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Development of Algorithmic Constructions

11:28:48 20.Apr 2024

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Polynom = x^2-40x+29

1. Algorithm

a:=1;
b:=-40;
c:=29;


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^2-40x+29

f(0)=29
f(1)=5
f(2)=47
f(3)=41
f(4)=23
f(5)=73
f(6)=7
f(7)=101
f(8)=227
f(9)=1
f(10)=271
f(11)=1
f(12)=307
f(13)=1
f(14)=67
f(15)=173
f(16)=71
f(17)=181
f(18)=367
f(19)=37
f(20)=53
f(21)=1
f(22)=1
f(23)=1
f(24)=1
f(25)=1
f(26)=1
f(27)=1
f(28)=1
f(29)=1
f(30)=1
f(31)=1
f(32)=1
f(33)=1
f(34)=1
f(35)=1
f(36)=1
f(37)=1
f(38)=1
f(39)=1
f(40)=1
f(41)=1
f(42)=113
f(43)=79
f(44)=1
f(45)=127
f(46)=61
f(47)=179
f(48)=59
f(49)=1
f(50)=1
f(51)=1
f(52)=653
f(53)=359
f(54)=157
f(55)=1
f(56)=1
f(57)=499
f(58)=1
f(59)=1
f(60)=1229
f(61)=131
f(62)=199
f(63)=739
f(64)=313
f(65)=827
f(66)=349
f(67)=919
f(68)=1933
f(69)=1
f(70)=2129
f(71)=223
f(72)=2333
f(73)=1
f(74)=509
f(75)=1327
f(76)=1
f(77)=1439
f(78)=1
f(79)=311
f(80)=3229
f(81)=1
f(82)=151
f(83)=257
f(84)=149
f(85)=1
f(86)=797
f(87)=1
f(88)=4253
f(89)=439
f(90)=647
f(91)=467
f(92)=4813
f(93)=1
f(94)=1021
f(95)=1
f(96)=1
f(97)=397
f(98)=197
f(99)=587

b) Substitution of the polynom
The polynom f(x)=x^2-40x+29  could be written as f(y)= y^2-371 with x=y+20

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-20
f'(x)>2x-41  with x > 19

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 : 29, 5, 47, 41, 23, 73, 7, 101, 227, 1, 271, 1, 307, 1, 67, 173, 71, 181, 367, 37,
Found in Database : 29, 5, 47, 41, 23, 73, 7, 101, 227, 271, 307, 67, 173, 71, 181, 367, 37, 53,
Found in Database : 5, 7, 23, 29, 37, 41, 47, 53, 59, 61, 67, 71, 73, 79, 101, 113, 127, 131, 149,