A222714
Smallest i such that prime(n) divides gcd(sigma(i), phi(i)) (cf. A009223).
Original entry on oeis.org
3, 14, 88, 116, 989, 477, 6901, 7067, 6439, 10207, 4976, 10877, 13529, 44461, 79523, 22577, 250277, 62023, 107869, 161027, 75008, 49769, 55277, 183296, 75077, 612463, 381923, 412163, 712423, 153679, 32576, 137549, 450181, 154289, 1776377, 1642577, 491723, 637981, 3903791, 239777, 642251, 1572889, 1608983, 1192739, 2791489, 316409, 888731, 4773091, 4942243, 1256293
Offset: 1
Given A009223 = 1, 1, 2, 1, 2, 2, 2, 1, 1, 2, 2, 4, 2, 6, 8, 1, 2, 3, ...
prime(1)=2 first divides A009223(3); prime(2)=3 first divides A009223(14)=6; prime(3)=5 first divides both sigma(88)=180 and phi(88)=40, so A222714(3)=88.
-
A009223_hunt(x)=local(n=0,g);while(n++,g=A009223(n);if(g%x,,return(n)));
for(x=1,50,print1(A009223_hunt(prime(x))", "))
A307640
Least number k such that n divides gcd(sigma(k), phi(k), tau(k)).
Original entry on oeis.org
1, 3, 18, 15, 3344, 45, 24128, 30, 882, 3344, 1012736, 126, 1953792, 24128, 16200, 168, 452263936, 2016, 1852571648, 3344, 40768, 1012736, 27007123456, 420, 1490000, 1953792, 103968, 24128, 2739920699392, 30096, 8348342681600, 840, 9114624, 452263936, 6163776, 2016
Offset: 1
For n = 2, sigma(3) = 4, phi(3) = 2, tau(3) = 4 are divisible by 2.
For n = 5, sigma(3344) = 7440, phi (3344) = 1440, tau (3344) = 20 are divisible by 5 and by 10.
For n = 11, sigma(1012736) = 2161632 = 11 * 196512, phi(1012736) = 11 * 43008, tau(1012736) = 11 * 4 are divisible by 11.
- Laurențiu Panaitopol, Alexandru Gica, Arithmetic problems and number theory. Ideas and methods of solving, Ed. Gil, Zalău, 2006, ch. 13, p. 79, pr. 18. (in Romanian).
-
for m in [1..16] do
for n in [1..2000000] do
if IsIntegral(SumOfDivisors(n)/m) and IsIntegral(EulerPhi(n)/m) and IsIntegral(NumberOfDivisors(n)/m) then
m,n;
break;
end if;
end for;
end for;
-
Array[Block[{i = 1}, While[Mod[GCD[DivisorSigma[1, i], EulerPhi@ i,DivisorSigma[0, i]], #] != 0, i++]; i] &, 16] (*Adaptation after A222713*)
-
isok(n,k) = ! frac(gcd(sigma(k), gcd(eulerphi(k), numdiv(k)))/n);
a(n) = my(k=1); while(!isok(n,k), k++); k; \\ Michel Marcus, Apr 20 2019
-
a(n) = {if(n==1,return(1)); my(res = oo, f = factor(n), hpf = f[#f~, 1]); forprime(p = 2, oo, if(p ^ (hpf - 1) > res, return(res)); forstep(i = p ^ (hpf - 1), res, p ^ (hpf - 1), if(isok(n, i), res = min(res, i); next(2) ) ) ) } \\ uses isok from above \\ David A. Corneth, Apr 22 2019
Showing 1-2 of 2 results.
Comments