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A331025
Products of terms of A232803.
2
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
1,2
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
EXAMPLE
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.
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 *)
PROG
(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
CROSSREFS
Cf. A232803.
Sequence in context: A247567 A271529 A070154 * A330403 A183211 A256469
KEYWORD
nonn
AUTHOR
Ali Sada, Jan 07 2020
STATUS
approved