cp's OEIS Frontend

A212863 Numbers k such that the sum of prime factors of k (counted with multiplicity) equals five times the largest prime divisor of k.

%I A212863 #28 Jan 31 2025 19:30:27
%S A212863 32,192,216,243,3125,3750,4000,4500,4800,5120,5400,5760,6075,6480,
%T A212863 7290,16807,24010,28812,34300,41160,43904,46305,49000,49392,55125,
%U A212863 55566,58800,62720,65625,66150,70000,70560,75264,78750,79380,84000,84672,89600,94500,95256
%N A212863 Numbers k such that the sum of prime factors of k (counted with multiplicity) equals five times the largest prime divisor of k.
%C A212863 The numbers prime(n)^5 are in the sequence.
%C A212863 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
%C A212863 From _David A. Corneth_, Apr 24 2020: (Start)
%C A212863 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).
%C A212863 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).
%C A212863 This enables us to search through the positive integers via a tree starting at 1. (End)
%H A212863 David A. Corneth, <a href="/A212863/b212863.txt">Table of n, a(n) for n = 1..10000</a> (first 200 terms from Robert Israel. Following 4800 terms from Amiram Eldar)
%H A212863 David A. Corneth, <a href="/A212863/a212863.gp.txt">PARI program</a>
%H A212863 David A. Corneth, <a href="/A212863/a212863_1.gp.txt">Terms <= 10^12</a>
%e A212863 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.
%p A212863 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:
%t A212863 Select[Range[2, 10^5], Plus @@ Times @@@ (f = FactorInteger[#]) == 5 * f[[-1, 1]] &] (* _Amiram Eldar_, Apr 24 2020 *)
%o A212863 (PARI) \\ See Corneth link. _David A. Corneth_, Apr 24 2020
%Y A212863 Cf. A001414, A163836, A212861, A212862.
%K A212863 nonn,easy
%O A212863 1,1
%A A212863 _Michel Lagneau_, May 29 2012