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Development of Algorithmic Constructions

16:17:20 19.Apr 2024

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Polynom = x^2+108x-281

1. Algorithm

a:=1;
b:=108;
c:=-281;


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+108x-281

f(0)=281
f(1)=43
f(2)=61
f(3)=13
f(4)=167
f(5)=71
f(6)=31
f(7)=131
f(8)=647
f(9)=193
f(10)=29
f(11)=257
f(12)=19
f(13)=17
f(14)=1427
f(15)=23
f(16)=1
f(17)=461
f(18)=1987
f(19)=41
f(20)=53
f(21)=607
f(22)=2579
f(23)=683
f(24)=2887
f(25)=761
f(26)=3203
f(27)=1
f(28)=3527
f(29)=1
f(30)=227
f(31)=1
f(32)=1
f(33)=1093
f(34)=4547
f(35)=1181
f(36)=4903
f(37)=1
f(38)=229
f(39)=47
f(40)=5639
f(41)=1
f(42)=463
f(43)=1553
f(44)=149
f(45)=127
f(46)=6803
f(47)=103
f(48)=7207
f(49)=109
f(50)=401
f(51)=1
f(52)=8039
f(53)=2063
f(54)=8467
f(55)=1
f(56)=307
f(57)=2281
f(58)=719
f(59)=2393
f(60)=239
f(61)=1
f(62)=10259
f(63)=1
f(64)=631
f(65)=2741
f(66)=659
f(67)=2861
f(68)=1
f(69)=157
f(70)=641
f(71)=1
f(72)=409
f(73)=1
f(74)=13187
f(75)=3361
f(76)=1
f(77)=3491
f(78)=347
f(79)=3623
f(80)=14759
f(81)=1
f(82)=15299
f(83)=1
f(84)=1
f(85)=139
f(86)=349
f(87)=97
f(88)=1
f(89)=1
f(90)=17539
f(91)=4457
f(92)=18119
f(93)=4603
f(94)=1439
f(95)=4751
f(96)=199
f(97)=1
f(98)=1171
f(99)=163

b) Substitution of the polynom
The polynom f(x)=x^2+108x-281  could be written as f(y)= y^2-3197 with x=y-54

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+54
f'(x)>2x+107

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 : 281, 43, 61, 13, 167, 71, 31, 131, 647, 193, 29, 257, 19, 17, 1427, 23, 1, 461, 1987, 41,
Found in Database : 281, 43, 61, 13, 167, 71, 31, 131, 647, 193, 29, 257, 19, 17, 1427, 23, 461, 1987, 41, 53, 607, 2579, 683, 2887, 761, 3203, 3527, 227, 1093, 4547, 1181, 4903, 229, 47,
Found in Database : 13, 17, 19, 23, 29, 31, 41, 43, 47, 53, 61, 71, 97, 103, 109, 127, 131, 139, 149,