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Development of Algorithmic Constructions

18:33:51 20.Apr 2024

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Polynom = x^2+192x-1

1. Algorithm

a:=1;
b:=192;
c:=-1;


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+192x-1

f(0)=1
f(1)=3
f(2)=43
f(3)=73
f(4)=29
f(5)=41
f(6)=1187
f(7)=1
f(8)=13
f(9)=113
f(10)=673
f(11)=31
f(12)=2447
f(13)=37
f(14)=1
f(15)=97
f(16)=1109
f(17)=1
f(18)=3779
f(19)=167
f(20)=157
f(21)=1
f(22)=523
f(23)=103
f(24)=71
f(25)=1
f(26)=1889
f(27)=739
f(28)=2053
f(29)=89
f(30)=6659
f(31)=1
f(32)=2389
f(33)=1
f(34)=197
f(35)=331
f(36)=283
f(37)=353
f(38)=971
f(39)=563
f(40)=1031
f(41)=199
f(42)=317
f(43)=421
f(44)=3461
f(45)=1
f(46)=1
f(47)=1
f(48)=11519
f(49)=1
f(50)=109
f(51)=1549
f(52)=4229
f(53)=541
f(54)=359
f(55)=1
f(56)=1543
f(57)=887
f(58)=179
f(59)=617
f(60)=1163
f(61)=643
f(62)=181
f(63)=251
f(64)=127
f(65)=1
f(66)=17027
f(67)=241
f(68)=83
f(69)=2251
f(70)=6113
f(71)=389
f(72)=229
f(73)=1
f(74)=1
f(75)=2503
f(76)=1
f(77)=863
f(78)=21059
f(79)=223
f(80)=7253
f(81)=691
f(82)=7489
f(83)=1
f(84)=239
f(85)=1
f(86)=613
f(87)=1
f(88)=191
f(89)=521
f(90)=619
f(91)=1
f(92)=2903
f(93)=3313
f(94)=1
f(95)=1
f(96)=27647
f(97)=1
f(98)=9473
f(99)=277

b) Substitution of the polynom
The polynom f(x)=x^2+192x-1  could be written as f(y)= y^2-9217 with x=y-96

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+96
f'(x)>2x+191

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 : 1, 3, 43, 73, 29, 41, 1187, 1, 13, 113, 673, 31, 2447, 37, 1, 97, 1109, 1, 3779, 167,
Found in Database : 3, 43, 73, 29, 41, 1187, 13, 113, 673, 31, 2447, 37, 97, 1109, 3779, 167, 157, 523, 103, 71, 1889, 739, 2053, 89, 6659, 2389, 197, 331, 283, 353, 971, 563,
Found in Database : 3, 13, 29, 31, 37, 41, 43, 71, 73, 83, 89, 97, 103, 109, 113, 127, 137,