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Development of Algorithmic Constructions

01:15:57 19.Apr 2024

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Polynom = x^2+1x+3

1. Algorithm

a:=1;
b:=1;
c:=3;


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+1x+3

f(0)=3
f(1)=5
f(2)=1
f(3)=1
f(4)=23
f(5)=11
f(6)=1
f(7)=59
f(8)=1
f(9)=31
f(10)=113
f(11)=1
f(12)=53
f(13)=37
f(14)=71
f(15)=1
f(16)=1
f(17)=103
f(18)=1
f(19)=383
f(20)=47
f(21)=1
f(22)=509
f(23)=1
f(24)=67
f(25)=653
f(26)=1
f(27)=1
f(28)=163
f(29)=97
f(30)=311
f(31)=199
f(32)=353
f(33)=1
f(34)=1193
f(35)=421
f(36)=89
f(37)=1409
f(38)=1
f(39)=521
f(40)=1
f(41)=1
f(42)=1
f(43)=379
f(44)=661
f(45)=691
f(46)=433
f(47)=251
f(48)=157
f(49)=223
f(50)=1
f(51)=1
f(52)=1
f(53)=191
f(54)=991
f(55)=3083
f(56)=1
f(57)=1103
f(58)=137
f(59)=1181
f(60)=1
f(61)=757
f(62)=1303
f(63)=269
f(64)=181
f(65)=1
f(66)=1
f(67)=1
f(68)=313
f(69)=179
f(70)=4973
f(71)=1
f(72)=1753
f(73)=1
f(74)=617
f(75)=1901
f(76)=1171
f(77)=2003
f(78)=1
f(79)=6323
f(80)=2161
f(81)=443
f(82)=619
f(83)=1
f(84)=2381
f(85)=1
f(86)=499
f(87)=1
f(88)=1567
f(89)=2671
f(90)=2731
f(91)=1
f(92)=317
f(93)=1
f(94)=8933
f(95)=3041
f(96)=1
f(97)=257
f(98)=647
f(99)=3301

b) Substitution of the polynom
The polynom f(x)=x^2+1x+3  could be written as f(y)= y^2+2.75 with x=y-0.5

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+0.5
f'(x)>2x

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 : 3, 5, 1, 1, 23, 11, 1, 59, 1, 31, 113, 1, 53, 37, 71, 1, 1, 103, 1, 383,
Found in Database : 3, 5, 23, 11, 59, 31, 113, 53, 37, 71, 103, 383, 47, 509, 67, 653, 163, 97, 311, 199, 353, 1193, 421, 89, 1409, 521,
Found in Database : 3, 5, 11, 23, 31, 37, 47, 53, 59, 67, 71, 89, 97, 103, 113, 137,