A212861
Numbers n such that the sum of prime factors of n (counted with repetition) equals three times the largest prime divisor.
Original entry on oeis.org
8, 24, 27, 125, 150, 160, 180, 343, 490, 588, 700, 840, 896, 945, 1008, 1134, 1331, 2197, 3388, 3718, 4840, 4913, 5445, 5808, 6292, 6534, 6859, 8085, 8624, 9464, 9625, 9702, 10647, 11550, 12167, 12274, 12320, 12675, 13520, 13750, 13860, 14784, 15015, 15028
Offset: 1
150 is in the sequence because 150 = 2*3*5^2 => sum of prime divisors = 2+3 + 5*2 = 15 = 3*5 where 5 is the greatest prime divisor.
-
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 20000 do: x:=factorset(m):n1:=nops(x):if A(m)=3*x[n1] then printf(`%d, `,m):else fi:od:
-
spfQ[n_]:=Module[{f=FactorInteger[n]},Total[Flatten[Table[#[[1]], #[[2]]]&/@ f]]==3*f[[-1,1]]]; Select[Range[16000],spfQ] (* Harvey P. Dale, Jul 26 2016 *)
-
is(n)=my(f=factor(n),k=#f[,1]); k && sum(i=1,k,f[i,1]*f[i,2]) == 3*f[k,1] \\ Charles R Greathouse IV, May 29 2012
A212862
Numbers k such that the sum of prime factors of k (counted with multiplicity) equals four times the largest prime divisor of k.
Original entry on oeis.org
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
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.
-
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
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.
-
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
A232241
Composites where the greatest prime factor is the sum of the other prime powers.
Original entry on oeis.org
4, 9, 25, 30, 49, 70, 84, 121, 169, 198, 264, 286, 289, 308, 361, 468, 520, 529, 646, 841, 884, 912, 961, 1224, 1369, 1566, 1672, 1681, 1748, 1798, 1849, 2209, 2576, 2809, 2900, 3135, 3348, 3481, 3526, 3570, 3721, 4489, 5041, 5329, 5704, 5920, 6032
Offset: 1
9 is in the sequence since prod(3^1)*sum(3^1)=(3^1)*(3^1)=3*3=9, and the gpf, 3 is prime.
1224 is in the sequence since (2^3*3^2)*(2^3+3^2)=(8*9)*(8+9)=72*17=1224, and the gpf, 17 is prime.
6032 is in the sequence since (2^4*13^1)*(2^4+13^1)=(16*13)*(16+13)=208*29=6032, and the gpf, 29 is prime.
-
public class Psfi {public static void main(String[] args) {String sequence = ""; for (int n = 2; sequence.length() < 250; n++) {int q = n; int s = 0; for (int d = 2; d <= Math.sqrt(q); d++) {int i = 0; while (q > d && q % d == 0) {i++; q = q / d;} if (i > 0) {s += Math.pow(d, i);} } if (s == q) {sequence += n + ", ";} } System.out.println(sequence);} }
-
seqQ[n_] := Module[{f = FactorInteger[n]}, If[Length[f] == 1, f[[1, 2]] == 2, f[[-1, 2]] == 1 && f[[-1, 1]] == Plus @@ Power @@@ Most[f]]]; Select[Range[6000], seqQ] (* Amiram Eldar, Apr 28 2020 *)
-
isok(n) = {if (n>1, my(fa = factor(n), gpf = fa[#fa~, 1], fb = factor(n/gpf)); gpf == sum(i=1, #fb~, fb[i, 1]^fb[i, 2])); } \\ Michel Marcus, Nov 21 2013; Apr 28 2020
Showing 1-4 of 4 results.
Comments