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%I #21 Mar 16 2018 08:22:55
%S 21,31,22,51,2131,71,23,32,2151,111,2231,131,2171,3151,24,171,2132,
%T 191,2251,3171,21111,231,2331,52,21131,33,2271,291,213151,311,25,
%U 31111,21171,5171,2232,371,21191,31131,2351,411,213171,431,22111,3251,21231
%N Decimal encoding of the prime factorization of n: concatenation of prime factors and exponents.
%C If n has prime factorization p_1^e_1 * ... * p_r^e_r with p_1 < ... < p_r, then its decimal encoding is p_1 e_1...p_r e_r. For example, 15 = 3^1 * 5^1, so has decimal encoding 3151.
%C Sequence A068633 is a duplicate, up to a conventional initial term a(1)=11.
%C a(31) = a(177147) = 311. Is there any solution to a(n) = n? - _Franklin T. Adams-Watters_, Dec 18 2006
%C The earliest duplicate is a(223) = 2231 = a(12). There is no fixed point below 3*10^6. - _M. F. Hasler_, Oct 06 2013
%H Reinhard Zumkeller, <a href="/A067599/b067599.txt">Table of n, a(n) for n = 2..10000</a>
%e The prime factorization of 24 = 2^3 * 3^1 has corresponding encoding 2331. So a(24) = 2331.
%e a(42) = 213171 since 42 = 2^1*3^1*7^1. - _Amarnath Murthy_, Feb 27 2002
%p with(ListTools): with(MmaTranslator[Mma]): seq(FromDigits(FlattenOnce(ifactors(n)[2])), n=2..46); # _Wolfdieter Lang_, Aug 16 2014
%p # second Maple program:
%p a:= n-> parse(cat(map(i-> i[], sort(ifactors(n)[2]))[])):
%p seq(a(n), n=2..60); # _Alois P. Heinz_, Mar 16 2018
%t f[n_] := FromDigits[ Flatten[ IntegerDigits[ FactorInteger[ n]]]]; Table[ f[n], {n, 2, 50} ]
%o (PARI) A067599(n)=eval(concat(concat([""],concat(Vec(factor(n)~))~))) \\ - _M. F. Hasler_, Oct 06 2013
%o (Haskell)
%o import Data.Function (on)
%o a067599 n = read $ foldl1 (++) $
%o zipWith ((++) `on` show) (a027748_row n) (a124010_row n) :: Integer
%o -- _Reinhard Zumkeller_, Oct 27 2013
%Y Cf. A037276, A080670, A112375.
%Y Cf. A027748, A124010.
%K base,easy,nonn
%O 2,1
%A _Joseph L. Pe_, Jan 31 2002
%E Edited by _Robert G. Wilson v_, Feb 02 2002
%E Merged contributions from A068633 to here, and minor edits by _M. F. Hasler_, Oct 06 2013
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