

A225395


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


3



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
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OFFSET

1,3


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


LINKS



FORMULA

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


EXAMPLE

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.


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}] (* JeanFrançois Alcover, Jun 07 2013 *)


PROG

(Perl) see link.
(Haskell)
a225395 n = product $ zipWith (^)
(map a049084 $ a027748_row n) (map a225395 $ a124010_row n)
(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


CROSSREFS



KEYWORD

nonn,mult,nice


AUTHOR



STATUS

approved



