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A284254
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Largest divisor of n such that all its prime factors are greater than the square of smallest prime factor of n, a(1) = 1.
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11
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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
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OFFSET
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1,10
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LINKS
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FORMULA
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Other identities. For all n >= 1:
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EXAMPLE
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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.
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MATHEMATICA
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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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PROG
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(Scheme, with memoization-macro definec)
(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)));
(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))
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CROSSREFS
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Cf. A251726 (gives the positions of ones after the initial a(1)=1).
Differs from related A284252 for the first time at n=50, where a(50) = 25, while A284252(50) = 5.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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