

A212863


Numbers k such that the sum of prime factors of k (counted with multiplicity) equals five times the largest prime divisor of k.


1



32, 192, 216, 243, 3125, 3750, 4000, 4500, 4800, 5120, 5400, 5760, 6075, 6480, 7290, 16807, 24010, 28812, 34300, 41160, 43904, 46305, 49000, 49392, 55125, 55566, 58800, 62720, 65625, 66150, 70000, 70560, 75264, 78750, 79380, 84000, 84672, 89600, 94500, 95256
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OFFSET

1,1


COMMENTS

The numbers prime(n)^5 are in the sequence.
Also contains 4^p p and 2^p p^3 for any prime p>2, and 3^p p^2 for any prime > 3.  Robert Israel, Jun 20 2017
Suppose we look for terms below (inclusive) u. Let maxp be the largest prime factor that is the multiple of at least one of those terms. Then maxp is the largest prime below (inclusive) u^(1/5).
Proof: The sum of prime factors counted with multiplicity of a term t divisible by maxp is 5*maxp. The smallest product of primes summing to 5*maxp where the largest prime factor of t is maxp is maxp^5 which must be <= u. Solving this gives maxp is the largest prime below (inclusive) u^(1/5).
This enables us to search through the positive integers via a tree starting at 1. (End)


LINKS



EXAMPLE

192 is in the sequence because 192 = 2^6 * 3 => sum of prime divisors = 2*6 + 3 = 15 = 5*3 where 3 is the greatest prime divisor.


MAPLE

with(numtheory):A:= proc(n) local e, j; e := ifactors(n)[2]: add (e[j][1]*e[j][2], j=1..nops(e)) end: for m from 2 to 500000 do: x:=factorset(m):n1:=nops(x):if A(m)=5*x[n1] then printf(`%d, `, m):else fi:od:


MATHEMATICA

Select[Range[2, 10^5], Plus @@ Times @@@ (f = FactorInteger[#]) == 5 * f[[1, 1]] &] (* Amiram Eldar, Apr 24 2020 *)


PROG



CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



