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Development of Algorithmic Constructions

02:26:26 20.Apr 2024

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Polynom = x^2+150x-47

1. Algorithm

a:=1;
b:=150;
c:=-47;


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+150x-47

f(0)=47
f(1)=13
f(2)=257
f(3)=103
f(4)=569
f(5)=7
f(6)=127
f(7)=263
f(8)=1217
f(9)=173
f(10)=1553
f(11)=431
f(12)=271
f(13)=37
f(14)=1
f(15)=607
f(16)=2609
f(17)=349
f(18)=229
f(19)=113
f(20)=479
f(21)=443
f(22)=101
f(23)=983
f(24)=4129
f(25)=541
f(26)=647
f(27)=1
f(28)=4937
f(29)=643
f(30)=53
f(31)=107
f(32)=109
f(33)=1
f(34)=887
f(35)=1607
f(36)=61
f(37)=859
f(38)=151
f(39)=1831
f(40)=83
f(41)=139
f(42)=8017
f(43)=2063
f(44)=653
f(45)=1091
f(46)=8969
f(47)=1
f(48)=193
f(49)=1213
f(50)=269
f(51)=2551
f(52)=10457
f(53)=1
f(54)=1567
f(55)=401
f(56)=11489
f(57)=1
f(58)=197
f(59)=1
f(60)=12553
f(61)=1
f(62)=1871
f(63)=3343
f(64)=13649
f(65)=1741
f(66)=1093
f(67)=3623
f(68)=2111
f(69)=1
f(70)=1181
f(71)=3911
f(72)=15937
f(73)=2029
f(74)=16529
f(75)=601
f(76)=2447
f(77)=2179
f(78)=17737
f(79)=347
f(80)=18353
f(81)=2333
f(82)=2711
f(83)=1
f(84)=19609
f(85)=1
f(86)=20249
f(87)=1
f(88)=20897
f(89)=379
f(90)=3079
f(91)=5471
f(92)=1709
f(93)=2819
f(94)=487
f(95)=5807
f(96)=1
f(97)=1
f(98)=191
f(99)=6151

b) Substitution of the polynom
The polynom f(x)=x^2+150x-47  could be written as f(y)= y^2-5672 with x=y-75

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+75
f'(x)>2x+149

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 : 47, 13, 257, 103, 569, 7, 127, 263, 1217, 173, 1553, 431, 271, 37, 1, 607, 2609, 349, 229, 113,
Found in Database : 47, 13, 257, 103, 569, 7, 127, 263, 1217, 173, 1553, 431, 271, 37, 607, 2609, 349, 229, 113, 479, 443, 101, 983, 4129, 541, 647, 4937, 643, 53, 107, 109, 887, 1607, 61, 859, 151, 1831,
Found in Database : 7, 13, 37, 47, 53, 61, 83, 101, 103, 107, 109, 113, 127, 131, 139,