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A127668 Concatenated indices of primes in prime factorization of n. 4

%I #14 Aug 29 2019 17:39:15

%S 1,2,11,3,21,4,111,22,31,5,211,6,41,32,1111,7,221,8,311,42,51,9,2111,

%T 33,61,222,411,10,321,11,11111,52,71,43,2211,12,81,62,3111,13,421,14,

%U 511,322,91,15,21111,44,331,72,611,16,2221,53,4111,82,101,17,3211,18,111

%N Concatenated indices of primes in prime factorization of n.

%C For each n>=2 the indices i of primes p(i), i>=1, in the prime number decomposition of n are ordered from right to left.

%C The mapping n->a(n) is from {2,3,...} onto {1,2,3,...}=N but not injective; hence not invertible.

%C There are at most pa(k):=A000041(k) (partition numbers) different numbers which map to any a(n) with k digits. 10 and 12 are the smallest numbers for which this is not equality; 10 because 1,0 is not a partition, and 12 because 1,2 lists partition parts in the wrong order.

%C For the invertible map onto lists of prime number indices see the W. Lang link; also A112798.

%H W. Lang: <a href="/A127668/a127668.txt">Unique representation as lists.</a>

%F If n=p_1^(n_1) p_2^(n_2)...p_k^(n_k), with n_j>=0, then a(n) = n_k times k followed by n_{k-1} times (k-1)... followed by n_1 times 1.

%e 111=a(2*2*2)=a(31*2)=a(607). 111 has k=3 digits, hence pa(3)=3 different numbers are mapped to it.

%e a(5)=3 because 5=p(3). a(4)=11 because 4=2*2=p(1)*p(1). Also a(31)=11 because p(11)=31.

%Y For numbers with no prime divisor > 23, the sum of digits gives A056239(n), n>=2.

%Y For numbers with no prime divisor > 23, the length of the digits gives A001222(n), n>=2, (number of prime divisors of n).

%Y The number of numbers mapped to a(n) gives A127669.

%Y Cf. A054841(n), n>=2: exponents in prime decomposition of n.

%Y See A112798 for another version of this data.

%K nonn,easy,base

%O 2,2

%A _Wolfdieter Lang_ Jan 23 2007

%E Edited by _Franklin T. Adams-Watters_, May 21 2014

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