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A284260
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Greatest prime dividing n which is less than A020639(n)^2, where A020639(n) is the smallest prime dividing n, a(1) = 1.
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8
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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
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OFFSET
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1,2
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LINKS
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FORMULA
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MATHEMATICA
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Table[Last[Function[s, Select[s, # < First[s]^2 &]]@ FactorInteger[n][[All, 1]] /. {} -> {1}], {n, 109}] (* Michael De Vlieger, Mar 24 2017 *)
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PROG
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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)));
gpf(n) = if(n>1, vecmax(factor(n)[, 1]), 1);
(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
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CROSSREFS
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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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