A308725 Number of steps to reach 6 or 7 when iterating x -> A227215(x) starting at x=n, where A227215(n) gives the smallest such sum a+b+c of three positive integers for which a*b*c = n.

3, 3, 2, 2, 1, 0, 0, 1, 1, 2, 4, 1, 3, 3, 2, 2, 7, 2, 6, 2, 5, 4, 6, 2, 5, 3, 2, 5, 5, 3, 4, 3, 3, 3, 4, 3, 9, 5, 8, 5, 7, 2, 6, 3, 5, 4, 4, 5, 3, 2, 6, 8, 9, 2, 8, 4, 7, 4, 6, 2, 5, 4, 4, 2, 7, 3, 4, 6, 3, 4, 6, 4, 5, 6, 4, 7, 7, 3, 4, 4, 3, 4, 8, 4, 7, 5, 4, 8, 7, 4, 6, 3, 5, 3, 6, 4, 9, 3, 8, 4

Offset: 1

Author

Ali Sada, Jun 20 2019

Keywords

• nonn

Comments

Starting from n, choose factorization n = m1*m2*m3 so that the sum x = m1+m2+m3 is minimal, then set n = x and repeat. a(n) gives the number of steps needed to reach either 6 or 7. The process is guaranteed to reach either term, because we only use factorization n = n*1*1 when n is either 1 or a prime number, that are the only cases (apart from A227215(4)=5) for which A227215(n) > n as then A227215(n) = n+2. Moreover, for n > 3, at least one of n, n+2, n+4 is composite, leading to a further significant drop in the trajectory after at most two consecutive +2 steps. - Comment clarified by Antti Karttunen, Jul 12 2019

Records: 3, 4, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17, ..., occur at: n = 1, 11, 17, 37, 107, 233, 307, 1289, 3986, 6637, 14347, 69029, .... - Antti Karttunen, Jul 12 2019

Examples

			    1 = 1*1*1  --> 1 + 1 + 1  = 3
    3 = 1*1*3  --> 1 + 1 + 3  = 5
    5 = 1*1*5  --> 1 + 1 + 5  = 7, thus a(1) = 3.
.
    4 = 1*2*2  --> 1 + 2 + 2  = 5,
    5 = 1*1*5  --> 1 + 1 + 5  = 7, thus a(4) = 2.
.
  560 = 7*8*10 --> 7 + 8 + 10 = 25
   25 = 1*5*5  --> 1 + 5 +  5 = 11
   11 = 1*1*11 --> 1 + 1 + 11 = 13
   13 = 1*1*13 --> 1 + 1 + 13 = 15
   15 = 1*3*5  --> 1 + 3 +  5 =  9
    9 = 1*3*3  --> 3 + 3 +  1 =  7, thus a(560) = 6.
.
   84 = 3*4*7  --> 3 + 4 + 7 = 14
   14 = 1*2*7  --> 1 + 2 + 7 = 10
   10 = 1*2*5  --> 1 + 2 + 5 =  8
    8 = 2*2*2  --> 2 + 2 + 2 =  6, thus a(84) = 4.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A227215, A308190.

Programs

Mathematica

maxTerm = 99 (* Should be increased if output -1 appears. *);
f[m_] := Module[{m1, m2, m3, factors}, factors = {m1, m2, m3} /. {ToRules[ Reduce[1 <= m1 <= m2 <= m3 && m == m1 m2 m3, {m1, m2, m3}, Integers]]}; SortBy[factors, Total] // First];
a[n_] := Module[{cnt = 0, m = n, fm, step}, While[!(m == 6 || m == 7), step = {fm = f[m], m = Total[fm]}; (* Print[n," ",step]; *) cnt++; If[cnt > maxTerm, Return[-1]]]; cnt];
Array[a, 100] (* Jean-François Alcover, Jul 03 2019 *)

PARI

A227215(n) = { my(ms=3*n); fordiv(n, i, for(j=i, (n/i), if(!(n%j),for(k=j, n/(i*j), if(i*j*k==n, ms = min(ms,(i+j+k))))))); (ms); }; \\ Like code in A227215.
A308725(n) = if((6==n)||(7==n),0,1+A308725(A227215(n)));
\\ Memoized implementation:
memoA308725 = Map();
A308725(n) = if((6==n)||(7==n), 0, my(v); if(mapisdefined(memoA308725,n,&v), v, v = 1+A308725(A227215(n)); mapput(memoA308725,n,v); (v))); \\ Antti Karttunen, Jul 12 2019

Formula

If n is 6 or 7, a(n) = 0, otherwise a(n) = 1 + a(A227215(n)). - Antti Karttunen, Jul 11 2019