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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
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
Giovanni Resta, Table of n, a(n) for n = 1..10000 (terms 1-388 from Marius A. Burtea, 389-2808 from David A. Corneth)
EXAMPLE
Given A009223 = 1, 1, 2, 1, 2, 2, 2, 1, 1, 2, 2, 4, 2, 6, 8, 1, 2, 3, ...
1 first divides A009223(1); 2 first divides A009223(3); 3 first divides A009223(14)=6.
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;
v // Marius A. Burtea, Mar 30 2019
CROSSREFS
Sequence in context: A291796 A155886 A319456 * A138959 A171653 A055435
KEYWORD
nonn
AUTHOR
Phil Carmody, Mar 01 2013
STATUS
approved