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Development of Algorithmic Constructions

14:17:29 20.Apr 2024

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Polynom = x^2+100x-2797

1. Algorithm

a:=1;
b:=100;
c:=-2797;


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+100x-2797

f(0)=2797
f(1)=337
f(2)=2593
f(3)=311
f(4)=2381
f(5)=71
f(6)=2161
f(7)=1
f(8)=1933
f(9)=227
f(10)=1697
f(11)=197
f(12)=1453
f(13)=83
f(14)=1201
f(15)=67
f(16)=941
f(17)=101
f(18)=673
f(19)=1
f(20)=397
f(21)=1
f(22)=113
f(23)=1
f(24)=179
f(25)=41
f(26)=479
f(27)=79
f(28)=787
f(29)=59
f(30)=1103
f(31)=1
f(32)=1427
f(33)=199
f(34)=1759
f(35)=241
f(36)=2099
f(37)=1
f(38)=2447
f(39)=1
f(40)=2803
f(41)=373
f(42)=3167
f(43)=419
f(44)=3539
f(45)=233
f(46)=3919
f(47)=257
f(48)=73
f(49)=563
f(50)=4703
f(51)=613
f(52)=5107
f(53)=1
f(54)=5519
f(55)=1
f(56)=5939
f(57)=769
f(58)=6367
f(59)=823
f(60)=6803
f(61)=439
f(62)=7247
f(63)=467
f(64)=7699
f(65)=991
f(66)=1
f(67)=1049
f(68)=8627
f(69)=277
f(70)=9103
f(71)=1
f(72)=9587
f(73)=1229
f(74)=10079
f(75)=1291
f(76)=149
f(77)=677
f(78)=11087
f(79)=709
f(80)=283
f(81)=1483
f(82)=181
f(83)=1549
f(84)=12659
f(85)=1
f(86)=1
f(87)=421
f(88)=1
f(89)=1753
f(90)=14303
f(91)=1823
f(92)=14867
f(93)=947
f(94)=15439
f(95)=983
f(96)=193
f(97)=2039
f(98)=16607
f(99)=2113

b) Substitution of the polynom
The polynom f(x)=x^2+100x-2797  could be written as f(y)= y^2-5297 with x=y-50

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+50
f'(x)>2x+99

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 : 2797, 337, 2593, 311, 2381, 71, 2161, 1, 1933, 227, 1697, 197, 1453, 83, 1201, 67, 941, 101, 673, 1,
Found in Database : 2797, 337, 2593, 311, 2381, 71, 2161, 1933, 227, 1697, 197, 1453, 83, 1201, 67, 941, 101, 673, 397, 113, 179, 41, 479, 79, 787, 59, 1103, 1427, 199, 1759, 241, 2099, 2447,
Found in Database : 41, 59, 67, 71, 73, 79, 83, 101, 113, 149,