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Development of Algorithmic Constructions

12:37:50 20.Apr 2024

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Polynom = x^2+80x-1657

1. Algorithm

a:=1;
b:=80;
c:=-1657;


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+80x-1657

f(0)=1657
f(1)=197
f(2)=1493
f(3)=11
f(4)=1321
f(5)=7
f(6)=163
f(7)=131
f(8)=953
f(9)=107
f(10)=757
f(11)=41
f(12)=79
f(13)=1
f(14)=31
f(15)=29
f(16)=1
f(17)=1
f(18)=1
f(19)=1
f(20)=1
f(21)=1
f(22)=587
f(23)=89
f(24)=839
f(25)=1
f(26)=157
f(27)=1
f(28)=1367
f(29)=47
f(30)=53
f(31)=223
f(32)=1
f(33)=37
f(34)=317
f(35)=1
f(36)=229
f(37)=167
f(38)=257
f(39)=373
f(40)=449
f(41)=59
f(42)=3467
f(43)=227
f(44)=1
f(45)=1
f(46)=4139
f(47)=1
f(48)=641
f(49)=1
f(50)=1
f(51)=1
f(52)=127
f(53)=337
f(54)=797
f(55)=103
f(56)=101
f(57)=769
f(58)=577
f(59)=409
f(60)=613
f(61)=1
f(62)=1021
f(63)=919
f(64)=7559
f(65)=971
f(66)=1
f(67)=1
f(68)=1201
f(69)=1
f(70)=239
f(71)=1
f(72)=251
f(73)=1
f(74)=9739
f(75)=1
f(76)=1
f(77)=1
f(78)=10667
f(79)=1
f(80)=1013
f(81)=1423
f(82)=151
f(83)=1
f(84)=12119
f(85)=773
f(86)=12619
f(87)=1609
f(88)=13127
f(89)=1
f(90)=1949
f(91)=1
f(92)=457
f(93)=1
f(94)=14699
f(95)=1871
f(96)=311
f(97)=277
f(98)=15787
f(99)=1

b) Substitution of the polynom
The polynom f(x)=x^2+80x-1657  could be written as f(y)= y^2-3257 with x=y-40

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+40
f'(x)>2x+79

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 : 1657, 197, 1493, 11, 1321, 7, 163, 131, 953, 107, 757, 41, 79, 1, 31, 29, 1, 1, 1, 1,
Found in Database : 1657, 197, 1493, 11, 1321, 7, 163, 131, 953, 107, 757, 41, 79, 31, 29, 587, 89, 839, 157, 1367, 47, 53, 223, 37, 317, 229, 167, 257, 373,
Found in Database : 7, 11, 29, 31, 37, 41, 47, 53, 59, 79, 89, 101, 103, 107, 127, 131,