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 A276044 Least k such that phi(k) has exactly n divisors. 2
 1, 3, 5, 7, 17, 13, 85, 31, 37, 65, 1285, 61, 4369, 193, 185, 143, 65537, 181, 327685, 241, 577, 3281, 5570645, 403, 1297, 12289, 1057, 1037, 286331153, 779, 1431655765, 899, 9509, 197633, 5629, 1333 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Least k such that A000005(A000010(k)) = n. From Jon E. Schoenfield, Nov 13 2016: (Start) For every n > 0, phi(2^n) = 2^(n-1) has exactly n divisors, so a(n) <= 2^n. For every prime p, since phi(a(p)) has exactly p divisors, phi(a(p)) must be of the form q^(p-1), where q is a prime number. If q >= 3, we would have phi(a(p)) >= 3^(p-1), and since k > phi(k) for every k > 1, we would have a(p) >= 3^(p-1)+1, which would be contradicted by the upper bound a(p) <= 2^p (see above) unless 3^(p-1)+1 <= 2^p, which is true only for p = 2. Thus, for every prime p > 2, phi(a(p)) = 2^(p-1), so a(p) > 2^(p-1). In summary, we can state that, for every prime p > 2: (1) a(p) is the least k such that phi(k) = 2^(p-1), and (2) 2^(p-1) < a(p) <= 2^p. After a(36)=1333, the next few known terms are a(38)=786433, a(39)=42653, a(40)=1763, and a(42)=2993; as shown above, known bounds on a(37) and a(41) are 2^36 < a(37) <= 2^37 and 2^40 < a(41) <= 2^41. For prime p < 37, a(p) = A001317(p-1). Observation: for prime p < 37, a(p) is the product of distinct Fermat primes 2^(2^j)+1 for j=0..4, i.e., 3, 5, 17, 257, and 65537 (see A019434), according to the locations of the 1-bits in p-1: .                                               p-1 in    p        a(p)   prime factorization of a(p)  binary   ==  ==========   ===========================  ======    2           3 =                        3          1    3           5 =                    5             10    5          17 =               17                100    7          85 =               17 * 5            110   11        1285 =         257      * 5           1010   13        4369 =         257 * 17               1100   17       65537 = 65537                         10000   19      327685 = 65537            * 5          10010   23     5570645 = 65537       * 17 * 5          10110   29   286331153 = 65537 * 257 * 17              11100   31  1431655765 = 65537 * 257 * 17 * 5          11110 . This pattern does not continue to p=37, since 2^(2^5)+1 is not prime. (See also A038183 and the observation there from Arkadiusz Wesolowski.) Does a(p) = 2^p for all primes p > 31? (End) LINKS EXAMPLE a(5) = 17 because phi(17) = 16 has 5 positive divisors. MATHEMATICA Table[k = 1; While[DivisorSigma[0, #] &@ EulerPhi@ k != n, k++]; k, {n, 28}] (* Michael De Vlieger, Aug 21 2016 *) PROG (PARI) a(n) = {my(k = 1); while(numdiv(eulerphi(k)) != n, k++); k; } CROSSREFS Cf. A000005, A000010, A001317, A019434, A038183, A062821. Sequence in context: A112092 A031441 A078150 * A114265 A258195 A110358 Adjacent sequences:  A276041 A276042 A276043 * A276045 A276046 A276047 KEYWORD nonn,more AUTHOR Altug Alkan, Aug 17 2016 EXTENSIONS a(31)-a(36) from Michel Marcus and Jon E. Schoenfield, Nov 13 2016 STATUS approved

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Last modified June 25 10:10 EDT 2019. Contains 324351 sequences. (Running on oeis4.)