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Development of Algorithmic Constructions

09:06:19 19.Apr 2024

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Polynom = x^2+11x-5

1. Algorithm

a:=1;
b:=11;
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+11x-5

f(0)=5
f(1)=7
f(2)=3
f(3)=37
f(4)=11
f(5)=1
f(6)=97
f(7)=1
f(8)=1
f(9)=1
f(10)=41
f(11)=79
f(12)=271
f(13)=307
f(14)=23
f(15)=1
f(16)=61
f(17)=157
f(18)=47
f(19)=113
f(20)=1
f(21)=29
f(22)=103
f(23)=1
f(24)=167
f(25)=179
f(26)=1
f(27)=1021
f(28)=1087
f(29)=1
f(30)=1
f(31)=1297
f(32)=457
f(33)=1447
f(34)=1
f(35)=107
f(36)=241
f(37)=1
f(38)=619
f(39)=389
f(40)=1
f(41)=709
f(42)=2221
f(43)=331
f(44)=1
f(45)=503
f(46)=2617
f(47)=907
f(48)=257
f(49)=587
f(50)=1
f(51)=1
f(52)=3271
f(53)=1129
f(54)=701
f(55)=1
f(56)=1249
f(57)=1
f(58)=571
f(59)=1
f(60)=1
f(61)=1
f(62)=137
f(63)=4657
f(64)=1
f(65)=1
f(66)=5077
f(67)=227
f(68)=1789
f(69)=1103
f(70)=1
f(71)=277
f(72)=853
f(73)=557
f(74)=419
f(75)=1289
f(76)=6607
f(77)=1
f(78)=991
f(79)=1
f(80)=1
f(81)=677
f(82)=7621
f(83)=1
f(84)=1
f(85)=233
f(86)=397
f(87)=8521
f(88)=8707
f(89)=593
f(90)=1
f(91)=9277
f(92)=1
f(93)=1381
f(94)=1973
f(95)=1
f(96)=10267
f(97)=283
f(98)=3559
f(99)=311

b) Substitution of the polynom
The polynom f(x)=x^2+11x-5  could be written as f(y)= y^2-35.25 with x=y-5.5

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+5.5
f'(x)>2x+10

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, 3, 37, 11, 1, 97, 1, 1, 1, 41, 79, 271, 307, 23, 1, 61, 157, 47, 113,
Found in Database : 5, 7, 3, 37, 11, 97, 41, 79, 271, 307, 23, 61, 157, 47, 113, 29, 103, 167, 179, 1021, 1087, 1297, 457, 1447, 107, 241, 619, 389,
Found in Database : 3, 5, 7, 11, 23, 29, 37, 41, 47, 61, 79, 97, 103, 107, 113, 137,