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Development of Algorithmic Constructions

13:48:17 20.Apr 2024

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Polynom = x^2+21x-5

1. Algorithm

a:=1;
b:=21;
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+21x-5

f(0)=5
f(1)=17
f(2)=41
f(3)=67
f(4)=19
f(5)=1
f(6)=157
f(7)=191
f(8)=227
f(9)=53
f(10)=61
f(11)=347
f(12)=23
f(13)=1
f(14)=97
f(15)=107
f(16)=587
f(17)=641
f(18)=1
f(19)=151
f(20)=163
f(21)=877
f(22)=941
f(23)=1
f(24)=43
f(25)=229
f(26)=1217
f(27)=1291
f(28)=1367
f(29)=1
f(30)=1
f(31)=1607
f(32)=89
f(33)=1777
f(34)=373
f(35)=1
f(36)=1
f(37)=2141
f(38)=2237
f(39)=467
f(40)=487
f(41)=59
f(42)=139
f(43)=1
f(44)=571
f(45)=593
f(46)=181
f(47)=3191
f(48)=3307
f(49)=137
f(50)=709
f(51)=193
f(52)=223
f(53)=3917
f(54)=809
f(55)=167
f(56)=73
f(57)=4441
f(58)=199
f(59)=1
f(60)=971
f(61)=263
f(62)=1
f(63)=311
f(64)=1087
f(65)=1117
f(66)=5737
f(67)=1
f(68)=6047
f(69)=1
f(70)=1
f(71)=1
f(72)=6691
f(73)=6857
f(74)=281
f(75)=1439
f(76)=1
f(77)=7541
f(78)=7717
f(79)=1579
f(80)=1
f(81)=359
f(82)=367
f(83)=8627
f(84)=1
f(85)=1801
f(86)=541
f(87)=9391
f(88)=9587
f(89)=103
f(90)=1997
f(91)=1
f(92)=10391
f(93)=10597
f(94)=2161
f(95)=2203
f(96)=109
f(97)=673
f(98)=11657
f(99)=1

b) Substitution of the polynom
The polynom f(x)=x^2+21x-5  could be written as f(y)= y^2-115.25 with x=y-10.5

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+10.5
f'(x)>2x+20

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, 17, 41, 67, 19, 1, 157, 191, 227, 53, 61, 347, 23, 1, 97, 107, 587, 641, 1, 151,
Found in Database : 5, 17, 41, 67, 19, 157, 191, 227, 53, 61, 347, 23, 97, 107, 587, 641, 151, 163, 877, 941, 43, 229, 1217, 1291, 1367, 1607, 89, 1777, 373, 2141, 2237, 467,
Found in Database : 5, 17, 19, 23, 41, 43, 53, 59, 61, 67, 73, 89, 97, 103, 107, 109, 113, 137, 139,