login
A369825
a(1,2) = 1,2; thereafter let i = a(n-2) and r(n) = a(1)*a(2)*...*a(n-1): a(n) is the smallest novel multiple m of w = rad(r(n))/rad(i), where rad is A007947.
2
1, 2, 4, 3, 6, 8, 5, 15, 12, 10, 20, 9, 18, 30, 25, 7, 42, 60, 35, 14, 24, 45, 70, 28, 21, 75, 40, 56, 63, 90, 50, 49, 84, 120, 55, 77, 126, 150, 110, 154, 105, 135, 22, 308, 210, 165, 11, 98, 420, 330, 33, 91, 910, 660, 66, 182, 455, 495, 132, 364, 1365, 825
OFFSET
1,2
COMMENTS
Rad(r(n)) is always a primorial (A002110), and there are two distinct ways a prime p > 2 can enter the sequence:
(i). Directly: rad(i) = rad(r(n)) implies a(n) is the least unused term, conjectured to be the smallest prime p not already in the sequence. In this case no prior term is divisible by p. This happens for 3, 5, 7. It is not known if this ever happens again (thought to be unlikely).
(ii). Indirectly: rad(i) and rad(r(n)) are consecutive primorials whose quotient is prime p = gpd(rad(r(n)). This implies that p has already entered the sequence as divisor of a previous (composite) term, since then p|rad(r(n)), and r(n) is the product of all prior terms, so p must be a factor of at least one of them.
In the first case m is the least novel multiple of 1, and rad(r(n)) increments to the next primorial at the point of entry a(n) = p. In the second case rad(r(n)) increments (to the next primorial; gpf = p) at the earliest term a(t); t < n where p | a(t). Example: a(35) = 55, the earliest term divisible by 11. We see six more composite terms divisible by 11, before finally a(47) = 11.
Prime terms occur in order as shown in Example, but only as far as 61. Primes do not appear in order for n > 667722.
For n < 667722, with r(A) = P(k+1) following prime(k+1) | a(A) for A < n, we admit a(n-2) = P(k) and a(n) = prime(k+1) where n < B such that r(B) = P(k+2) following prime(k+2) | a(B). For sufficiently small n, primes a(n) = prime(k+1) follow primorials a(n-2) = P(k) in the sequence.
Primorials and primes decouple such that a(676472) = P(18), but a(676474) = 4757 = prime(19)*prime(20). This is the result of r(n) increasing twice (at n = 253724 and 667722), offering a greater degree of freedom for a(676474) than for terms that follow previous instances of primorials in the sequence. We also have a(9061722) = 73, following a(9061720) = P(23)/73. Primorials a(n) = P(k) do not appear in order for n > 681764.
Powerful numbers in the sequence are sparse, since they require w to appear m times such that m*w is powerful. Only 10 powerful numbers appear for n <= 2^24. Even after 2^24 terms, 36 is still missing. Therefore the sequence is predominantly of weak numbers.
It is not known if this sequence could be a permutation of A000027.
LINKS
Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^20.
Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^14, showing primes in red, composite prime powers in gold, squarefree composites in green, and numbers neither squarefree nor prime powers in blue and purple, accentuating primorials in large green, and powerful numbers that are not prime powers in purple.
Michael De Vlieger, Plot p^k | a(n) at (x,y) = (n, pi(p)) for n = 1..2^11, 12X vertical exaggeration, with a color function showing k = 1 in black, k = 2 in red, ... maximum value of k in reference range in magenta. The color bar under the plot indicates numbers as immediately above, red = prime, etc.
Michael De Vlieger, Plot a(n) at (x,y) = (x mod 1024, -floor(x/1024)) for n = 1..2^20, showing primes in red, composite prime powers in gold, squarefree composites in green, and numbers neither squarefree nor prime powers in blue and purple, the latter color indicating powerful numbers that are not prime powers. Indicates large-scale pattern of prime power decomposition.
EXAMPLE
a(3) = 4 because rad(1*2)/rad(1) = 2 and 4 is the least novel multiple of 2.
a(4) = 3 because rad(1*2*4)/rad(2) = 1, and 3 is the least novel multiple of 1.
a(5) = 6 since rad(1*2*4*3)/rad(4) = 3 and 6 is the least novel multiple of 3.
a(6) = 8, the least novel multiple of 2, since rad(1*2*4*3*6)/rad(3) = 2.
a(7) = 5 since rad(1*2*4*3*6*8)/rad(6) = 1 and 5 is the least novel multiple of 1.
a(8) = 15 since rad(m)/rad(i) = 30/2 = 15, which has not appeared previously.
Table of n, a(n) for primes (in 3rd column, i means rad(i) = rad(a(r)); ii means rad(i) properly divides rad(r(n)), and both rad(i) and rad(r(n)) are (for the data shown here) consecutive primorials:
2 2 (given)
4 3 i
7 5 i
16 7 i
47 11 ii
96 13 ii
193 17 ii
476 19 ii
697 23 ii
1168 29 ii
1349 31 ii
4613 37 ii
8898 41 ii
19728 43 ii
40553 47 ii
49054 53 ii
63802 59 ii
240925 61 ii
681766 71 ii
2191325 79 ii
9061722 73 ii
13178788 89 ii
26120340 97 ii
MATHEMATICA
nn = 1000; c[_] := False; m[_] := 1;
f[x_] := f[x] = Times @@ FactorInteger[x][[All, 1]];
Array[Set[{a[#], c[#], m[#]}, {#, True, 2}] &, 2]; i = 1; j = p = 2;
Do[(While[c[Set[k, # m[#]]], m[#]++]) &[p/i];
Set[{a[n], c[k], i, j, p}, {k, True, f[j], k, f[p*k]}], {n, 3, nn}];
Array[a, nn] (* Michael De Vlieger, Feb 03 2024 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved