A284254 Largest divisor of n such that all its prime factors are greater than the square of smallest prime factor of n, 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, 25, 17, 13, 1, 1, 1, 7, 19, 29, 1, 5, 1, 31, 1, 1, 1, 11, 1, 17, 23, 35, 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, 49, 11, 25, 1, 17, 1, 13, 1, 53, 1, 1, 1, 55
Offset: 1
Keywords
Examples
For n = 15 = 3*5, no prime factor is larger than 3^2, thus a(15) = 1. In this case the largest divisor satisfying the condition has no prime factors at all. For n = 50 = 2*5*5, the primes larger than 2^2 are 5 and 5, thus a(50) = 5*5 = 25.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..10001
Crossrefs
Programs
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Mathematica
Table[If[n == 1, 1, Function[d, Last[Select[Reverse@ First@ d, Times @@ Boole@ Map[# > Last[d]^2 &, FactorInteger[#][[All, 1]]] == 1 &] /. {} -> {1}]]@ {#, First@ Select[#, PrimeQ]} &@ Divisors@ n], {n, 108}] (* Michael De Vlieger, Mar 24 2017 *)
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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))); for(n=1, 150, print1(a(n),", ")) \\ Indranil Ghosh, after David A. Corneth, Mar 24 2017
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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)) print([a(n) for n in range(1, 151)]) # Indranil Ghosh, Mar 24 2017