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Development of Algorithmic Constructions

08:28:35 20.Apr 2024

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Polynom = x^2+1x-19

1. Algorithm

a:=1;
b:=1;
c:=-19;


liste_max:=100000;

sieving:=proc (stelle, p)
begin
   while (stelle<=liste_max) do
     erg:=liste[stelle];
     while(erg mod p=0) do
        // Divison of the stored f(x) by the prime
        erg:=erg /p;
     end_while;
     liste[stelle]:=erg;
     stelle:=stelle+p;
   end_while;
end_proc;

// Calculation of the values of the polynom for x from 0 to liste_max
for x from 0 to liste_max do
   p:=abs (a*x^2+b*x+c);
   while (p mod 2=0) p:=p/2;
   liste [x]:=p;
end_for;

for x from 0 to liste_max do
    p:=liste[x];

    if (p>1) then

      // Printing the Primes
      print (x, p);

      // 1. Sieving
      sieving (x+p,  p);

      t:=(-x-b/a) mod p;
      if t=0 then t:=p; end_if;

      // 2. Sieving
      sieving (t, p);

    end_if;

end_for;

 2. Mathematical background
Lemma: If p | f(x) then also p | f(x+p) and  p | f(-x-b/a)

a)  p | f(x)   <=> ax^2 + bx + c = 0                       mod p
    p | f(x+p) <=> a(x+p)^2 + b(x+p) + c = 0               mod p
               <=> ax^2 + 2axp + ap^2 + bx + bp + c = 0    mod p
               <=> ax^2 + bx + c = 0                       mod p

    Thus if p | f(x) then p | f(x+p)

b)  if b = 0 mod a
    p | f(x)       <=> ax^2 + bx + c = 0                         mod p
    p | f(-x-b/a)  <=> a(-x-b/a)^2 + b(-x-b/a) + c = 0           mod p
                   <=> ax^2 + 2bx + b^2/a - bx - b^2/a + c = 0   mod p
                   <=> ax^2 + bx + c = 0                         mod p

    Thus if p | f(x)  then p | f(-x-b/a)



 3. Correctness of the algorithm
The proof for this polynom is similar to the proof for the polynom f(x)=x^2-4x+1.

a) First terms for the polynom f(x) = x^2+1x-19

f(0)=19
f(1)=17
f(2)=13
f(3)=7
f(4)=1
f(5)=11
f(6)=23
f(7)=37
f(8)=53
f(9)=71
f(10)=1
f(11)=113
f(12)=137
f(13)=163
f(14)=191
f(15)=1
f(16)=1
f(17)=41
f(18)=1
f(19)=1
f(20)=401
f(21)=443
f(22)=487
f(23)=1
f(24)=83
f(25)=631
f(26)=683
f(27)=67
f(28)=61
f(29)=1
f(30)=911
f(31)=139
f(32)=1
f(33)=1103
f(34)=1171
f(35)=73
f(36)=101
f(37)=1
f(38)=1
f(39)=1
f(40)=1621
f(41)=131
f(42)=1787
f(43)=1873
f(44)=1
f(45)=293
f(46)=2143
f(47)=2237
f(48)=2333
f(49)=1
f(50)=2531
f(51)=2633
f(52)=1
f(53)=2843
f(54)=227
f(55)=3061
f(56)=167
f(57)=173
f(58)=1
f(59)=503
f(60)=331
f(61)=1
f(62)=1
f(63)=4013
f(64)=1
f(65)=4271
f(66)=1
f(67)=349
f(68)=4673
f(69)=283
f(70)=4951
f(71)=463
f(72)=5237
f(73)=769
f(74)=5531
f(75)=1
f(76)=307
f(77)=5987
f(78)=6143
f(79)=6301
f(80)=1
f(81)=179
f(82)=617
f(83)=409
f(84)=7121
f(85)=317
f(86)=439
f(87)=1091
f(88)=601
f(89)=1
f(90)=8171
f(91)=8353
f(92)=8537
f(93)=1
f(94)=1
f(95)=479
f(96)=9293
f(97)=1
f(98)=421
f(99)=241

b) Substitution of the polynom
The polynom f(x)=x^2+1x-19  could be written as f(y)= y^2-19.25 with x=y-0.5

c) Backsubstitution

Beside by backsubstitution you get an estimation for the huge of the primes with p | f(x) and p < f(x)

f'(y)>(2y-1) with with y=x+0.5
f'(x)>2x

4. Infinity of the sequence

5. Sequence of the polynom with 1

6. Sequence of the polynom (only primes)

7. Distribution of the primes

8. Check for existing Integer Sequences by OEIS

Found in Database : 19, 17, 13, 7, 1, 11, 23, 37, 53, 71, 1, 113, 137, 163, 191, 1, 1, 41, 1, 1,
Found in Database : 19, 17, 13, 7, 11, 23, 37, 53, 71, 113, 137, 163, 191, 41, 401, 443, 487, 83, 631, 683, 67, 61, 911, 139, 1103, 1171, 73, 101,
Found in Database : 7, 11, 13, 17, 19, 23, 37, 41, 53, 61, 67, 71, 73, 83, 101, 113, 131, 137, 139,