A304752 Beginning with a(1) = 1, for n > 1, a(n) = the least divisor of a(n-1) not included earlier, otherwise a(n) = the least multiple m*a(n-1) such that m is not a divisor of a(n-1) and m*a(n-1) is not included earlier.

1, 2, 6, 3, 12, 4, 20, 5, 10, 30, 15, 60, 420, 7, 14, 42, 21, 84, 28, 140, 35, 70, 210, 105, 630, 9, 18, 72, 8, 24, 120, 40, 240, 16, 48, 336, 56, 168, 840, 280, 1680, 80, 480, 32, 96, 672, 112, 560, 3360, 160, 960, 64, 192, 1344, 224, 1120, 6720, 320, 1920, 128, 384, 2688, 448, 2240, 13440, 640, 3840, 256, 768, 5376, 896, 4480, 26880, 1280, 7680, 512

Offset: 1

Author

Antti Karttunen, Michael De Vlieger, and Robert G. Wilson v, May 19 2018

Keywords

• nonn

Comments

This is a variant of A113552.

From Michael De Vlieger, May 20 2018: (Start)

In the table below, we note a cycle that subtends for 41 <= n <= 2^14.

Let e = floor(n/8). We write multiple k if the condition is false, or the parity of divisor d if d does not occur in a. We can express a(n) as the product of the smallest four primes as shown below.

n (mod 8) k or d 2 3 5 7

-------------------------------------------

0 5 2^(e-2) 5 7

1 6 2^(e-1) 3 5 7

2 EVEN 2^(e-1) 5

3 6 2^(e-1) 3 5

4 EVEN 2^e

5 3 2^e 3

6 7 2^e 3 7

7 EVEN 2^(e-1) 7

Conjectures:

1. All terms are divisible only by some combination of the smallest 4 primes.

2. Powers 2^e, positive integer e, are at n = {1, 2, 6, 29, 34, 44, 52, 60, 68, ...}; first differences are {1, 4, 23, 5, 10, 8, 8, 8, ...}, and 8 thereafter.

3. For n > 41 such that n (mod 8) = 4, a(n) = 2^((n-4)/8).

4. For n > 26 all terms are even. Odd terms are {1, 3, 5, 15, 7, 21, 35, 105, 9} at indices {1, 4, 8, 11, 14, 17, 21, 24, 26}. (End)

Examples

			After a(27) = 18 = 2 * 3^2, the next term a(28) is neither 2*18 = 2^2 * 3^2, nor 3*18 = 2 * 3^3 as both 2 and divide 18. But 4 does not divide 18, and 4*18 = 72 haven't yet been used in the sequence, thus a(28) = 72.
		

Links

• Antti Karttunen & Michael De Vlieger, Table of n, a(n) for n = 1..16384

Crossrefs

Differs from A113552 for the first time at n=28, where a(28) = 72, while A113552(28) = 90.

Programs

Maple

lim:=60: with(numtheory): membera := proc(val) global a, n: local j: for j from 1 to n-1 do if(a[j]=val)then return true: fi: od: return false: end: a[1]:=1:for n from 2 to lim do d:=sort([divisors(a[n-1])[]]): s:=true: for k from 1 to nops(d) do if(not membera(d[k]))then a[n]:=d[k]:s:=false: break:fi:od: if(s)then for j from 2 do if(not member(j, d) and not membera(j*a[n-1]))then a[n]:=j*a[n-1]:break: fi:od:fi:od: seq(a[n], n=1..lim); # Nathaniel Johnston, May 10 2011, given originally for A113552
# second Maple program:
b:= proc(n) is(n=1) end:
a:= proc(n) option remember; local j, l, i, m;
      j:= a(n-1): l:= sort([numtheory[divisors](j)[]]);
      for i to nops(l) do if not b(l[i])
        then b(l[i]):=true; return l[i]
      fi od;
      for m while m in l or b(m*j) do od;
      b(m*j):=true; m*j
    end: a(1):=1:
seq(a(n), n=1..100);  # Alois P. Heinz, May 22 2018

Mathematica

f[s_] := Append[s, d = Divisors[ s[[ -1]]]; If[ Complement[d, s] != {}, Complement[d, s][[1]], k = 2; While[ Mod[ s[[ -1]], k] == 0 || MemberQ[s, k*s[[ -1]]], k++ ]; k*s[[ -1]] ]]; Nest[f, {1}, 60] (* Robert G. Wilson v, Aug 20 2006, given originally for A113552 *)

PARI

up_to = (2^14)+1;
v304752 = vector(up_to);
m_occurrences = Map();
k=0; prev=1; for(n=1,up_to,fordiv(prev,d,if(!mapisdefined(m_occurrences,d),v304752[n] = d;mapput(m_occurrences,d,n);break)); if(!v304752[n], m = 1; try = prev; while(!(prev%m) || mapisdefined(m_occurrences,try), m++; try = prev*m); mapput(m_occurrences,v304752[n] = try,n)); prev = v304752[n]);
A304752(n) = v304752[n];

PARI

A304752(n,a=1,list=List(a)/*set to 0 to get just a(n)*/,U=[])={ for(i=2,n, U=setunion(U,[a]); fordiv(a,d,setsearch(U,d)||[a=-d,break]); if(a>0, for(m=2,oo, a%m && !setsearch(U,m*a)&& (a*=m)&& break),a=-a);list&& listput(list,a); /*a%2&&printf("a(%d)=%d, ",i,a)*/);if(list,list,a)} \\ M. F. Hasler, Dec 26 2020