A336485 The least positive integer k for which there exist primes p, q, r such that phi(p*q*s^k) = phi(r*s^(2k+1)) and sigma(p*q*s^k) = sigma(r*s^(2k+1)), where s is the n-th prime.

1, 1, 1, 1, 17, 3, 29, 4, 4, 4, 1, 5, 4, 1, 20, 32, 2, 38, 12, 29, 9, 4, 26, 20, 8, 14, 2, 14, 8, 41, 4

Offset: 1

Author

Vladimir Letsko, Jul 23 2020

Keywords

• nonn
• more

Comments

This sequence is an inversion of A336486.

A236255 contains the primes s = prime(m) for which a(m) = 1.

Let "a and b are similar" mean that for positive integers a and b we have phi(a) = phi(b), tau(a) = tau(b) and sigma(a) = sigma(b).

Conjecture: For each prime s there are infinitely many positive integers k such that numbers p*q*s^k and r*s^(2k+1) are similar for some primes p, q, r.

Examples

			a(6) = 3 because:
1. For the 6th prime, s = 13, k = 3 and with primes p = 62807837, q = 57149, r = 125672849 we have phi(p*q*s^k) = phi(r*s^(2k+1)) and sigma(p*q*s^k) = sigma(r*s^(2k+1)).
2. There is no such equality for s = 13 and k less than 3.
		

Crossrefs

Cf. A000005, A000010, A000203, A134922, A336486.

Programs

Maple

with(NumberTheory):
SK := []; for ii to 31 do s := ithprime(ii); tf := false;
for k do c := 2*s^(k+1)+1; cc := (c^2-1)*(1/2); Q := Divisors(cc);
for d in Q do q := d+c; if isprime(q) then p := c+cc/(q-c); if p < q then break end if;
if isprime(p) then r := 2*(p+q)-c; if isprime(r) then print([s, [p, q], r], k); SK := [op(SK), [s, k]]; tf := true; break end if end if end if end do;
if tf then break end if end do end do; SK

PARI

is(t, u, x, y) = ispseudoprime(t*x+1) && ispseudoprime(u*y/t+1) && ispseudoprime(x*y+1);
a(n) = {my(s=prime(n), t, u); for(k=1, oo, for(i=0, 1+k\2, t=s^i; fordiv(2*(1+u=s^(k+1)), d, if(is(t, u, 2*u/t+d, 2*t+(2*u+2)/d) || is(t, u, 2*u/t-d, 2*t-(2*u+2)/d), return(k))))); } \\ Jinyuan Wang, Sep 30 2020