A331025 Products of terms of A232803.

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, 24, 64, 54, 14, 60, 17, 243, 28, 30, 30, 144, 19, 33, 32, 135, 22, 72, 23, 63, 80, 39, 26, 324, 36, 75, 40, 72, 29, 192, 35, 162, 44, 42, 31, 180, 34, 51, 96, 729

Offset: 1

Author

Ali Sada, Jan 07 2020

Keywords

• nonn

Comments

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".

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

Examples

			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.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000
• Robert Dougherty-Bliss, The Number 2 Does Not Exist and other p-removed primes

Crossrefs

Cf. A232803.

Programs

Mathematica

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 *)

PARI

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