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Development of Algorithmic Constructions

12:59:03 19.Apr 2024

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Polynom = x^2-184x+367

1. Algorithm

a:=1;
b:=-184;
c:=367;


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^2-184x+367

f(0)=367
f(1)=23
f(2)=3
f(3)=11
f(4)=353
f(5)=1
f(6)=701
f(7)=109
f(8)=347
f(9)=151
f(10)=1373
f(11)=1
f(12)=1697
f(13)=29
f(14)=61
f(15)=271
f(16)=211
f(17)=103
f(18)=2621
f(19)=173
f(20)=971
f(21)=191
f(22)=139
f(23)=1
f(24)=1
f(25)=41
f(26)=43
f(27)=1
f(28)=4001
f(29)=1
f(30)=4253
f(31)=547
f(32)=1499
f(33)=577
f(34)=4733
f(35)=101
f(36)=1
f(37)=317
f(38)=157
f(39)=661
f(40)=5393
f(41)=229
f(42)=193
f(43)=89
f(44)=1931
f(45)=1
f(46)=5981
f(47)=1
f(48)=1
f(49)=71
f(50)=2111
f(51)=401
f(52)=73
f(53)=137
f(54)=6653
f(55)=1
f(56)=2267
f(57)=859
f(58)=631
f(59)=1
f(60)=643
f(61)=223
f(62)=2399
f(63)=907
f(64)=1
f(65)=307
f(66)=181
f(67)=467
f(68)=1
f(69)=1
f(70)=331
f(71)=1
f(72)=179
f(73)=967
f(74)=2591
f(75)=1
f(76)=7841
f(77)=1
f(78)=7901
f(79)=991
f(80)=241
f(81)=997
f(82)=727
f(83)=167
f(84)=277
f(85)=503
f(86)=2687
f(87)=1009
f(88)=8081
f(89)=337
f(90)=8093
f(91)=1
f(92)=2699
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^2-184x+367  could be written as f(y)= y^2-8097 with x=y+92

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-92
f'(x)>2x-185  with x > 90

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 : 367, 23, 3, 11, 353, 1, 701, 109, 347, 151, 1373, 1, 1697, 29, 61, 271, 211, 103, 2621, 173,
Found in Database : 367, 23, 3, 11, 353, 701, 109, 347, 151, 1373, 1697, 29, 61, 271, 211, 103, 2621, 173, 971, 191, 139, 41, 43, 4001, 4253, 547, 1499, 577, 4733, 101, 317, 157, 661,
Found in Database : 3, 11, 23, 29, 41, 43, 61, 71, 73, 89, 101, 103, 109, 137, 139,