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A353125
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).
2
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
OFFSET
1,1
COMMENTS
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).
LINKS
Michael De Vlieger, Annotated log-log scatterplot of a(n), n = 1..2^12, records in red and likely local minima in blue.
Michael De Vlieger, Log-log scatterplot of a(n), n = 1..2^16.
EXAMPLE
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.
MATHEMATICA
nn = 1000; c[_] = 0; j = a[1] = 2; c[2]++; Do[If[c[j] == 1, Set[k, Total@ Flatten[ConstantArray[#1, #2] & @@@ FactorInteger[j]]], Set[k, c[j] j]]; j = a[i] = k; c[k]++, {i, 2, nn}]; Array[a, nn] (* Michael De Vlieger, Apr 24 2022 *)
PROG
(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
CROSSREFS
Cf. A001414 (sopfr), A046363, A000607, A000040.
Sequence in context: A300123 A175359 A336125 * A330807 A008330 A191234
KEYWORD
nonn
AUTHOR
EXTENSIONS
More terms from Michael De Vlieger, Apr 24 2022
STATUS
approved