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a(1)=1, a(2)=2. Thereafter, if there are prime divisors p | a(n-1) that are coprime to a(n-2), a(n) is the least novel multiple of the product of these primes. Otherwise a(n) is the least novel multiple of the squarefree kernel of a(n-1). See comments.
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1, 2, 4, 6, 3, 9, 12, 8, 10, 5, 15, 18, 14, 7, 21, 24, 16, 20, 25, 30, 36, 42, 28, 56, 70, 35, 105, 27, 33, 11, 22, 26, 13, 39, 45, 40, 32, 34, 17, 51, 48, 38, 19, 57, 54, 44, 55, 50, 46, 23, 69, 60, 80, 90, 63, 49, 77, 66, 72, 78, 52, 104, 130, 65, 195, 75, 120
COMMENTS
Let k be the greatest common squarefree divisor of a(n-2) and a(n-1) and let s = A007947(a(n-1)). If k = 1, then a(n) = m_s*s, else a(n) = m_k*k, where m_i is the smallest multiple of i such that m*i does not appear in a(1..n-1).
LINKS
Michael De Vlieger, Log-log scatterplot of a(n), n = 1..2^14, showing records in red, local minima in blue, highlighting primes in green and other prime powers in gold.
EXAMPLE
a(1) = 1, a(2) = 2 and 2 divides 2 but does not divide 1. Since 2 is the only prime divisor of 2, a(3) = 4, the smallest unused multiple of 2.
Since every prime divisor of a(3)=4 also divides a(2) = 2, a(4) = 6, the least novel multiple of the squarefree kernel of 4.
a(19,20) = (25,30); 2|30 and 3|30 but 2 and 3 do not divide 25. The smallest multiple of 2*3 = 6 not already in the sequence is 36. Therefore a(21) = 36.
MATHEMATICA
nn = 67; c[_] = False; q[_] = 1; Array[Set[{a[#], c[#]}, {#, True}] &, 2]; q[2] = 2; Do[m = FactorInteger[a[n - 1]][[All, 1]]; If[Length@ # == 0, While[Set[k, #* q[#]]; c[k], q[#]++] &[Times @@ m], While[Set[k, #* q[#]]; c[k], q[#]++] &[Times @@ #]] &@ Select[m, CoprimeQ[#, a[n - 2]] &]; Set[{a[n], c[k]}, {k, True}], {n, 3, nn}]; Array[a, nn]
a(1)=1,a(2)=2,a(3)=3. Thereafter, if there are prime divisors p of a(n-2) which do not divide a(n-1), a(n) is the least novel multiple of any such p. Otherwise a(n) is the least novel multiple of the squarefree kernel of a(n-2).
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1, 2, 3, 4, 6, 8, 9, 10, 12, 5, 14, 15, 7, 18, 21, 16, 24, 20, 27, 22, 30, 11, 25, 33, 35, 36, 28, 39, 26, 42, 13, 32, 52, 34, 65, 17, 40, 51, 38, 45, 19, 48, 57, 44, 54, 55, 46, 50, 23, 56, 69, 49, 60, 63, 58, 66, 29, 62, 87, 31, 72, 93, 64, 75, 68, 70, 85, 74
COMMENTS
In other words, if a(n-2) has k prime divisors p_j; 1 <= j <= k which do not divide a(n-1), where 1 <= k <= omega(a(n-2)), and if m_j*p_j is the least multiple of p_j which is not already a term, then a(n) = Min{m_j*p_j; 1 <= j <= k}. Otherwise, every prime divisor of a(n-2) also divides a(n-1), in which case a(n) is the least multiple of the squarefree kernel of a(n-2) which is not already a term.
Unlike in A064413, a prime p occurrence here is not directly flanked by multiples of p, but by numbers x, y sharing divisors other than p. The terms preceding and following x and y are divisible by p. Typically we observe m*p, x, p, y, r*p, where gcd(x, y) > 1, for multiples m,r of p which do not always follow the (2,3) pattern observed in A064413 and elsewhere. The case of p = 13 is remarkable for being preceded and followed by two multiples of itself (the first five multiples of 13 occur within the span of eight consecutive terms). Conjecture 1: A permutation of the positive integers with primes in natural order.
Conjecture 2: The primes are the slowest numbers to appear (see also A352187).
LINKS
Michael De Vlieger, Log-log scatterplot of a(n), n = 1..2^12 labeling records in red, local minima in blue, highlighting primes in green and other prime powers in gold.
EXAMPLE
With a(2)=2, and a(3)=3, a(4) must be 4, the least unused multiple of 2.
Likewise, with a(3),a(4) = 3,4 a(5) must be the 6, the least unused multiple of 3.
Since every divisor of 4 also divides 6 a(6) = 8, the least unused multiple of 2, (squarefree kernel of 4).
Since a(8),a(9) = 10,12 and 5 is the only prime dividing 10 but not 12, it follows that a(10) = 5.
MATHEMATICA
nn = 68; c[_] = False; q[_] = 1; Array[Set[{a[#], c[#]}, {#, True}] &, 2]; q[2] = 2; Do[m = FactorInteger[a[n - 1]][[All, 1]]; f = Select[m, CoprimeQ[#, a[n - 2]] &]; If[Length[f] == 0, While[Set[k, # * q[#]]; c[k], q[#]++] &[Times @@ m], Set[{k, q[#1]}, {#2, #2/#1}] & @@ First@ MaximalBy[Map[{#, Set[g, q[#]]; While[c[g #], g++]; # g} &, f], Last] ]; Set[{a[n], c[k]}, {k, True}], {n, 3, nn}]; Array[a, nn] (* Michael De Vlieger, Oct 23 2022 *)
a(1)=1, a(2)=2. Thereafter, if there are prime divisors p of a(n-1) which do not divide a(n-2), a(n) is the greatest least multiple of any such p which has not already occurred. Otherwise a(n) is the least novel multiple of the squarefree kernel of a(n-1). (see comments).
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1, 2, 4, 6, 3, 9, 12, 8, 10, 5, 15, 18, 14, 7, 21, 24, 16, 20, 25, 30, 27, 33, 11, 22, 26, 13, 39, 36, 28, 35, 40, 32, 34, 17, 51, 42, 49, 56, 38, 19, 57, 45, 50, 44, 55, 60, 48, 54, 66, 77, 63, 69, 23, 46, 52, 65, 70, 84, 72, 78, 91, 98, 58, 29, 87, 75, 80, 62
COMMENTS
In other words, if a(n-1) has k prime divisors p_j; 1 <= j <= k which do not divide a(n-2), where 1 <= k <= omega(a(n-1)), and if m_j*p_j is the least multiple of p_j which is not already a term, then a(n) = Max{m_j*p_j; 1 <= j <= k}. Otherwise every prime divisor of a(n-1) also divides a(n-2), in which case a(n) is the least multiple of the squarefree kernel of a(n-1) which has not already occurred.
Companion to A357963, (which uses "Min" rather than "Max" in selection of a(n)). The first departure from A357963 occurs at a(21) because a(19),a(20) = 25,30, and 30 has two divisors (2,3) which do not divide 25. Of these the least multiples not occurring already are 22, and 27 respectively. At this point in A357963 22 is the chosen term, whereas here a(21) = 27. This has the effect of temporarily reversing (for the next prime = 11) the normal way primes seem to arrive in this sequence (2p,p,3p, as in A064413). Thus we see 30,27,33,11,22 (3p,p,2p). This may happen elsewhere in the data, consequent to choice of "Max" over "Min". Conjecture: Permutation of the positive integers; primes being in natural order, and the slowest numbers to appear (as in A352187).
LINKS
Michael De Vlieger, Scatterplot of a(n), n = 1..120, showing primes p in red, 2p in blue, and 3p in green, for comparison to A064413. Michael De Vlieger, Log-log scatterplot of a(n), n = 1..2^14, labeling records in red and local minima in blue, highlighting primes in green, prime squares in light blue, and other prime powers in gold. Demonstrates delayed 17^2 and 19^2, early records such as a(268) = 455.
EXAMPLE
a(1),a(2)=1,2 and 2 is the only prime dividing 2 which does not divide 1, therefore a(3)=4, the least multiple of 2 which has not occurred already. In this case (as in all terms up to and including a(20), "Max" gives the same term as "Min".
MATHEMATICA
nn = 68; c[_] = False; q[_] = 1; Array[Set[{a[#], c[#]}, {#, True}] &, 2]; q[2] = 2; Do[m = FactorInteger[a[n - 1]][[All, 1]]; f = Select[m, CoprimeQ[#, a[n - 2]] &]; If[Length[f] == 0, While[Set[k, # q[#]]; c[k], q[#]++] &[Times @@ m], Set[{k, q[#1]}, {#2, #2/#1}] & @@ First@ MaximalBy[Map[{#, Set[g, q[#]]; While[c[g #], g++]; # g} &, f], Last] ]; Set[{a[n], c[k]}, {k, True}], {n, 3, nn}]; Array[a, nn] (* Michael De Vlieger, Oct 23 2022 *)
Let G(n) = gcd(a(n-2),a(n-1)), a(1)=1, a(2)=2, a(3)=3. Thereafter if G(n) = 1, a(n) is the least novel m sharing a divisor with both a(n-2) and a(n-1). If G(n) > 1 and every prime divisor of a(n-1) also divides a(n-2), a(n) is the least m prime to both a(n-1) and a(n-2). Otherwise a(n) is the least novel multiple of any prime divisor of a(n-1) which does not divide a(n-2).
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1, 2, 3, 6, 4, 5, 10, 8, 7, 14, 12, 9, 11, 33, 15, 20, 16, 13, 26, 18, 21, 28, 22, 44, 17, 34, 24, 27, 19, 57, 30, 25, 23, 115, 35, 42, 32, 29, 58, 36, 39, 52, 38, 76, 31, 62, 40, 45, 48, 46, 69, 51, 68, 50, 55, 66, 54, 37, 74, 56, 49, 41, 287, 63, 60, 64, 43
COMMENTS
Conjectured to be a permutation of the positive integers with the primes in natural order, and primes are the slowest numbers to appear (as in A352187).
LINKS
Michael De Vlieger, Log-log scatterplot of a(n), n = 1..2^14, showing records in red and local minima in blue, highlighting primes in green and other prime powers in gold.
EXAMPLE
a(4) = 6, the least novel number sharing a factor with both 2 and 3.
a(5) = 4, the least novel multiple of 2, which divides a(4) but does not divide a(3).
Since every prime dividing a(5)=4 also divides a(4)=6, a(6)=5 the least novel term prime to 3 and 6.
MATHEMATICA
nn = 67; c[_] = False; q[_] = 1; u = 4; Do[(Set[{a[n], c[n]}, {n, True}]; q[n]++), {n, u - 1}]; Do[m = FactorInteger[a[n - 1]][[All, 1]]; f = Select[m, CoprimeQ[#, a[n - 2]] &]; Which[Length[f] == PrimeNu[a[n - 1]], Set[{k, q[#1]}, {#2, #2/#1}] & @@ First@ MinimalBy[Map[{#, Set[g, q[#]]; While[c[g #], g++]; # g} &, Flatten@ Outer[Times, m, FactorInteger[a[n - 2]][[All, 1]] ] ], Last], Length[f] == 0, k = u; While[Nand[! c[k], CoprimeQ[a[n - 2], k], CoprimeQ[a[n - 1], k]], k++]; If[k == u, While[c[u], u++]], True, 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}], {n, 4, nn}]; Array[a, nn] (* Michael De Vlieger, Oct 25 2022 *)
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, 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
COMMENTS
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
LINKS
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.
EXAMPLE
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.
MATHEMATICA
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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