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%I #45 Feb 22 2018 20:19:09
%S 0,3,2,1,1,2,3,2,5,4,3,4,3,2,4,3,3,4,3,2,3,4,4,3,7,6,4,3,5,4,5,2,5,4,
%T 3,4,5,4,6,5,4,5,4,3,2,3,5,3,3,2,3,6,3,6,6,5,7,6,6,2,4,5,6,5,4,4,4,3,
%U 7,6,4,3,4,4,4,3,3,5,6,5,3,5,5,4,6,5,5,4,3,6,5,4,4,3,4,4,5,4,6,5,3,4,4,3,5,4,5,5,6,5,7,5,5
%N a(n) is the number of iterations of A008474 needed to reach 5, starting at n, where A008474(Product p_i^e_i) = Sum(p_i+e_i).
%C If the canonical representation of m is Product(p(i)^k(i)), where p(i) is the i-th prime and k(i) is its exponent, then A008474(m) equals Sum(p(i)+k(i)).
%C Conjecture 1: for m > 4, by iterating the map m -> A008474(m) one always reaches 5. Tested for m up to 320000.
%C a(n) = -1 in the cases (if such exist) where the iteration acting on n does not reach 5. - _Ivan N. Ianakiev_, Jun 03 2015
%C a(A247095(n)) = n and a(m) != n for m < A247095(n). - _Reinhard Zumkeller_, Nov 18 2014
%C Conjecture 2: Let S(n) be the sum of the first a(n) numbers in the n-to-5 orbit, i.e., the sum of all the numbers in the orbit without 5. Except for n in [8,9], S(n) is not divisible by n. Verified for n up to 10^6. - _Ivan N. Ianakiev_, May 06 2015
%C From _Ivan N. Ianakiev_, Aug 12 2015: (Start)
%C Proof outline for Conjecture 1:
%C 1. a(n) < n is true for:
%C a) all composite numbers n such that A001221(n) = 1 in the cases where the exponent of the prime is greater than 2,
%C b) all composite numbers n such that A001221(n) = 2 in the cases where one of the exponents of the primes is greater than 1, and
%C c) all composite numbers n such that A001221(n) > 2, the smallest of which is 30 = 2^1*3^1*5^1.
%C 2. a(a(n)) < n is true for all primes p such that p+1 equals a composite number handled in item 1 above.
%C 3. The only thing needed to complete the proof is to exhaustively confirm the conjecture for all numbers in [5, 29] that were not handled in items 1 and 2 above, i.e., 5, 6, 8, 9, 10, 11, 13, 14, 15, 21, 22, 25, 26. (End)
%H Reinhard Zumkeller, <a href="/A250030/b250030.txt">Table of n, a(n) for n = 5..10000</a>
%e 3 iterations are needed to start at 11 and finish at 5 (11 -> 12 -> 8 -> 5), therefore a(11) is 3.
%e Conjecture 2: For n = 11, S(n) = 11 + 12 + 8 = 31, which is not divisible by 11. - _Ivan N. Ianakiev_, May 06 2015
%p f:= n -> convert(map(convert,ifactors(n)[2],`+`),`+`):
%p a:= proc(n) option remember;
%p if n = 5 then 0 else 1+procname(f(n)) fi
%p end proc:
%p 4, seq(a(n),n=6..200); # _Robert Israel_, May 11 2015
%t trajLen={0};Do[lst={};g[n_]:=Total[Flatten[FactorInteger[n]]];While[n>5,n=g[n];AppendTo[lst,n]];AppendTo[trajLen,Length[lst]],{n,6,105}];trajLen
%o (Haskell)
%o a250030 n = snd $ until ((== 5) . fst)
%o (\(x, s) -> (a008474 x, s + 1)) (a008474 n, 1)
%o -- _Reinhard Zumkeller_, Nov 18 2014
%o (PARI) A008474(n)=my(f=factor(n)); sum(i=1,#f~,f[i,1]+f[i,2])
%o a(n)=my(k); while(n!=5, n=A008474(n); k++); k \\ _Charles R Greathouse IV_, Jun 03 2015
%Y Cf. A008474, A247095.
%K nonn
%O 5,2
%A _Ivan N. Ianakiev_, Nov 11 2014
%E a(5) in b-file corrected by _Andrew Howroyd_, Feb 22 2018
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