How to resolve the algorithm Count in factors step by step in the PARI/GP programming language

Published on 12 May 2024 09:40 PM

How to resolve the algorithm Count in factors step by step in the PARI/GP programming language

Table of Contents

Problem Statement

Write a program which counts up from   1,   displaying each number as the multiplication of its prime factors. For the purpose of this task,   1   (unity)   may be shown as itself.

2   is prime,   so it would be shown as itself.       6   is not prime;   it would be shown as

2 × 3

{\displaystyle 2\times 3}

. 2144   is not prime;   it would be shown as

2 × 2 × 2 × 2 × 2 × 67

{\displaystyle 2\times 2\times 2\times 2\times 2\times 67}

.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Count in factors step by step in the PARI/GP programming language

Source code in the pari/gp programming language

fnice(n)={
	my(f,s="",s1);
	if (n < 2, return(n));
	f = factor(n);
	s = Str(s, f[1,1]);
	if (f[1, 2] != 1, s=Str(s, "^", f[1,2]));
	for(i=2,#f[,1], s1 = Str(" * ", f[i, 1]); if (f[i, 2] != 1, s1 = Str(s1, "^", f[i, 2])); s = Str(s, s1)); 
    s
};

n=0;while(n++<21, printf("%2s: %s\n",n,fnice(n)))

  

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