2 - 79.x + 1601	Escott (1899)
r1(x) = 4.x2 - 154.x + 1523	Rivera (2003)
r2(x) = 4.x2 - 158.x + 1601	Rivera (2003)
smr(x) = 9.x2 - 231.x + 1523	SMR (2003)
r3(x) = 9.x2 -471.x + 6203	Rivera (2003)

• The first four polynomials (e, es, r1 & r2) all have the same primes, just in distinct order
• The two last polynomials (smr, r3) have the same primes just in reverse order
• The primes in es are in reverse order than in e
• The primes in r2 are in reverse order than in r1
• The primes in r3 are in reverse order than in smr; both share only 27 primes with the first four polynomials.
• if x=x-40 in e you obtain es
• if x=x-1/2 in r1 you obtain r2
• if x=x-40/3 in smr you obtain r3
• es(2.n)=r2(n), for 0<=n<=19; es(79-2.n)=r2(n), for 20<=n<=39

Conjecture: There is not any quadratic polynomial having positive and distinct primes for all x, 0<=x<=k, for k>39.

Question: Do you know if these six polynomials have beer reported together before?

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Jon Wharf devised the analytic way these six polynomials are linked, so mystery solved...:

The Euler parabola equation is y = x.x+x+41

 

The es series samples the points on the parabola from -40 to -1. So in order to shift this series to occupy x=0 to 39, substitute (x-40) for x in the original equation:

x.x - 80x + 1600 + x - 40 + 41

 = x.x - 79x +1601

 

The r1 series samples every 2nd point starting at x = -39. Substitute (2x-39) for x in the Euler equation:

4x.x - 156x + 1521 + 2x - 39 + 41

 = 4x.x - 154x + 1523

 

The r2 series samples every 2nd point starting at x = -40. Substitute (2x-40) for x in the Euler equation:

4x.x - 160x + 1600 + 2x - 40 + 41

 = 4x.x - 158x + 1601

 

The smr series samples every 3rd point starting at x = -39. Substitute (3x-39) for x in the Euler equation:

9x.x - 234x + 1521 + 3x - 39 + 41

 = 9x.x - 231x + 1523

 

The r3 series samples every 3rd point starting at x = -79. Substitute (3x-79) for x in the Euler equation:

9x.x - 474x + 6241 + 3x - 79 + 41

 = 9x.x - 471x + 6203

 

You can also substitute so as to sample the Euler parabola "backwards" for all of these, for example substitute (39-x) for x to get es again. Same final equation in each case of course.

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Seiji Tomita wrote (April 2005):

I found better polynomials than CR ones.

There are the polynomials with D>=30.


a b c d T N P D
2 -28 -2334 46153 36 0 36 36

2 -22 -2384 43793 35 0 35 35

2 -16 -2422 41389 34 0 34 34

2 -10 -2448 38953 33 0 33 33

2 -4 -2462 36497 32 0 32 32

2 2 -2464 34033 31 0 31 31

2 8 -2454 31573 30 0 30 30

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Gobbo Franck wrote on Dec. 2005:

I just want indicate a new polynomial who is generating 62 prime numbers consecutively (for n=0 to n=61) :
8n² - 488n + 7243
This polynomial must be considered in absolute value.

Note : This polynomial generate 31 different primes (the first 31 numbers.

I found this polynomial on 12/27/2005

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Franck Gobbo wrote (Janary, 2006):

I just want indicate a new cubic polynomial who is generating 47 prime numbers consecutively (for n=0 to n=46) :
3n^3 - 231n^2 + 5526n - 38651

In the 47 prime numbers generated :
P=32
N=15
Diff. = 43

I found this polynomial on 25/01/2006.

Later he added

I just want to add a precision about the cubic polynomial I discovered
P(n) = 3n^3 - 231n^2 + 5526n - 38651

This polynomial is linked with the S.M Ruiz cubic polynomial
K(n) = 3n^3 - 183n^2 + 3318n - 18757
as follow :

P(-n+46) = - K(n)
and
K(-n+46) = -P(n) (suggested by S.M Ruiz)

***

Parviz Afereidoon wrote:

I'm ms-student of ferdowsi university of mashhad _iran. Now I am researching about polynomials that have a long string of prime values and found the below

 

 Polynomials:

 

  f(x)=2x^3-212x^2+5778x-42841

 

( I found this polynomial on March 6, 2006 )

 

a b c d T N P D  

2  -212  5778  -42841  42  29  13  41

5  -185  1220   4157  40  14  26  36

 

I found another cubic polynomials which for them D>30.

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Franck Gobbo comment about the last contribution is: