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Development of Algorithmic Constructions

18:00:04 16.Apr 2024

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Polynom = x^2-156x+131

1. Algorithm

a:=1;
b:=-156;
c:=131;


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^2-156x+131

f(0)=131
f(1)=3
f(2)=59
f(3)=41
f(4)=53
f(5)=13
f(6)=769
f(7)=19
f(8)=1
f(9)=149
f(10)=443
f(11)=61
f(12)=1597
f(13)=1
f(14)=619
f(15)=31
f(16)=37
f(17)=1
f(18)=181
f(19)=103
f(20)=863
f(21)=1
f(22)=313
f(23)=1
f(24)=3037
f(25)=1
f(26)=1
f(27)=419
f(28)=1151
f(29)=1
f(30)=89
f(31)=1
f(32)=1279
f(33)=491
f(34)=1
f(35)=1
f(36)=71
f(37)=1
f(38)=1451
f(39)=277
f(40)=167
f(41)=191
f(42)=4657
f(43)=197
f(44)=1
f(45)=1
f(46)=1
f(47)=1
f(48)=163
f(49)=1
f(50)=1723
f(51)=653
f(52)=1759
f(53)=1
f(54)=283
f(55)=113
f(56)=1823
f(57)=1
f(58)=617
f(59)=233
f(60)=433
f(61)=1
f(62)=211
f(63)=179
f(64)=101
f(65)=241
f(66)=157
f(67)=1
f(68)=1951
f(69)=367
f(70)=151
f(71)=1
f(72)=97
f(73)=1
f(74)=1979
f(75)=743
f(76)=661
f(77)=1
f(78)=5953
f(79)=1
f(80)=1
f(81)=1
f(82)=1
f(83)=1
f(84)=1
f(85)=1
f(86)=1
f(87)=1
f(88)=1
f(89)=1
f(90)=1
f(91)=1
f(92)=1
f(93)=1
f(94)=1
f(95)=1
f(96)=1
f(97)=1
f(98)=1
f(99)=1

b) Substitution of the polynom
The polynom f(x)=x^2-156x+131  could be written as f(y)= y^2-5953 with x=y+78

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-78
f'(x)>2x-157  with x > 77

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 : 131, 3, 59, 41, 53, 13, 769, 19, 1, 149, 443, 61, 1597, 1, 619, 31, 37, 1, 181, 103,
Found in Database : 131, 3, 59, 41, 53, 13, 769, 19, 149, 443, 61, 1597, 619, 31, 37, 181, 103, 863, 313, 3037, 419, 1151, 89, 1279, 491, 71, 1451, 277,
Found in Database : 3, 13, 19, 31, 37, 41, 53, 59, 61, 71, 89, 97, 101, 103, 113, 131, 149,