%I #25 Mar 30 2017 22:36:56

%S 1,1,2,1,3,2,4,1,2,3,5,2,6,4,6,1,7,2,8,3,8,5,9,2,3,6,4,4,10,6,11,1,10,

%T 7,12,2,12,8,12,3,13,8,14,5,6,9,15,2,4,3,14,6,16,4,15,4,16,10,17,6,18,

%U 11,8,1,18,10,19,7,18,12,20,2,21,12,6,8,20,12,22,3,2,13,23,8,21,14,20,5,24,6,24,9,22,15,24,2,25,4,10,3,26,14,27,6,24,16,28

%N Replace each prime number with its rank in the recursive prime factorization of n.

%C a(A000040(n)) = n, hence all natural numbers appear in this sequence.

%C a(2n) = n.

%C It appears that a(35) = 12 is the only instance where a composite index yields a larger value than any smaller index. Checked to 10^7. - _Charles R Greathouse IV_, Jul 30 2016

%H Paul Tek, <a href="/A225395/b225395.txt">Table of n, a(n) for n = 1..10000</a>

%H Paul Tek, <a href="/A225395/a225395.txt">Perl program for this sequence</a>

%F Multiplicative, with a(prime(i)^j) = i^a(j).

%F a(n) = prod(A049084(A027748(k))^a(A124010(k)): k=1..A001221(n)). - _Reinhard Zumkeller_, May 10 2013

%e The number 9967 is the 1228th prime number.

%e Hence a(9967) = 1228.

%e The recursive prime factorization of 31250 is 2*5^(2*3).

%e The numbers 2, 3 and 5 are respectively the 1st, 2nd and 3rd prime numbers.

%e Hence a(31250) = a(2*5^(2*3)) = 1*3^(1*2) = 9.

%t a[1] = 1; a[p_?PrimeQ] := a[p] = PrimePi[p]; a[n_] := a[n] = Times @@ (PrimePi[#[[1]]]^a[#[[2]]]& /@ FactorInteger[n]); Table[a[n], {n, 1, 100}] (* _Jean-François Alcover_, Jun 07 2013 *)

%o (Perl) see link.

%o (Haskell)

%o a225395 n = product $ zipWith (^)

%o (map a049084 $ a027748_row n) (map a225395 $ a124010_row n)

%o -- _Reinhard Zumkeller_, May 10 2013

%o (PARI) a(n)=if(n<3, return(1)); my(f=factor(n)); prod(i=1,#f~, primepi(f[i,1])^a(f[i,2])) \\ _Charles R Greathouse IV_, Jul 30 2016

%Y Cf. A000040, A000026, A156061.

%K nonn,mult,nice

%O 1,3

%A _Paul Tek_, May 06 2013