A229027 Numbers k such that k/rad(k) equals the greatest prime dividing k.

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

Offset: 1

Author

Michel Lagneau, Sep 11 2013

Keywords

• nonn

Comments

Numbers k such that k/A007947(k) = A006530(k) where A007947 is the product of the distinct prime factors of k and A006530 is the greatest prime dividing k.

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

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A007947, A006530, A228957.

Programs

Maple

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:

Mathematica

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 *)

PARI

isok(n) = my(f = factor(n)); n/factorback(f[, 1]) == f[#f~, 1]; \\ Michel Marcus, Aug 16 2014