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Development of Algorithmic Constructions

06:53:39 18.Apr 2024

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Polynom = x^2+1x+5

1. Algorithm

a:=1;
b:=1;
c:=5;


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+5

f(0)=5
f(1)=7
f(2)=11
f(3)=17
f(4)=1
f(5)=1
f(6)=47
f(7)=61
f(8)=1
f(9)=19
f(10)=23
f(11)=137
f(12)=1
f(13)=1
f(14)=43
f(15)=1
f(16)=277
f(17)=311
f(18)=347
f(19)=1
f(20)=1
f(21)=467
f(22)=73
f(23)=557
f(24)=1
f(25)=131
f(26)=101
f(27)=761
f(28)=1
f(29)=1
f(30)=1
f(31)=997
f(32)=1061
f(33)=1
f(34)=239
f(35)=1
f(36)=191
f(37)=83
f(38)=1487
f(39)=313
f(40)=1
f(41)=157
f(42)=1811
f(43)=271
f(44)=397
f(45)=1
f(46)=197
f(47)=1
f(48)=2357
f(49)=491
f(50)=1
f(51)=2657
f(52)=251
f(53)=1
f(54)=1
f(55)=617
f(56)=139
f(57)=1
f(58)=149
f(59)=709
f(60)=733
f(61)=541
f(62)=3911
f(63)=367
f(64)=1
f(65)=859
f(66)=233
f(67)=4561
f(68)=1
f(69)=967
f(70)=199
f(71)=1
f(72)=5261
f(73)=5407
f(74)=1
f(75)=163
f(76)=5857
f(77)=6011
f(78)=881
f(79)=1
f(80)=1297
f(81)=1
f(82)=1
f(83)=6977
f(84)=1429
f(85)=1
f(86)=7487
f(87)=1
f(88)=461
f(89)=229
f(90)=1
f(91)=8377
f(92)=1223
f(93)=8747
f(94)=1787
f(95)=1
f(96)=1
f(97)=9511
f(98)=571
f(99)=283

b) Substitution of the polynom
The polynom f(x)=x^2+1x+5  could be written as f(y)= y^2+4.75 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 : 5, 7, 11, 17, 1, 1, 47, 61, 1, 19, 23, 137, 1, 1, 43, 1, 277, 311, 347, 1,
Found in Database : 5, 7, 11, 17, 47, 61, 19, 23, 137, 43, 277, 311, 347, 467, 73, 557, 131, 101, 761, 997, 1061, 239, 191, 83, 1487, 313,
Found in Database : 5, 7, 11, 17, 19, 23, 43, 47, 61, 73, 83, 101, 131, 137, 139, 149,