|
|
A242152
|
|
Numbers n such that the sum of their unitary prime divisors divides sigma(n).
|
|
2
|
|
|
15, 24, 28, 35, 40, 42, 54, 60, 66, 95, 96, 110, 114, 117, 119, 120, 132, 135, 140, 143, 147, 168, 195, 198, 209, 224, 240, 250, 252, 258, 280, 287, 290, 315, 319, 322, 323, 360, 375, 377, 380, 384, 408, 460, 468, 470, 476, 480, 486, 496, 506, 507, 510, 520
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
LINKS
|
|
|
EXAMPLE
|
Divisors of 315 are 1, 3, 5, 7, 9, 15, 21, 35, 45, 63, 105, 315. Its unitary prime divisors are 5 and 7. Finally, sigma(315) = 624 and 624 / (5 + 7) = 52.
|
|
MAPLE
|
with(numtheory): P:=proc(q) local a, b, k, n; for n from 1 to q do a:=divisors(n); b:=0;
for k from 1 to nops(a) do if isprime(a[k]) then if gcd(a[k], n/a[k])=1 then b:=b+a[k]; fi; fi; od;
if b>0 then if type(sigma(n)/b, integer) then print(n); fi; fi; od; end: P(10^10);
|
|
MATHEMATICA
|
unitaryPrimeSum[1]=0; unitaryPrimeSum[n_] := Total[(f = FactorInteger[n])[[;; , 1]] * (Boole[# == 1]& /@ f[[;; , 2]])]; Select[Range[500], (ups = unitaryPrimeSum[#]) > 0 && Divisible[DivisorSigma[1, #], ups] &] (* Amiram Eldar, Nov 26 2019 *)
|
|
PROG
|
(PARI) isok(n) = (v = sumdiv(n, d, d*isprime(d)*(gcd(d, n/d)==1))) && ! (sigma(n) % v); \\ Michel Marcus, May 05 2014
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|