A051119 a(n) = n/p^k, where p = largest prime dividing n and p^k = highest power of p dividing n.

1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 4, 1, 2, 3, 1, 1, 2, 1, 4, 3, 2, 1, 8, 1, 2, 1, 4, 1, 6, 1, 1, 3, 2, 5, 4, 1, 2, 3, 8, 1, 6, 1, 4, 9, 2, 1, 16, 1, 2, 3, 4, 1, 2, 5, 8, 3, 2, 1, 12, 1, 2, 9, 1, 5, 6, 1, 4, 3, 10, 1, 8, 1, 2, 3, 4, 7, 6, 1, 16, 1, 2, 1, 12, 5, 2, 3, 8, 1, 18, 7, 4, 3, 2, 5, 32, 1, 2, 9, 4, 1

Offset: 1

Author

Leroy Quet

Keywords

• nonn
• easy
• nice

Examples

			a(36) = 4 because 36/3^2 = 4, 3^2 is highest power dividing 36 of largest prime dividing 36.
a(50) = 50/5^2 = 2.
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A053585.

Programs

Mathematica

f[n_]:=Module[{c=Last[FactorInteger[n]]},n/First[c]^Last[c]]; Array[ f, 110] (* Harvey P. Dale, Oct 14 2011 *)

PARI

a(n) = if(n>1, my(f=factor(n)); n/f[#f~, 1]^f[#f~, 2], 1); \\ Michel Marcus, Jan 10 2025

Python

from sympy import factorint, primefactors
def a053585(n):
    if n==1: return 1
    p = primefactors(n)[-1]
    return p**factorint(n)[p]
def a(n): return n/a053585(n) # Indranil Ghosh, May 19 2017

Formula

a(n) = n/A053585(n).

Extensions

More terms from James Sellers, Jan 21 2000