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A331025
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Products of terms of A232803.
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2
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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
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OFFSET
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1,2
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COMMENTS
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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
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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