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 A277254 Numbers n such that p = n - phi(n) < q = n - lambda(n), and p and q are both primes, where phi(n) = A000010(n) and lambda(n) = A002322(n). 1
 15, 33, 35, 65, 77, 87, 91, 95, 119, 123, 143, 185, 215, 221, 247, 255, 259, 287, 329, 341, 377, 395, 407, 427, 437, 455, 473, 485, 511, 515, 537, 573, 595, 635, 705, 713, 717, 721, 749, 767, 779, 793, 795, 803, 805, 815, 817, 843, 869, 871, 885, 899, 923, 965, 1001 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Numbers n such that p = A051953(n) < q = A277127(n), and p and q are both primes. If n is such number, then b^p == b^q (mod n) for every integer b. Problem: are there infinitely many such numbers? Suppose p^2 divides n. Then p divides n - phi(n), and so the only way n - phi(n) can be prime is if n = p^2. But then n - phi(n) = n - A002322(n). Hence all terms in this sequence are squarefree. - Charles R Greathouse IV, Oct 08 2016 All terms are odd composites. - Robert Israel, Oct 09 2016 It seems that gpf(n) < p = n - phi(n). - Thomas Ordowski, Oct 09 2016 LINKS Robert Israel, Table of n, a(n) for n = 1..10000 EXAMPLE For n=15, A051953(15) = 7, A277127(15) = 11, 7 < 11 and both are primes, thus 15 is included in the sequence. MAPLE filter:= proc(n) uses numtheory;   local p, q;   p:= n-phi(n);   q:= n-lambda(n);   p

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Last modified May 12 16:02 EDT 2021. Contains 343825 sequences. (Running on oeis4.)