A284260 Greatest prime dividing n which is less than A020639(n)^2, where A020639(n) is the smallest prime dividing n, a(1) = 1.

1, 2, 3, 2, 5, 3, 7, 2, 3, 2, 11, 3, 13, 2, 5, 2, 17, 3, 19, 2, 7, 2, 23, 3, 5, 2, 3, 2, 29, 3, 31, 2, 3, 2, 7, 3, 37, 2, 3, 2, 41, 3, 43, 2, 5, 2, 47, 3, 7, 2, 3, 2, 53, 3, 11, 2, 3, 2, 59, 3, 61, 2, 7, 2, 13, 3, 67, 2, 3, 2, 71, 3, 73, 2, 5, 2, 11, 3, 79, 2, 3, 2, 83, 3, 17, 2, 3, 2, 89, 3, 13, 2, 3, 2, 19, 3, 97, 2, 3, 2, 101, 3, 103, 2, 7, 2, 107, 3, 109

Offset: 1

Author

Antti Karttunen, Mar 24 2017

Keywords

• nonn

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10001

Crossrefs

Cf. A006530, A020639, A284252, A284253, A284254, A284255, A284256, A284257, A284258, A284259.

Cf. A251726 (gives n > 1 such that a(n) = A006530(n)).

Programs

Mathematica

Table[Last[Function[s, Select[s, # < First[s]^2 &]]@ FactorInteger[n][[All, 1]] /. {} -> {1}], {n, 109}] (* Michael De Vlieger, Mar 24 2017 *)

PARI

A(n) = if(n<2, return(1), my(f=factor(n)[, 1]); for(i=2, #f, if(f[i]>f[1]^2, return(f[i]))); return(1));
a(n) = if(A(n)==1, 1, A(n)*a(n/A(n)));
gpf(n) = if(n>1, vecmax(factor(n)[,1]),1);
for(n=1, 150, print1(gpf(n/a(n)),", ")) \\ Indranil Ghosh, Mar 24 2017, after David A. Corneth

Python

from sympy import primefactors
def A(n):
    for i in primefactors(n):
        if i>min(primefactors(n))**2: return i
    return 1
def a(n): return 1 if A(n)==1 else A(n)*a(n//A(n))
def gpf(n): return 1 if n<2 else max(primefactors(n))
print([gpf(n//a(n)) for n in range(1, 151)]) # Indranil Ghosh, Mar 24 2017

Scheme

(define (A284260 n) (A006530 (A284255 n)))

Formula

a(n) = A006530(A284255(n)).