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Development of Algorithmic Constructions

18:36:26 18.Apr 2024

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Polynom = x^2+4x-139

1. Algorithm

a:=1;
b:=4;
c:=-139;


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+4x-139

f(0)=139
f(1)=67
f(2)=127
f(3)=59
f(4)=107
f(5)=47
f(6)=79
f(7)=31
f(8)=43
f(9)=11
f(10)=1
f(11)=13
f(12)=53
f(13)=41
f(14)=113
f(15)=73
f(16)=181
f(17)=109
f(18)=257
f(19)=149
f(20)=1
f(21)=193
f(22)=433
f(23)=241
f(24)=1
f(25)=293
f(26)=641
f(27)=349
f(28)=757
f(29)=409
f(30)=881
f(31)=1
f(32)=1013
f(33)=541
f(34)=1153
f(35)=613
f(36)=1301
f(37)=1
f(38)=1
f(39)=769
f(40)=1621
f(41)=853
f(42)=163
f(43)=941
f(44)=1973
f(45)=1033
f(46)=2161
f(47)=1129
f(48)=2357
f(49)=1229
f(50)=197
f(51)=1
f(52)=1
f(53)=131
f(54)=1
f(55)=1553
f(56)=3221
f(57)=1669
f(58)=3457
f(59)=1789
f(60)=3701
f(61)=1913
f(62)=1
f(63)=157
f(64)=383
f(65)=1
f(66)=4481
f(67)=2309
f(68)=71
f(69)=1
f(70)=1
f(71)=2593
f(72)=5333
f(73)=2741
f(74)=1
f(75)=263
f(76)=457
f(77)=3049
f(78)=6257
f(79)=3209
f(80)=6581
f(81)=3373
f(82)=223
f(83)=3541
f(84)=7253
f(85)=1
f(86)=691
f(87)=3889
f(88)=1
f(89)=313
f(90)=1
f(91)=4253
f(92)=8693
f(93)=4441
f(94)=211
f(95)=1
f(96)=9461
f(97)=439
f(98)=9857
f(99)=1

b) Substitution of the polynom
The polynom f(x)=x^2+4x-139  could be written as f(y)= y^2-143 with x=y-2

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+2
f'(x)>2x+3

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 : 139, 67, 127, 59, 107, 47, 79, 31, 43, 11, 1, 13, 53, 41, 113, 73, 181, 109, 257, 149,
Found in Database : 139, 67, 127, 59, 107, 47, 79, 31, 43, 11, 13, 53, 41, 113, 73, 181, 109, 257, 149, 193, 433, 241, 293, 641, 349, 757, 409, 881, 1013, 541, 1153, 613, 1301, 769,
Found in Database : 11, 13, 31, 41, 43, 47, 53, 59, 67, 71, 73, 79, 107, 109, 113, 127, 131, 139, 149,