---

Development of Algorithmic Constructions

05:34:47 18.Apr 2024

---

Polynom = x^2+136x-1249

1. Algorithm

a:=1;
b:=136;
c:=-1249;


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+136x-1249

f(0)=1249
f(1)=139
f(2)=7
f(3)=13
f(4)=53
f(5)=17
f(6)=397
f(7)=31
f(8)=97
f(9)=1
f(10)=211
f(11)=23
f(12)=1
f(13)=43
f(14)=37
f(15)=127
f(16)=1
f(17)=1
f(18)=1523
f(19)=1
f(20)=1871
f(21)=1
f(22)=131
f(23)=1
f(24)=2591
f(25)=347
f(26)=2963
f(27)=197
f(28)=3343
f(29)=1
f(30)=41
f(31)=491
f(32)=4127
f(33)=541
f(34)=1
f(35)=1
f(36)=4943
f(37)=1
f(38)=173
f(39)=1
f(40)=5791
f(41)=751
f(42)=479
f(43)=1
f(44)=953
f(45)=431
f(46)=419
f(47)=919
f(48)=7583
f(49)=977
f(50)=83
f(51)=1
f(52)=8527
f(53)=137
f(54)=9011
f(55)=89
f(56)=1
f(57)=1
f(58)=1429
f(59)=641
f(60)=457
f(61)=673
f(62)=11027
f(63)=1
f(64)=11551
f(65)=1
f(66)=281
f(67)=193
f(68)=971
f(69)=1
f(70)=13171
f(71)=1
f(72)=1
f(73)=103
f(74)=461
f(75)=911
f(76)=167
f(77)=947
f(78)=15443
f(79)=1
f(80)=1
f(81)=157
f(82)=1279
f(83)=1
f(84)=17231
f(85)=1
f(86)=2549
f(87)=2269
f(88)=499
f(89)=2347
f(90)=1123
f(91)=1213
f(92)=19727
f(93)=179
f(94)=1567
f(95)=199
f(96)=21023
f(97)=1
f(98)=21683
f(99)=1

b) Substitution of the polynom
The polynom f(x)=x^2+136x-1249  could be written as f(y)= y^2-5873 with x=y-68

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+68
f'(x)>2x+135

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 : 1249, 139, 7, 13, 53, 17, 397, 31, 97, 1, 211, 23, 1, 43, 37, 127, 1, 1, 1523, 1,
Found in Database : 1249, 139, 7, 13, 53, 17, 397, 31, 97, 211, 23, 43, 37, 127, 1523, 1871, 131, 2591, 347, 2963, 197, 3343, 41, 491, 4127, 541, 4943, 173,
Found in Database : 7, 13, 17, 23, 31, 37, 41, 43, 53, 83, 89, 97, 103, 127, 131, 137, 139,