A225395 Replace each prime number with its rank in the recursive prime factorization of n.

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, 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, 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

Offset: 1

Author

Paul Tek, May 06 2013

Keywords

• nonn
• mult
• nice

Comments

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

a(2n) = n.

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

Examples

			The number 9967 is the 1228th prime number.
Hence a(9967) = 1228.
The recursive prime factorization of 31250 is 2*5^(2*3).
The numbers 2, 3 and 5 are respectively the 1st, 2nd and 3rd prime numbers.
Hence a(31250) = a(2*5^(2*3)) = 1*3^(1*2) = 9.
		

Links

• Paul Tek, Table of n, a(n) for n = 1..10000
• Paul Tek, Perl program for this sequence

Crossrefs

Cf. A000040, A000026, A156061.

Programs

Haskell

a225395 n = product $ zipWith (^)
    (map a049084 $ a027748_row n) (map a225395 $ a124010_row n)
-- Reinhard Zumkeller, May 10 2013

Mathematica

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 *)

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

Perl

# See Tek link.

Formula

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

a(n) = prod(A049084(A027748(k))^a(A124010(k)): k=1..A001221(n)). - Reinhard Zumkeller, May 10 2013