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A330807
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a(1) = 2; a(n+1) = Sum_{k=1..n} {S(a(k)): S(a(k)) = S(a(n))}, where S is sopfr (A001414).
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2
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2, 2, 4, 4, 8, 6, 5, 10, 7, 14, 9, 12, 21, 10, 28, 11, 22, 13, 26, 15, 8, 18, 16, 24, 18, 32, 20, 27, 36, 30, 40, 33, 14, 45, 44, 30, 50, 12, 35, 24, 54, 55, 16, 40, 66, 32, 60, 36, 70, 28, 77, 18, 48, 88, 17, 34, 19, 38, 21, 80, 39, 48, 99, 51, 20, 63, 52, 68, 42, 48, 110, 36, 90, 65, 54, 121, 22, 78, 72, 60, 72, 84, 42, 96
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OFFSET
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1,1
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COMMENTS
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a(n+1) is k(n)*sopfr(a(n)), where k(n) is the number of times (up to and including a(n)) that a term having the same sopfr as a(n) has occurred in the sequence so far.
The smallest possible initial term is a(1)=2, since 2 is the smallest number having a prime divisor. Not every integer appears; for example, 3 cannot occur unless chosen as a(1) (in which case 2 cannot appear).
The primes do not appear in their natural order (e.g., 31 precedes 29). If the lesser of twin primes p is a(k) and the greater twin has not already occurred, then a(k+2) is the greater twin. Whereas if a composite number appears, it may appear more than once, a prime > 2, if it appears, can appear once only. Open question: Do all primes (except for 3) eventually appear?
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LINKS
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EXAMPLE
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a(2)=2 because it is the sopfr of a(1) and there are no prior terms which could contribute to the sum.
a(3)=2+2=4 because 2 has occurred twice already as sopf of prior terms.
a(7)=5 because a(6)=6 and has sopfr 5 which has not been seen before.
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MATHEMATICA
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s[1] = 0; s[n_] := Plus @@ Times @@@ FactorInteger[n]; a[1] = 2; a[n_] := a[n] = (s1 = s[a[n - 1]])*(1 + Sum[Boole[s[a[k]] == s1], {k, 1, n - 2}]); Array[a, 100] (* Amiram Eldar, Jan 01 2020 *)
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PROG
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(Magma) sopfr:=func<n|&+[j[1]*j[2]: j in Factorization(n)]>; a:=[2]; for n in [2..90] do Append(~a, &+[sopfr(a[k]):k in [1..n-1]|sopfr(a[k]) eq sopfr(a[n-1])]); end for; a; // Marius A. Burtea, Jan 01 2020
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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