A284252 a(n) = smallest prime dividing n which is larger than the square of smallest prime dividing n, or 1 if no such prime exists, a(1) = 1.

1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 7, 1, 1, 1, 1, 1, 5, 1, 11, 1, 1, 1, 13, 1, 7, 1, 5, 1, 1, 11, 17, 1, 1, 1, 19, 13, 5, 1, 7, 1, 11, 1, 23, 1, 1, 1, 5, 17, 13, 1, 1, 1, 7, 19, 29, 1, 5, 1, 31, 1, 1, 1, 11, 1, 17, 23, 5, 1, 1, 1, 37, 1, 19, 1, 13, 1, 5, 1, 41, 1, 7, 1, 43, 29, 11, 1, 5, 1, 23, 31, 47, 1, 1, 1, 7, 11, 5, 1, 17, 1, 13, 1, 53, 1, 1, 1, 5, 37

Offset

1,10

Links

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

Formula

a(n) = A020639(A284254(n)).

a(k) > 1 iff k is in A251727. - David A. Corneth, Mar 25 2017

Example

For n=10 = 2*5, the smallest prime divisor > 2^2 is 5, thus a(10) = 5.
For n=15 = 3*5, there are no prime divisors > 3^2, thus a(15) = 1.
For n=165 = 3*5*11, the smallest prime divisor > 3^2 is 11, thus a(165) = 11.

Mathematica

a[n_] := Block[{p = First /@ FactorInteger[n]}, SelectFirst[p, # > p[[1]]^2 &, 1]]; Array[a, 120] (* Giovanni Resta, Mar 24 2017 *)

Prog

(Scheme) (define (A284252 n) (let ((spf1 (A020639 n))) (let loop ((n (/ n spf1))) (let ((spf2 (A020639 n))) (cond ((= 1 spf2) 1) ((> spf2 (* spf1 spf1)) spf2) (else (loop (/ n spf2))))))))
(PARI) a(n) = if(n==1, return(1), my(f=factor(n)[, 1]); s = f[1]; for(i=2, #f, if(f[i]>s^2, return(f[i]))); return(1)) \\ David A. Corneth, Mar 24 2017
(Python)
from sympy import primefactors
def a(n):
    for i in primefactors(n):
        if i>min(primefactors(n))**2: return i
    return 1
print([a(n) for n in range(1, 151)]) # Indranil Ghosh, Mar 24 2017

Crossrefs

Cf. A020639, A284253, A284254, A284255, A284256, A284257, A284258, A284259, A284260.

Cf. A251726 (gives the positions of ones after the initial a(1) = 1), A251727 (positions of terms > 1).

Sequence in context: A293718 A068316 A359945 * A284254 A309206 A358016

Adjacent sequences: A284249 A284250 A284251 * A284253 A284254 A284255

Keyword

nonn

Author

Antti Karttunen, Mar 24 2017

Status

approved