How to resolve the algorithm Numbers which are the cube roots of the product of their proper divisors step by step in the Pascal programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Numbers which are the cube roots of the product of their proper divisors step by step in the Pascal programming language
Table of Contents
Problem Statement
Consider the number 24. Its proper divisors are: 1, 2, 3, 4, 6, 8 and 12. Their product is 13,824 and the cube root of this is 24. So 24 satisfies the definition in the task title. Compute and show here the first 50 positive integers which are the cube roots of the product of their proper divisors. Also show the 500th and 5,000th such numbers. Compute and show the 50,000th such number. OEIS considers 1 to be the first number in this sequence even though, strictly speaking, it has no proper divisors. Please therefore do likewise.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Numbers which are the cube roots of the product of their proper divisors step by step in the Pascal programming language
Source code in the pascal programming language
program Root3rd_divs_n.pas;
{$IFDEF FPC}
{$MODE DELPHI} {$OPTIMIZATION ON,ALL} {$COPERATORS ON}
{$ENDIF}
{$IFDEF WINDOWS}
{$APPTYPE CONSOLE}
{$ENDIF}
uses
sysutils
{$IFDEF WINDOWS},Windows{$ENDIF}
;
const
limit = 110*1000 *1000;
var
sol : array [0..limit] of byte;
primes : array of Uint32;
gblCount: Uint64;
procedure SievePrimes(lmt:Uint32);
var
sieve : array of byte;
p,i,delta : NativeInt;
Begin
setlength(sieve,lmt DIV 2);
//estimate count of prime
i := trunc(lmt/(ln(lmt)-1.1));
setlength(primes,i);
p := 1;
repeat
delta := 2*p+1;
// ((2*p+1)^2 ) -1)/ 2 = ((4*p*p+4*p+1) -1)/2 = 2*p*(p+1)
i := 2*p*(p+1);
if i>High(sieve) then
BREAK;
while i <= High(sieve) do
begin
sieve[i] := 1;
i += delta;
end;
repeat
inc(p);
until sieve[p] = 0;
until false;
primes[0] := 2;
i := 1;
For p := 1 to High(sieve) do
if sieve[p] = 0 then
begin
primes[i] := 2*p+1;
inc(i);
end;
setlength(primes,i);
end;
procedure Get_a7;
var
q3,n : UInt64;
i : nativeInt;
begin
sol[1] := 1;
gblCount +=1;
For i := 0 to High(primes) do
begin
q3 := primes[i];
n := sqr(sqr(sqr(q3))) DIV q3;
if n > limit then
break;
sol[n] := 1;
gblCount +=1;
end;
end;
procedure Get_a3_b;
var
i,j,q3,n : nativeInt;
begin
For i := 0 to High(primes) do
begin
q3 := primes[i];
q3 := q3*q3*q3;
if q3 > limit then
BREAK;
For j := 0 to High(primes) do
begin
if j = i then
continue;
n := Primes[j]*q3;
if n > limit then
BREAK;
sol[n] := 1;
gblCount +=1;
end;
end;
end;
procedure Get_a_b_c;
var
i,j,k,q1,q2,n : nativeInt;
begin
For i := 0 to High(primes)-2 do
begin
q1 := primes[i];
For j := i+1 to High(primes)-1 do
Begin
q2:= q1*Primes[j];
if q2 > limit then
BREAK;
For k := j+1 to High(primes) do
Begin
n:= q2*Primes[k];
if n > limit then
BREAK;
sol[n] := 1;
gblCount +=1;
end;
end;
end;
end;
var
i,cnt,lmt : Int32;
begin
SievePrimes(limit DIV 6);// 2*3*c * (c> 3 prime)
gblCount := 0;
Get_a7;
Get_a3_b;
Get_a_b_c;
Writeln('First 50 numbers which are the cube roots of the products of their proper divisors:');
cnt := 0;
i := 1;
while cnt < 50 do
begin
if sol[i] <> 0 then
begin
write(i:5);
cnt +=1;
if cnt mod 10 = 0 then writeln;
end;
inc(i);
end;
dec(i);
lmt := 500;
repeat
while cnt < lmt do
begin
inc(i);
if sol[i] <> 0 then
cnt +=1;
if i > limit then
break;
end;
if i > limit then
break;
writeln(lmt:8,'.th:',i:12);
lmt *= 10;
until lmt> limit;
writeln('Total found: ', gblCount, ' til ',limit);
end.
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