%I #35 Sep 14 2020 11:19:57

%S 1,3,4,9,5,12,6,27,16,15,7,36,8,18,20,81,10,48,11,45,24,21,13,108,25,

%T 24,64,54,14,60,17,243,28,30,30,144,19,33,32,135,22,72,23,63,80,39,26,

%U 324,36,75,40,72,29,192,35,162,44,42,31,180,34,51,96,729

%N Products of terms of A232803.

%C If 2 were not a prime factor, the prime numbers sequence would change. 4,8, and twice odd primes would become "primes". The new "prime numbers" sequence would be 3, 4, 5, 6, 7, 8, 10, 11, 13, 14, 17, 19, 22, 23, 26, 29, ... (A232803). The products of the terms of A232803 would become the new "natural numbers".

%C In order to compute a(n), one must write the prime factorization of n and replace each prime(k) with A232803(k). - _Michel Marcus_, Sep 14 2020

%H Michael De Vlieger, <a href="/A331025/b331025.txt">Table of n, a(n) for n = 1..10000</a>

%H Robert Dougherty-Bliss, <a href="https://rwdb.xyz/files/prime_removal.pdf">The Number 2 Does Not Exist and other p-removed primes</a>

%e In the natural numbers sequence, a(15)=prime(2)*prime(3). If we use the terms of A232803 as prime factors, then prime(2)=4 and prime(3)=5. So, a(15) will be 4*5 = 20.

%t With[{s = Select[Range[37], And[# != 2, Or[Log2[#] == 3, PrimeQ@#, PrimeQ[#/2]]] &]}, Array[Times @@ Map[If[#[[1]] == 1, 1, # /. {p_, e_} :> s[[PrimePi@ p]]^e] &, FactorInteger[#]] &, Prime@ Length@ s]] (* _Michael De Vlieger_, Aug 21 2020 *)

%o (PARI) isp(n) = (isprime(n) && (n%2)) || (n==8) || (!(n%2) && isprime(n/2)); \\ A232803

%o lista(nn) = {my(vall = [1..nn]); my(vp = select(x->isp(x), vall)); for (n=2, nn, my(f=factor(n)); for (k=1, #f~, f[k,1] = vp[primepi(f[k,1])]); vall[n] = factorback(f);); vall;} \\ _Michel Marcus_, Sep 14 2020

%Y Cf. A232803.

%K nonn

%O 1,2

%A _Ali Sada_, Jan 07 2020