

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).


3



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
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OFFSET

5,2


COMMENTS

If the canonical representation of m is Product(p(i)^k(i)), where p(i) is the ith 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 nto5 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)


LINKS

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


EXAMPLE

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


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


PROG

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


CROSSREFS

Cf. A008474, A247095.
Sequence in context: A326613 A288627 A232096 * A316456 A083716 A231820
Adjacent sequences: A250027 A250028 A250029 * A250031 A250032 A250033


KEYWORD

nonn


AUTHOR

Ivan N. Ianakiev, Nov 11 2014


EXTENSIONS

a(5) in bfile corrected by Andrew Howroyd, Feb 22 2018


STATUS

approved



