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A229027
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Numbers n such that n/rad(n) equals the greatest prime dividing n.
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1
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4, 9, 18, 25, 49, 50, 75, 98, 121, 147, 150, 169, 242, 245, 289, 294, 338, 361, 363, 490, 507, 529, 578, 605, 722, 726, 735, 841, 845, 847, 867, 961, 1014, 1058, 1083, 1183, 1210, 1369, 1445, 1470, 1587, 1681, 1682, 1690, 1694, 1734, 1805, 1815, 1849, 1859
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OFFSET
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1,1
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COMMENTS
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Numbers n such that n/A007947(n)= A006530(n) where A007947 is the product of the distinct prime factors of n and A006530 is the greatest prime dividing n.
The numbers of the form p^2 with p prime, or of the form p_1*p_2*...*p_k*p^2 with p_i primes < p prime are in the sequence.
All these numbers are round numbers (their greatest prime factor is <= their square root. - Emmanuel Vantieghem, Feb 22 2017
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LINKS
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MAPLE
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with(numtheory) :for n from 2 to 2000 do:x:=factorset(n):n1:=nops(x): p:= product('x[i]', 'i'=1..n1):m:=n/p:if m=x[n1] then printf(`%d, `, n):else fi:od:
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MATHEMATICA
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rad[n_]:=Times@@(First@#&/@FactorInteger@n); Select[Range[2, 2000], FactorInteger[#][[-1, 1]]==#/rad[#]&]
gpQ[n_]:=Module[{pf=Transpose[FactorInteger[n]][[1]]}, n/Times@@pf == Last[ pf]]; Select[Range[2, 2000], gpQ] (* Harvey P. Dale, Aug 16 2014 *)
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PROG
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(PARI) isok(n) = my(f = factor(n)); n/factorback(f[, 1]) == f[#f~, 1]; \\ Michel Marcus, Aug 16 2014
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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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