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).




Primes 


a 
b 
c 
d 
T 
N 
P 
D 
Author 
Comment 
1 
34 
381 
1511 
26 
17 
9 
24 
Goetgheluck 
Cfr. P. 203, Ribenboim 
2 
45 
331 
3191 
26 
19 
7 
24 
Goetgheluck 
Cfr. P. 203, Ribenboim 
1 
35 
308 
73 
29 
0 
29 
25 
CR 
More only positive 
3 
183 
3318 
18757 
47 
32 
15 
43 
SMR 
More total primes and more distinct primes 










0 
1 
1 
41 
40 
0 
40 
40 
Euler 
Cfr. P. 200, Ribenboim 
0 
36 
810 
2753 
45 
14 
31 
45 
Ruby 
Cfr. P. 203, Ribenboim 
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.
***
Adam Stinchcombe:
A slight improvement over CR in the
distinct column is 20*x^3683*x^2+7553*x26881 produces only distinct
primes numbering D=26, but some negatives, N=11 and P=15...
I also get
poly
n p d
x^343*x^2+572*x2143 7 23 26
x^343*x^2+654*x3701
19 12 31
x^348*x^2+629*x1213
3 27 26
x^344*x^2+563*x1811 11 19 26
x^344*x^2+601*x2341
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^365x^2+908x617
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 (TD=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 = T1.
In each of the two polynomials (of CR
& SMR) two primes is repeated and each of them is repeated three times
(TD=2(31)). In the first polynomial of Goetgheluck one prime is repeated
three times (TD=2).In the second polynomial of Goetgheluck two primes is
repeated (each of them is repeated two times (TD=2(21)).
...
b) Note that in cubic polynomial that
is found by you ,if we replace x by x6 we obtain the polynomial
x^353x+8363251 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 TD=4.
***
Sebastián Martín Ruiz found a
quadratic polynomial that has 40 distinct and positive primes matching the
Euler's one in productivity but using nonunitary 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)=9n^{2}231n+1523
(SMR)
Q(n)=n^{2}+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(383n) for n=0,1,...,12
P(n)=Q(3n39) 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.x^{2}
 b.x + c, trying to find other polynomials as primesproductive
(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) = x^{2}
+ x + 41 
Euler (?) 
es(x) = x^{2}
 79.x + 1601 
Escott (1899) 
r1(x) = 4.x^{2}
 154.x + 1523 
Rivera (2003) 
r2(x) = 4.x^{2}
 158.x + 1601 
Rivera (2003) 
smr(x) = 9.x^{2}
 231.x + 1523 
SMR (2003) 
r3(x) = 9.x^{2}
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=x40 in e you obtain es
 if x=x1/2 in r1 you obtain r2
 if x=x40/3 in smr you obtain r3
 es(2.n)=r2(n), for 0<=n<=19;
es(792.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?
***
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 (x40) 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 (2x39) 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 (2x40) 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 (3x39) 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 (3x79) 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 (39x) for x to get es again. Same final equation
in each case of course.
***
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
***
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) :
8n˛  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
***
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 msstudent 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^3212x^2+5778x42841
( 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.
***
Franck Gobbo comment about the last contribution is:
I want to add a precision about the cubic polynomial that Parviz
Afereidoon discovered :
P(n) = 2n^3  212n^2 + 5778n  42841
This polynomial is linked with a polynomial I discovered on January 2006
(I spoke about this polynomial in a mail to S.M Ruiz, February 23,
2006). This polynomial is the following :
G(n) =
2n^3 – 34n^2 – 1520n + 24473
(see
my website)
And here is the link :
If F(n) = 2n^3 – 212n^2 + 5778n – 42841 (generating 42 prime numbers 0
> 41), then G(n) = F(n+41) = 2n^3 + 34n^2 + 1520n  24473, so
G(n) =
2n^3 – 34n^2 – 1520n + 24473 (generating 42 primes too,
T=42 , P=29, N=13, D=42).
More generally (and for cubic polynomials) I would say that :
Let F(n) = (An^3Bn^2+CnD) generating P successive prime numbers, then
–G(n) = F(n + (P1)) and
G(n) is generating P successive prime numbers too.Here are some others
little results :
2n^3  113n^2 + 1967n – 9949 (T=39 P=25 N=14 Diff.
= 38)
n^3  171n^2 + 5858n – 47711 (T=40 P=18 N=22 Diff.
= 39)
2n^3  115n^2 + 2043n – 11369 (T=39 P=14 N=25 Diff.
= 38)
4n^3  246n^2 + 4184n – 19913 (T=40 P=11 N=29 Diff.
= 38)
8n^3  487n^2 + 9449n – 56813 (T=40 P=28 N=12 Diff.
= 40)
2n^3  115n^2 + 2043n – 11369 (T=39 P=14 N=25 Diff.
= 38)
5n^3  291n^2 + 5266n – 30347 (T=39 P=11 N=28 Diff.
= 37)
5n^3  263n^2 + 4338n – 21841 (T=40 P=24 N=16 Diff.
= 39)
6n^3  316n^2 + 5604n – 33811 (T=39 P=18 N=21 Diff.
= 39)
5n^3  279n^2 + 4810n – 23917 (T=39 P=28 N=11 Diff.
= 37)
8n^3  449n^2 + 7967n – 45523 (T=40 P=12 N=28 Diff.
= 40)
***
On June 21, 06, J. C. Meyrignac
reported this:
Following our
programming contest http://www.recmath.org/contest/index.php
New records for N=3 (puzzle 232):
42.x^3 + 270.x^2  26436.x + 250703 gives 40 positive primes (discovered by Jaroslaw Wroblewski & JeanCharles Meyrignac)
 66.x^3 + 3845.x^2  60897.x + 251831 abs() of this function gives 46 distinct primes (discovered by Ivan Kazmenko & Vadim Trofimov)
BTW, Shyam Sunder Gupta found:
abs(1/4x^{5}  133/4x^{4} + 6729/4x^{3} 
158379/4x^{2} + 860147/2x  1705829)
which provides 57 primes !!!
1705829 1313701 991127 729173 519643 355049 228581 134077 65993 19373
10181 26539 33073 32687 27847 20611 12659 5323 383 3733 4259 1721 3923
12547 23887 37571 53149 70123 87977 106207 124351 142019 158923 174907
189977 204331 218389 232823 248587 266947 289511 318259 355573 404267
467617 549391 653879 785923 950947 1154987 1404721 1707499 2071373
2505127 3018307 3621251 4325119
***
