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Development of Algorithmic Constructions

18:07:39 20.Apr 2024

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Polynom = x^2-220x+1499

1. Algorithm

a:=1;
b:=-220;
c:=1499;


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^2-220x+1499

f(0)=1499
f(1)=5
f(2)=1063
f(3)=53
f(4)=127
f(5)=1
f(6)=43
f(7)=1
f(8)=197
f(9)=1
f(10)=601
f(11)=1
f(12)=997
f(13)=149
f(14)=277
f(15)=1
f(16)=353
f(17)=61
f(18)=2137
f(19)=29
f(20)=41
f(21)=67
f(22)=2857
f(23)=379
f(24)=641
f(25)=211
f(26)=709
f(27)=1
f(28)=3877
f(29)=101
f(30)=4201
f(31)=109
f(32)=4517
f(33)=73
f(34)=193
f(35)=311
f(36)=1
f(37)=659
f(38)=5417
f(39)=139
f(40)=5701
f(41)=1
f(42)=1
f(43)=191
f(44)=1249
f(45)=797
f(46)=1301
f(47)=829
f(48)=233
f(49)=1
f(50)=7001
f(51)=89
f(52)=7237
f(53)=919
f(54)=1493
f(55)=947
f(56)=1
f(57)=487
f(58)=1
f(59)=1
f(60)=8101
f(61)=1
f(62)=8297
f(63)=1049
f(64)=1697
f(65)=1
f(66)=1733
f(67)=547
f(68)=8837
f(69)=223
f(70)=9001
f(71)=227
f(72)=9157
f(73)=577
f(74)=1861
f(75)=293
f(76)=1889
f(77)=1
f(78)=157
f(79)=241
f(80)=1
f(81)=1
f(82)=9817
f(83)=617
f(84)=397
f(85)=1
f(86)=401
f(87)=1259
f(88)=151
f(89)=1
f(90)=1
f(91)=1
f(92)=239
f(93)=1289
f(94)=2069
f(95)=1297
f(96)=2081
f(97)=163
f(98)=10457
f(99)=131

b) Substitution of the polynom
The polynom f(x)=x^2-220x+1499  could be written as f(y)= y^2-10601 with x=y+110

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-110
f'(x)>2x-221  with x > 103

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 : 1499, 5, 1063, 53, 127, 1, 43, 1, 197, 1, 601, 1, 997, 149, 277, 1, 353, 61, 2137, 29,
Found in Database : 1499, 5, 1063, 53, 127, 43, 197, 601, 997, 149, 277, 353, 61, 2137, 29, 41, 67, 2857, 379, 641, 211, 709, 3877, 101, 4201, 109, 4517, 73, 193, 311, 659, 5417, 139,
Found in Database : 5, 29, 41, 43, 53, 61, 67, 73, 89, 101, 109, 127, 131, 139, 149,