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%I #21 Sep 30 2020 09:05:16
%S 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,
%T 41,4
%N 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.
%C This sequence is an inversion of A336486.
%C A236255 contains the primes s = prime(m) for which a(m) = 1.
%C 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).
%C 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.
%e a(6) = 3 because:
%e 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)).
%e 2. There is no such equality for s = 13 and k less than 3.
%p with(NumberTheory):
%p SK := []; for ii to 31 do s := ithprime(ii); tf := false;
%p for k do c := 2*s^(k+1)+1; cc := (c^2-1)*(1/2); Q := Divisors(cc);
%p for d in Q do q := d+c; if isprime(q) then p := c+cc/(q-c); if p < q then break end if;
%p 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;
%p if tf then break end if end do end do; SK
%o (PARI) is(t, u, x, y) = ispseudoprime(t*x+1) && ispseudoprime(u*y/t+1) && ispseudoprime(x*y+1);
%o 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
%Y Cf. A000005, A000010, A000203, A134922, A336486.
%K nonn,more
%O 1,5
%A _Vladimir Letsko_, Jul 23 2020
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