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%I #25 Mar 20 2022 12:40:34
%S 37,53,67,97,127,137,149,157,163,191,211,223,251,257,263,277,293,307,
%T 331,337,347,367,373,397,409,419,431,457,479,487,499,521,541,547,557,
%U 563,577,587,593,613,631,641,653,673,691,701,709,719,727,751,757,769,787,797,809,821,839,853,877,907,919,929,937,953,967,977
%N Primes with index k >= 3 such that A288189(prime(k)) < A295185(prime(k)).
%C Let A,B,X represent A288189, A295185, A056240 respectively. A(p) is defined for every prime, B(p) is defined for primes >= 5. For a prime p with index k >= 3, A(p) = X(t)(rp-t) for some multiple r of p, and some integer t such that rp-t is prime. Then Sopfr(A(p)) = Sopfr(X(t))+(rp-t) = t+rp-t = rp. B(p) = X(g)(p-g) where g = p-q for some prime q = p-g < p. q is the greatest prime divisor of A295185(p), so Sopfr(B(p)) = p. A(p) < B(p) if r and t exist such that (rp-t) is prime, with X(t)(rp-t) < X(g)(p-g). A(p) is computed from the list of possible values in the list of inequalities: 3(2p-3) < 2(3p-2) < 5(2p-5) < 2(5p-2) < ... < X(g)(p-g), selecting the first (smallest) value of (rp-t) which is prime. If such a term exists and is < X(p)(p-g), then A(p) < B(p) and p is in this sequence. Otherwise A(p) = B(p) = X(p)(p-g) and p is in A299760.
%e k=12, prime(12)=37, A288189(37) = 213 < 248 = A295185(37). 37 is the smallest prime with this property, so a(1)=37.
%o (PARI) sopfr(k) = my(f=factor(k)); sum(j=1, #f~, f[j, 1]*f[j, 2]);
%o ap288189(p) = forcomposite(c=p, , if (!(sopfr(c) % p), return(c)));
%o ap295185(p) = forcomposite(c=p, , if (sopfr(c) == p, return(c)));
%o isokp(p) = (ap288189(p) < ap295185(p));
%o lista(nn) = forprime(p=5, nn, if (isokp(p), print1(p, ", "))); \\ _Michel Marcus_, May 13 2018
%Y Cf. A288189, A295185, A299760, A299760.
%K nonn
%O 1,1
%A _David James Sycamore_, Apr 12 2018
%E a(53) corrected by _Georg Fischer_, Mar 20 2022
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