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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.

%I A308725 #67 Jul 29 2019 13:56:24
%S A308725 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,
%T A308725 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,
%U A308725 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
%N 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.
%C A308725 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
%C A308725 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
%H A308725 Antti Karttunen, <a href="/A308725/b308725.txt">Table of n, a(n) for n = 1..10000</a>
%F A308725 If n is 6 or 7, a(n) = 0, otherwise a(n) = 1 + a(A227215(n)). - _Antti Karttunen_, Jul 11 2019
%e A308725     1 = 1*1*1  --> 1 + 1 + 1  = 3
%e A308725     3 = 1*1*3  --> 1 + 1 + 3  = 5
%e A308725     5 = 1*1*5  --> 1 + 1 + 5  = 7, thus a(1) = 3.
%e A308725 .
%e A308725     4 = 1*2*2  --> 1 + 2 + 2  = 5,
%e A308725     5 = 1*1*5  --> 1 + 1 + 5  = 7, thus a(4) = 2.
%e A308725 .
%e A308725   560 = 7*8*10 --> 7 + 8 + 10 = 25
%e A308725    25 = 1*5*5  --> 1 + 5 +  5 = 11
%e A308725    11 = 1*1*11 --> 1 + 1 + 11 = 13
%e A308725    13 = 1*1*13 --> 1 + 1 + 13 = 15
%e A308725    15 = 1*3*5  --> 1 + 3 +  5 =  9
%e A308725     9 = 1*3*3  --> 3 + 3 +  1 =  7, thus a(560) = 6.
%e A308725 .
%e A308725    84 = 3*4*7  --> 3 + 4 + 7 = 14
%e A308725    14 = 1*2*7  --> 1 + 2 + 7 = 10
%e A308725    10 = 1*2*5  --> 1 + 2 + 5 =  8
%e A308725     8 = 2*2*2  --> 2 + 2 + 2 =  6, thus a(84) = 4.
%t A308725 maxTerm = 99 (* Should be increased if output -1 appears. *);
%t A308725 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];
%t A308725 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];
%t A308725 Array[a, 100] (* _Jean-François Alcover_, Jul 03 2019 *)
%o A308725 (PARI)
%o A308725 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.
%o A308725 A308725(n) = if((6==n)||(7==n),0,1+A308725(A227215(n)));
%o A308725 \\ Memoized implementation:
%o A308725 memoA308725 = Map();
%o A308725 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
%Y A308725 Cf. A227215, A308190.
%K A308725 nonn
%O A308725 1,1
%A A308725 _Ali Sada_, Jun 20 2019