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Development of Algorithmic Constructions

08:40:24 19.Apr 2024

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Polynom = x^2+1x-109

1. Algorithm

a:=1;
b:=1;
c:=-109;


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-109

f(0)=109
f(1)=107
f(2)=103
f(3)=97
f(4)=89
f(5)=79
f(6)=67
f(7)=53
f(8)=37
f(9)=19
f(10)=1
f(11)=23
f(12)=47
f(13)=73
f(14)=101
f(15)=131
f(16)=163
f(17)=197
f(18)=233
f(19)=271
f(20)=311
f(21)=353
f(22)=397
f(23)=443
f(24)=491
f(25)=541
f(26)=593
f(27)=647
f(28)=1
f(29)=761
f(30)=821
f(31)=883
f(32)=947
f(33)=1013
f(34)=1
f(35)=1151
f(36)=1223
f(37)=1297
f(38)=1373
f(39)=1451
f(40)=1531
f(41)=1613
f(42)=1697
f(43)=1783
f(44)=1871
f(45)=1
f(46)=2053
f(47)=113
f(48)=2243
f(49)=2341
f(50)=2441
f(51)=2543
f(52)=2647
f(53)=2753
f(54)=2861
f(55)=2971
f(56)=3083
f(57)=139
f(58)=3313
f(59)=1
f(60)=1
f(61)=3673
f(62)=3797
f(63)=3923
f(64)=4051
f(65)=1
f(66)=227
f(67)=4447
f(68)=4583
f(69)=4721
f(70)=4861
f(71)=5003
f(72)=5147
f(73)=1
f(74)=5441
f(75)=5591
f(76)=5743
f(77)=5897
f(78)=6053
f(79)=6211
f(80)=277
f(81)=1
f(82)=181
f(83)=6863
f(84)=1
f(85)=379
f(86)=1
f(87)=7547
f(88)=7723
f(89)=7901
f(90)=8081
f(91)=8263
f(92)=8447
f(93)=1
f(94)=8821
f(95)=9011
f(96)=9203
f(97)=9397
f(98)=1
f(99)=9791

b) Substitution of the polynom
The polynom f(x)=x^2+1x-109  could be written as f(y)= y^2-109.25 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 : 109, 107, 103, 97, 89, 79, 67, 53, 37, 19, 1, 23, 47, 73, 101, 131, 163, 197, 233, 271,
Found in Database : 109, 107, 103, 97, 89, 79, 67, 53, 37, 19, 23, 47, 73, 101, 131, 163, 197, 233, 271, 311, 353, 397, 443, 491, 541, 593, 647, 761, 821, 883, 947, 1013, 1151, 1223, 1297, 1373, 1451,
Found in Database : 19, 23, 37, 47, 53, 67, 73, 79, 89, 97, 101, 103, 107, 109, 113, 131, 139,