A250030 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).

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, 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, 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

Offset: 5

Author

Ivan N. Ianakiev, Nov 11 2014

Keywords

• nonn

Comments

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)).

Conjecture 1: for m > 4, by iterating the map m -> A008474(m) one always reaches 5. Tested for m up to 320000.

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

a(A247095(n)) = n and a(m) != n for m < A247095(n). - Reinhard Zumkeller, Nov 18 2014

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

From Ivan N. Ianakiev, Aug 12 2015: (Start)

Proof outline for Conjecture 1:

1. a(n) < n is true for:

a) all composite numbers n such that A001221(n) = 1 in the cases where the exponent of the prime is greater than 2,

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) all composite numbers n such that A001221(n) > 2, the smallest of which is 30 = 2^1*3^1*5^1.

2. a(a(n)) < n is true for all primes p such that p+1 equals a composite number handled in item 1 above.

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)

Examples

			3 iterations are needed to start at 11 and finish at 5 (11 -> 12 -> 8 -> 5), therefore a(11) is 3.
Conjecture 2: For n = 11, S(n) = 11 + 12 + 8 = 31, which is not divisible by 11. - _Ivan N. Ianakiev_, May 06 2015
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 5..10000

Crossrefs

Cf. A008474, A247095.

Programs

Haskell

a250030 n = snd $ until ((== 5) . fst)
                  (\(x, s) -> (a008474 x, s + 1)) (a008474 n, 1)
-- Reinhard Zumkeller, Nov 18 2014

Maple

f:= n -> convert(map(convert,ifactors(n)[2],`+`),`+`):
a:= proc(n) option remember;
if n = 5 then 0 else 1+procname(f(n)) fi
end proc:
4, seq(a(n),n=6..200); # Robert Israel, May 11 2015

Mathematica

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

PARI

A008474(n)=my(f=factor(n)); sum(i=1,#f~,f[i,1]+f[i,2])
a(n)=my(k); while(n!=5, n=A008474(n); k++); k \\ Charles R Greathouse IV, Jun 03 2015

Extensions

a(5) in b-file corrected by Andrew Howroyd, Feb 22 2018