|
|
A222713
|
|
Least number k such that n divides gcd(sigma(k), phi(k)) (A009223).
|
|
3
|
|
|
1, 3, 14, 12, 88, 14, 116, 15, 190, 88, 989, 35, 477, 116, 209, 105, 6901, 190, 7067, 88, 196, 989, 6439, 35, 15049, 477, 2754, 172, 10207, 209, 4976, 336, 989, 6901, 1189, 190, 10877, 7067, 477, 248, 13529, 377, 44461, 989, 418, 6439, 79523, 105, 10244, 15049
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
For each n there are infinitely many numbers k for which n divides sigma(k) and phi(k). - Marius A. Burtea, Mar 28 2019
|
|
LINKS
|
|
|
EXAMPLE
|
Given A009223 = 1, 1, 2, 1, 2, 2, 2, 1, 1, 2, 2, 4, 2, 6, 8, 1, 2, 3, ...
|
|
MATHEMATICA
|
Array[Block[{i = 1}, While[Mod[GCD[DivisorSigma[1, i], EulerPhi@ i], #] != 0, i++]; i] &, 50] (* Michael De Vlieger, Mar 28 2019 *)
|
|
PROG
|
(PARI) A009223_hunt(x)=local(n=0, g); while(n++, g=A009223(n); if(g%x, , return(n)));
for(x=1, 50, print1(A009223_hunt(x)", "))
(Magma) [Min([n: n in [1..300000] | IsIntegral(SumOfDivisors(n)/m) and IsIntegral(EulerPhi(n)/m) ]): m in [1..70]]; // Marius A. Burtea, Mar 28 2019
(Magma) v:=[];
for n in [1..60] do
m:=1;
while not EulerPhi(m) mod n eq 0 or not SumOfDivisors(m) mod n eq 0 do
v[n]:=0;
m:=m+1;
end while;
v[n]:=m;
end for;
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|