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Development of Algorithmic Constructions

23:04:01 19.Apr 2024

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Polynom = x^2-200x+103

1. Algorithm

a:=1;
b:=-200;
c:=103;


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^2-200x+103

f(0)=103
f(1)=3
f(2)=293
f(3)=61
f(4)=227
f(5)=109
f(6)=1061
f(7)=13
f(8)=1433
f(9)=101
f(10)=599
f(11)=19
f(12)=2153
f(13)=97
f(14)=41
f(15)=167
f(16)=947
f(17)=47
f(18)=1
f(19)=139
f(20)=269
f(21)=457
f(22)=31
f(23)=1
f(24)=317
f(25)=89
f(26)=4421
f(27)=571
f(28)=1571
f(29)=607
f(30)=263
f(31)=107
f(32)=5273
f(33)=1
f(34)=1847
f(35)=709
f(36)=5801
f(37)=1
f(38)=6053
f(39)=193
f(40)=2099
f(41)=401
f(42)=1
f(43)=277
f(44)=6761
f(45)=859
f(46)=179
f(47)=443
f(48)=7193
f(49)=1
f(50)=569
f(51)=937
f(52)=2531
f(53)=1
f(54)=251
f(55)=1
f(56)=419
f(57)=503
f(58)=2711
f(59)=79
f(60)=8297
f(61)=349
f(62)=1
f(63)=1
f(64)=1
f(65)=271
f(66)=8741
f(67)=367
f(68)=467
f(69)=1117
f(70)=2999
f(71)=283
f(72)=701
f(73)=191
f(74)=9221
f(75)=1
f(76)=239
f(77)=1171
f(78)=9413
f(79)=197
f(80)=9497
f(81)=149
f(82)=3191
f(83)=1201
f(84)=311
f(85)=1
f(86)=1
f(87)=1
f(88)=3251
f(89)=1
f(90)=1
f(91)=409
f(92)=9833
f(93)=1231
f(94)=173
f(95)=617
f(96)=241
f(97)=1
f(98)=761
f(99)=1237

b) Substitution of the polynom
The polynom f(x)=x^2-200x+103  could be written as f(y)= y^2-9897 with x=y+100

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-100
f'(x)>2x-201  with x > 99

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 : 103, 3, 293, 61, 227, 109, 1061, 13, 1433, 101, 599, 19, 2153, 97, 41, 167, 947, 47, 1, 139,
Found in Database : 103, 3, 293, 61, 227, 109, 1061, 13, 1433, 101, 599, 19, 2153, 97, 41, 167, 947, 47, 139, 269, 457, 31, 317, 89, 4421, 571, 1571, 607, 263, 107, 5273, 1847, 709, 5801, 6053, 193,
Found in Database : 3, 13, 19, 31, 41, 47, 61, 79, 89, 97, 101, 103, 107, 109, 139, 149,