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A358267
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a(1) = 1, a(2) = 2. Thereafter:(i). If no prime divisor of a(n-1) divides a(n-2), a(n) is the least novel multiple of the squarefree kernel of a(n-1). (ii). If some (but not all) prime divisors of a(n-1) do not divide a(n-2), a(n) is the least of the least novel multiples of all such primes. (iii). If every prime divisor of a(n-1) also divides a(n-2), a(n) = u, the least unused number.
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1
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1, 2, 4, 3, 6, 8, 5, 10, 12, 9, 7, 14, 16, 11, 22, 18, 15, 20, 24, 21, 28, 26, 13, 17, 34, 30, 25, 19, 38, 32, 23, 46, 36, 27, 29, 58, 40, 35, 42, 33, 44, 48, 39, 52, 50, 45, 51, 68, 54, 57, 76, 56, 49, 31, 62, 60, 55, 66, 63, 70, 64, 37, 74, 72, 69, 92, 78, 65
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OFFSET
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1,2
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COMMENTS
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Let a(n-2) = i, a(n-1) = j. The sequence is generated from divisor relationships j->i, ranging from coprime: gcd(i,j) =1, to partial: 1 < gcd(i,j) < j, to total: gcd(i,j) = j, using conditions described in the definition.
A prime cannot occur consequent to condition (i). a(n) = prime p either because p|a(n-1) but not a(n-2); see (ii), or because every prime divisor of a(n-1) also divides a(n-2), as when for example a(n-1) is a prime power q^k and q|a(n-2), which forces a(n) = u prime, see (iii).
If a(n) = u = p from condition (iii), a(n+1) = 2*p. If p|a(n-1)-> a(n) = p we see m*p->p->u (and u may of course be prime, as in ...,13,17,...). 13 is the first prime to appear consequent to condition (ii), see Example. Consecutive primes appear often: (13,17); (53,59); (61,67); ... Sequence is conjectured to be a permutation of the positive integers with primes appearing slowest, and in natural order.
Local minima consist of 1 and the primes p, while 4p dominates the maxima as n increases. - Michael De Vlieger, Nov 06 2022
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LINKS
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Michael De Vlieger, Log-log scatterplot of a(n) n = 1..2^14, showing maxima in red, local minima in blue, highlighting primes p in green and other prime powers in gold.
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EXAMPLE
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a(1) = 1, a(2) = 2 and since 2|a(2) but not a(1), and no other primes are involved, a(3) = 4, the least novel multiple of 2, the squarefree kernel of 2 (by (i)).
Every prime divisor of a(3) = 4 also divides a(2) = 2, thus a(4) = 3, the least unused number (by (iii)).
a(23) = 13 because 13|a(22) = 26, but does not divide a(21) = 28 (by (ii)). Then since every prime divisor of a(23) also divides a(22), a(24) = 17, the least unused term (by (iii)). This is the first occasion of consecutive primes.
a(25) = 34, a(26) = 30 and there are two primes (3,5) which divide 30 but not 34. At this point the least novel multiples of 3 and 5 are 27 and 25 respectively, so a(27) = 25 (by (ii)). This is the first departure from A280864/A280866, which both have a(27) = 45.
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MATHEMATICA
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nn = 120; c[_] = False; q[_] = 1; Array[Set[{a[#], c[#]}, {#, True}] &, 3]; q[2] = 2; u = 3; Do[m = FactorInteger[a[n - 1]][[All, 1]]; f = Select[m, CoprimeQ[#, a[n - 2]] &]; If[AllTrue[m, Mod[a[n - 2], #] == 0 &], k = u, Set[{k, q[#1]}, {#2, #2/#1}] & @@ First@ MinimalBy[Map[{#, Set[g, q[#]]; While[c[g #], g++]; # g} &, f], Last] ]; Set[{a[n], c[k]}, {k, True}]; If[k == u, While[c[u], u++]], {n, 3, nn}]; Array[a, nn] (* Michael De Vlieger, Nov 06 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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