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Development of Algorithmic Constructions

06:27:13 20.Apr 2024

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Polynom = x^2-140x-3

1. Algorithm

a:=1;
b:=-140;
c:=-3;


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^2-140x-3

f(0)=3
f(1)=71
f(2)=31
f(3)=23
f(4)=547
f(5)=113
f(6)=269
f(7)=467
f(8)=353
f(9)=197
f(10)=1303
f(11)=79
f(12)=19
f(13)=827
f(14)=1
f(15)=313
f(16)=1987
f(17)=349
f(18)=733
f(19)=1151
f(20)=89
f(21)=139
f(22)=1
f(23)=449
f(24)=929
f(25)=1439
f(26)=43
f(27)=509
f(28)=73
f(29)=179
f(30)=367
f(31)=1
f(32)=1153
f(33)=1
f(34)=3607
f(35)=613
f(36)=1249
f(37)=1907
f(38)=431
f(39)=1
f(40)=4003
f(41)=677
f(42)=1373
f(43)=2087
f(44)=1409
f(45)=1
f(46)=4327
f(47)=1
f(48)=491
f(49)=97
f(50)=1
f(51)=757
f(52)=241
f(53)=769
f(54)=1549
f(55)=2339
f(56)=523
f(57)=263
f(58)=4759
f(59)=797
f(60)=1601
f(61)=2411
f(62)=1613
f(63)=809
f(64)=157
f(65)=271
f(66)=181
f(67)=2447
f(68)=1
f(69)=1
f(70)=4903
f(71)=1
f(72)=1
f(73)=1
f(74)=1
f(75)=1
f(76)=1
f(77)=1
f(78)=1
f(79)=1
f(80)=1
f(81)=1
f(82)=1
f(83)=1
f(84)=1
f(85)=1
f(86)=1
f(87)=1
f(88)=1
f(89)=1
f(90)=1
f(91)=1
f(92)=1
f(93)=1
f(94)=1
f(95)=1
f(96)=1
f(97)=1
f(98)=1
f(99)=1

b) Substitution of the polynom
The polynom f(x)=x^2-140x-3  could be written as f(y)= y^2-4903 with x=y+70

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-70
f'(x)>2x-141  with x > 70

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 : 3, 71, 31, 23, 547, 113, 269, 467, 353, 197, 1303, 79, 19, 827, 1, 313, 1987, 349, 733, 1151,
Found in Database : 3, 71, 31, 23, 547, 113, 269, 467, 353, 197, 1303, 79, 19, 827, 313, 1987, 349, 733, 1151, 89, 139, 449, 929, 1439, 43, 509, 73, 179, 367, 1153, 3607, 613, 1249, 1907, 431,
Found in Database : 3, 19, 23, 31, 43, 71, 73, 79, 89, 97, 113, 139,