OFFSET
1,2
COMMENTS
Any prime p may be used to generate a sequence D(p) of this kind. The present sequence is D(3), and D(2) is the Doudna sequence, A005940.
Conjectured to be a permutation of the positive integers in which the primes appear in order.
From Antti Karttunen, Sep 16 2023: (Start)
The conjecture is true: Sequence is a permutation of natural numbers. By definition it is injective, and the surjectivity is guaranteed by the fact that there are infinitely many such n > k encountered by the greedy algorithm that a(n) will be a multiple of a(k), and "the smallest prime multiple" condition guarantees that all multiples of a(k) will eventually appear. That the primes and A100484 appear in order follows from the formulas a(3^m + 1) = prime(m+2), and a(3^m + 2) = 2*prime(m+2).
If the base-3 representation of n-1 has the base-3 representation of k-1 as its suffix, then a(n) is a multiple of a(k). For example, A007089(16-1) = 120, and A007089(43-1) = 1120, thus the former is the suffix of the latter, and a(16) = 50 indeed divides a(43) = 250.
(End)
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..19683 (19683 = 3^9)
Michael De Vlieger, Fan style ternary tree showing a(n) for n = 1..3^9, with a heat map color function for level m where 3^m is blue, smaller values are bluer, and larger are yellow-green. The smallest value in level m is shown in purple and largest is shown in red.
FORMULA
a(3^m + 1) = prime(m+2) for m >= 1.
Conjectures from Jianing Song, Nov 23 2022: (Start)
(1) a(3^m+2) = 2*prime(m+2) for m >= 2. - [The conjecture is true because a(2) = 2 and 3^m + 2 < 3^(1+m) + (3^m) + 1 for all m - Antti Karttunen, Sep 16 2023]
(2) For n > m >= 1, a(3^n+3^m+1) = prime(m+2)^2 for n = m+1; prime(n+2)*prime(m+2)^2 for n >= m+2.
(3) For n > m >= 1, a(3^n+3^m+2) = 4*prime(n+2) for n >= 3, m = 1; 2*prime(m+2)^2 for n = m+1, m >= 2; 2*prime(m+2)*prime(m+3) for n = m+2, m >= 2; 2*prime(n+2)*prime(m+2)^2 for n >= m+3, m >= 2. (End)
From Antti Karttunen, Sep 17 2023: (Start)
For n >= 1, a(3^n - 1) = 2^(2n - 1), a(A048473(n)) = 2^(2*(n-1)).
These are conjectures so far:
For n >= 1, a(3^n - 2) = 10^(n-1).
For n >= 2, a(3^n - 3) = A002023(n-2) = 6*4^(n-2).
(End)
EXAMPLE
n=1=3^0+0 so a(1)=1. n=2=3^0+1 so k=1 and a(2)=2. Similarly a(3)=3 and a(9)=9.
n=10=3^2+1, therefore k=1 and a(1)=1 so a(10)=1*7=7 (since 2 and 5 have already occurred).
MATHEMATICA
nn = 64; m = 1; i = 2; p = Prime[i]; c[_] = False; Do[Set[{m, k}, {1, n - p^Floor[Log[p, n]]}]; If[k == 0, Set[{a[n], c[n]}, {n, True}], While[Set[t, Prime[m] a[k]]; Or[m == i, c[t]], m++]; Set[{a[n], c[t]}, {t, True}]], {n, nn}]; Array[a, nn] (* Michael De Vlieger, Sep 01 2022 *)
PROG
(Python)
from sympy import nextprime
from sympy.ntheory import digits
from itertools import count, islice
def b(n): return n - 3**(len(digits(n, 3)) - 2)
def agen():
aset, alst = set(), [None]
for n in count(1):
k = b(n)
if k == 0: an = n
else:
ak, p = alst[k], 2
while p == 3 or p*ak in aset: p = nextprime(p)
an = p*ak
yield an; aset.add(an); alst.append(an)
print(list(islice(agen(), 64))) # Michael S. Branicky, Sep 02 2022
(PARI)
up_to = 19683;
A356867list(up_to) = { my(v=vector(up_to), met=Map(), h=0, ak); for(i=1, #v, if(1==vecsum(digits(i, 3)), v[i] = i; h = i, ak = v[i-h]; forprime(p=2, , if(3!=p && !mapisdefined(met, p*ak), v[i] = p*ak; break))); mapput(met, v[i], i)); (v); };
v356867 = A356867list(up_to);
A356867(n) = v356867[n]; \\ Antti Karttunen, Sep 15 2023
CROSSREFS
Cf. A007089, A007949, A011655, A048473, A100484, A053735, A364958 (fixed points), A365390 (inverse permutation), A365424, A365459, A365462 [= a(n)-n], A365463 [= gcd(a(n),n)], A365464, A365465, A365717 [= A348717(a(1+n))], A365719 [= A046523(a(1+n))], A365721 [= omega(a(1+n))], A365722 [= bigomega(a(1+n))].
KEYWORD
nonn,look
AUTHOR
David James Sycamore, Sep 01 2022
EXTENSIONS
More terms from Michael De Vlieger, Sep 01 2022
STATUS
approved