I implemented a solution to the problem below in Mathematica, but it takes a very long time (hours) to compute f of kis or the set B for large numbers.

Somebody suggested that implementing this in C++ resulted in a solution in less than 10 minutes. Would C++ be a good language to learn to solve these problems, or can my Mathematica code be improved to fix the performance issues?

I don't know anything about C or C++ and it should be difficult to start to learn this languages. I prefer to improve or write new code in mathematica.

Problem Description

Let $f$ be an arithmetic function and A={k1,k2,...,kn} are integers in increasing order.

Now I want to start with k1 and compare f(ki) with f(k1). If f(ki)>f(k1), put ki as k1.

Now start with ki, and compare f(kj) with f(ki), for j>i. If f(kj)>f(ki), put kj as ki, and repeat this procedure.

At the end we will have a sub sequence B={L1,...,Lm} of A by this property: f(L(i+1))>f(L(i)), for any 1<=i<=m-1

For example, let f is the divisor function of integers.

Here I put some part of my code and this is just a sample and the question in my program could be more larger than these:

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f[n_] := DivisorSigma[0, n];

g[n_] := Product[Prime[i], {i, 1, PrimePi[n]}];


k1 = g[67757] g[353] g[59] g[19] g[11] g[7] g[5]^2 6^3 2^7;

k2 = g[67757] g[353] g[59] g[19] g[11] g[7] g[5] 6^5 2^7;

k3 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5] 6^4 2^7;

k4 = g[67759] g[349] g[53] g[19] g[11] g[7] g[5] 6^5 2^6;

k5 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5] 6^4 2^8;

k6 = g[67759] g[349] g[53] g[19] g[11] g[7] g[5]^2 6^3 2^7;

k7 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5] 6^5 2^6;

k8 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5] 6^4 2^9;

k9 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5]^2 6^3 2^7;

k10 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5] 6^5 2^7;

k11 = g[67759] g[349] g[53] g[19] g[11] g[7] g[5]^2 6^4 2^6;

k12 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5]^2 6^3 2^8;

k13 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5]^2 6^4 2^6;

k14 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5]^2 6^3 2^9;

k15 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5]^2 6^4 2^7;

k16 = g[67757] g[359] g[53] g[23] g[11] g[7] g[5] 6^4 2^8;

k17 = g[67757] g[359] g[59] g[19] g[11] g[7] g[5] 6^4 2^7;

k18 = g[67757] g[359] g[53] g[23] g[11] g[7] g[5] 6^4 2^9;

k19 = g[67759] g[353] g[53] g[19] g[11] g[7] g[5] 6^4 2^6;

k20 = g[67763] g[347] g[53] g[19] g[11] g[7] g[5] 6^4 2^7;

k = Table[k1, k2, k3, k4, k5, k6, k7, k8, k9, k10, k11, k12, k13, k14, k15, k16, k17, k18, k19, k20];

i = 1;

count = 0;

For[j = i, j <= 20, j++, 
  If[f[k[[j]]] - f[k[[i]]] > 0, i = j; Print["k",i];
   count = count + 1]];

Print["count= ", count]

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