A212861 -id:A212861 - cp's OEIS frontend

A212862 Numbers k such that the sum of prime factors of k (counted with multiplicity) equals four times the largest prime divisor of k.

16, 72, 81, 625, 750, 800, 900, 960, 1080, 1215, 2401, 3430, 4116, 4900, 5880, 6272, 6615, 7000, 7056, 7875, 7938, 8400, 8960, 9450, 10080, 10752, 11340, 12096, 13608, 14641, 15309, 28561, 37268, 48334, 53240, 59895, 63888, 71874, 81796, 83521, 88935, 94864

Offset: 1

Michel Lagneau, May 29 2012

nonn

The numbers prime(n)^4 are in the sequence.

			750 is in the sequence because 750 = 2*3*5^3 => sum of prime divisors = 2+3 + 5*3 = 20 = 4*5 where 5 is the greatest prime divisor.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A001414, A163836, A212861.

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 100000 do: x:=factorset(m):n1:=nops(x):if A(m)=4*x[n1] then printf(`%d, `,m):else fi:od:

Select[Range[2, 10^5], Plus @@ Times @@@ (f = FactorInteger[#]) == 4 * f[[-1, 1]] &] (* Amiram Eldar, Apr 24 2020 *)

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

Original entry on oeis.org

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

Offset: 1

Michel Lagneau, May 29 2012

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
From David A. Corneth, Apr 24 2020: (Start)
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)

			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.

David A. Corneth, Table of n, a(n) for n = 1..10000 (first 200 terms from Robert Israel. Following 4800 terms from Amiram Eldar)
David A. Corneth, PARI program
David A. Corneth, Terms <= 10^12

Cf. A001414, A163836, A212861, A212862.

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:

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

\\ See Corneth link. David A. Corneth, Apr 24 2020

cp's OEIS Frontend

A212862 Numbers k such that the sum of prime factors of k (counted with multiplicity) equals four times the largest prime divisor of k.

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