A356867 For n >= 1, write n = 3^m + k, where m >= 0 is the greatest power of 3 <= n, and k is in the range 0 <= k < 3^(m+1) - 3^m, then for n such that k=0, a(n)=n, and for n such that k > 0, a(n) is the smallest prime multiple p*a(k), p != 3, that is not already a term.
1, 2, 3, 5, 4, 6, 10, 8, 9, 7, 14, 15, 25, 20, 12, 50, 16, 18, 35, 28, 30, 125, 40, 24, 100, 32, 27, 11, 22, 21, 55, 44, 42, 70, 56, 45, 49, 98, 75, 175, 140, 60, 250, 80, 36, 245, 196, 150, 625, 200, 48, 500, 64, 54, 77, 110, 105, 275, 88, 84, 350, 112, 90, 343
Offset: 1
Examples
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).
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.
- Index entries for sequences that are permutations of the natural numbers
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))].
Programs
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Mathematica
Block[{a, c, i, j, k, m, t, nn}, 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 *) -
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
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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
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)
Extensions
More terms from Michael De Vlieger, Sep 01 2022
Comments