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Development of Algorithmic Constructions

06:37:34 20.Apr 2024

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Polynom = x^2+19x+3

1. Algorithm

a:=1;
b:=19;
c:=3;


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+19x+3

f(0)=3
f(1)=23
f(2)=5
f(3)=1
f(4)=19
f(5)=41
f(6)=17
f(7)=37
f(8)=73
f(9)=1
f(10)=293
f(11)=1
f(12)=1
f(13)=419
f(14)=31
f(15)=1
f(16)=563
f(17)=1
f(18)=223
f(19)=29
f(20)=1
f(21)=281
f(22)=181
f(23)=1
f(24)=1
f(25)=1103
f(26)=1
f(27)=83
f(28)=1319
f(29)=1
f(30)=491
f(31)=1553
f(32)=109
f(33)=191
f(34)=1
f(35)=631
f(36)=661
f(37)=1
f(38)=241
f(39)=151
f(40)=139
f(41)=821
f(42)=1
f(43)=157
f(44)=1
f(45)=1
f(46)=1
f(47)=1
f(48)=1
f(49)=1
f(50)=1151
f(51)=397
f(52)=739
f(53)=67
f(54)=263
f(55)=4073
f(56)=467
f(57)=1
f(58)=1
f(59)=307
f(60)=1
f(61)=257
f(62)=1
f(63)=1723
f(64)=1063
f(65)=607
f(66)=1871
f(67)=1153
f(68)=1973
f(69)=1
f(70)=271
f(71)=2131
f(72)=1
f(73)=6719
f(74)=1
f(75)=2351
f(76)=233
f(77)=1
f(78)=1
f(79)=1549
f(80)=1
f(81)=1
f(82)=1657
f(83)=941
f(84)=577
f(85)=239
f(86)=3011
f(87)=1
f(88)=9419
f(89)=641
f(90)=3271
f(91)=1
f(92)=227
f(93)=1
f(94)=1
f(95)=1
f(96)=409
f(97)=2251
f(98)=3823
f(99)=1

b) Substitution of the polynom
The polynom f(x)=x^2+19x+3  could be written as f(y)= y^2-87.25 with x=y-9.5

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+9.5
f'(x)>2x+18

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 : 3, 23, 5, 1, 19, 41, 17, 37, 73, 1, 293, 1, 1, 419, 31, 1, 563, 1, 223, 29,
Found in Database : 3, 23, 5, 19, 41, 17, 37, 73, 293, 419, 31, 563, 223, 29, 281, 181, 1103, 83, 1319, 491, 1553, 109, 191, 631, 661, 241, 151,
Found in Database : 3, 5, 17, 19, 23, 29, 31, 37, 41, 67, 73, 83, 109, 139,