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Development of Algorithmic Constructions

22:05:47 19.Apr 2024

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Polynom = x^2-36x+5

1. Algorithm

a:=1;
b:=-36;
c:=5;


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^2-36x+5

f(0)=5
f(1)=3
f(2)=7
f(3)=47
f(4)=41
f(5)=1
f(6)=1
f(7)=11
f(8)=73
f(9)=17
f(10)=1
f(11)=1
f(12)=283
f(13)=1
f(14)=101
f(15)=31
f(16)=1
f(17)=53
f(18)=29
f(19)=1
f(20)=1
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)=61
f(40)=1
f(41)=1
f(42)=257
f(43)=1
f(44)=1
f(45)=1
f(46)=1
f(47)=1
f(48)=83
f(49)=107
f(50)=1
f(51)=1
f(52)=1
f(53)=151
f(54)=977
f(55)=1
f(56)=1
f(57)=601
f(58)=1
f(59)=227
f(60)=1
f(61)=1
f(62)=1
f(63)=853
f(64)=599
f(65)=1
f(66)=397
f(67)=347
f(68)=727
f(69)=163
f(70)=1
f(71)=1
f(72)=1
f(73)=1
f(74)=313
f(75)=293
f(76)=1
f(77)=1
f(78)=193
f(79)=1
f(80)=1
f(81)=1
f(82)=1259
f(83)=1
f(84)=367
f(85)=139
f(86)=1
f(87)=2221
f(88)=509
f(89)=787
f(90)=1
f(91)=167
f(92)=191
f(93)=379
f(94)=1
f(95)=1
f(96)=1153
f(97)=1
f(98)=2027
f(99)=3121

b) Substitution of the polynom
The polynom f(x)=x^2-36x+5  could be written as f(y)= y^2-319 with x=y+18

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-18
f'(x)>2x-37  with x > 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 : 5, 3, 7, 47, 41, 1, 1, 11, 73, 17, 1, 1, 283, 1, 101, 31, 1, 53, 29, 1,
Found in Database : 5, 3, 7, 47, 41, 11, 73, 17, 283, 101, 31, 53, 29, 61,
Found in Database : 3, 5, 7, 11, 17, 29, 31, 41, 47, 53, 61, 73, 83, 101, 107, 139,