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A353125
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a(1)=2. If a(n) is a novel term, a(n+1) = sopfr(a(n)), else if there are k occurrences of a(j)=a(n), (1<=j<=n), a(n+1)=k*a(n).
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1
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2, 2, 4, 4, 8, 6, 5, 5, 10, 7, 7, 14, 9, 6, 12, 7, 21, 10, 20, 9, 18, 8, 16, 8, 24, 9, 27, 9, 36, 10, 30, 10, 40, 11, 11, 22, 13, 13, 26, 15, 8, 32, 10, 50, 12, 24, 48, 11, 33, 14, 28, 11, 44, 15, 30, 60, 12, 36, 72, 12, 48, 96, 13, 39, 16, 32, 64, 12, 60, 120
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OFFSET
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1,1
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COMMENTS
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1 and 3 cannot be terms. Let conditions 1, 2 refer to the effects of a novel and repeat term respectively, and let M(m) be the multiplicity of m in the sequence. M(m) >= 2 for all m in N \ {1,3}. All first occurrences of a prime p > 2 follow a novel term in A046363 (by condition 1).
If the first occurrence of composite m arises from condition 1, then so does the second. Proof: Suppose not, then the second m must be a multiple w*t of a prior term t (condition 2); w >= 2. The term following the second m must be (w+1)*t, and this must equal 2*m (condition 2). Thus 2*w*t = (w+1)*t, so then w=1. But w >= 2; contradiction. Corollary: Once composite m has occurred by condition 1, then all subsequent occurrences of m occur the same way.
Most composite terms appear first by condition 2 (e.g., 12 as 2 copies of 6), and then subsequently by condition 1. Thus for all k in the sequence M(k) >= A000607(k), with equality when k is prime > 2, or certain composite numbers.
Conjecture: The 10 composite numbers 6,9,15,25,35,49,77,121,143,169 behave as primes in this sequence, namely for any such m, M(m) = A000607(m). For all other composite m, M(m) > A000607(m), i.e., at least one (up front) copy by condition 2.
The first occurrences of primes appear in natural order initially, but this is not sustained (e.g., 61 appears before 59).
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LINKS
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EXAMPLE
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a(1)=2, a novel term, so a(2)=sopfr(2)=2. 2 occurs twice and is the only prime p whose multiplicity is not A000607(p), simply because it is the seed term.
Since 2 has now appeared twice, a(3)=2*2=4, a novel term, so a(4)=sopfr(4)=4.
a(25)=24 (3 occurrences of 8), a(46)=24 (2 occurrences of 12). Subsequently all occurrences of 24 are from condition 1. Therefore M(24) = 2 + A000607(24) = 48.
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MATHEMATICA
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nn = 1000; c[_] = 0; a[1] = 2; c[2]++; Do[If[c[#] == 1, Set[k, Total@ Flatten[ConstantArray[#1, #2] & @@@ FactorInteger[#]]], Set[k, c[#] #]] &@ a[i - 1]; a[i] = k; c[k]++, {i, Length[s] + 1, nn}], i]; Array[a, nn] (* Michael De Vlieger, Apr 24 2022 *)
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PROG
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(PARI) sopfr(n) = (n=factor(n))[, 1]~*n[, 2]; \\ A001414
lista(nn) = {my(v=vector(nn), k); v[1] = 2; for (n=2, nn, if ((k=#select(x->(x==v[n-1]), Vec(v, n-1))) == 1, v[n] = sopfr(v[n-1]), v[n] = k*v[n-1]); ); v; } \\ Michel Marcus, May 16 2022
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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