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Development of Algorithmic Constructions

05:57:44 19.Apr 2024

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Polynom = x^2-136x+5

1. Algorithm

a:=1;
b:=-136;
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^2-136x+5

f(0)=5
f(1)=13
f(2)=263
f(3)=197
f(4)=523
f(5)=1
f(6)=31
f(7)=449
f(8)=1019
f(9)=569
f(10)=251
f(11)=137
f(12)=1483
f(13)=797
f(14)=131
f(15)=181
f(16)=383
f(17)=1009
f(18)=163
f(19)=1109
f(20)=463
f(21)=241
f(22)=2503
f(23)=1297
f(24)=2683
f(25)=277
f(26)=571
f(27)=113
f(28)=3019
f(29)=1549
f(30)=127
f(31)=1
f(32)=3323
f(33)=1697
f(34)=3463
f(35)=353
f(36)=719
f(37)=59
f(38)=3719
f(39)=1889
f(40)=1
f(41)=389
f(42)=3943
f(43)=1997
f(44)=311
f(45)=409
f(46)=827
f(47)=2089
f(48)=4219
f(49)=2129
f(50)=859
f(51)=433
f(52)=4363
f(53)=1
f(54)=4423
f(55)=89
f(56)=179
f(57)=173
f(58)=4519
f(59)=2269
f(60)=911
f(61)=457
f(62)=4583
f(63)=2297
f(64)=4603
f(65)=461
f(66)=71
f(67)=2309
f(68)=149
f(69)=1
f(70)=1
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^2-136x+5  could be written as f(y)= y^2-4619 with x=y+68

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-68
f'(x)>2x-137  with x > 68

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, 13, 263, 197, 523, 1, 31, 449, 1019, 569, 251, 137, 1483, 797, 131, 181, 383, 1009, 163, 1109,
Found in Database : 5, 13, 263, 197, 523, 31, 449, 1019, 569, 251, 137, 1483, 797, 131, 181, 383, 1009, 163, 1109, 463, 241, 2503, 1297, 2683, 277, 571, 113, 3019, 1549, 127, 3323, 1697, 3463, 353, 719, 59, 3719, 1889,
Found in Database : 5, 13, 31, 59, 71, 89, 113, 127, 131, 137, 149,