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 A350069 Semiprimes k such that 1+(2^(1+A243055(k))) is a Fermat prime, where A243055(k) gives the difference between the indices of the smallest and the largest prime divisor of k. 1
 4, 6, 9, 14, 15, 25, 33, 35, 38, 49, 65, 69, 77, 106, 119, 121, 143, 145, 169, 177, 209, 217, 221, 289, 299, 305, 323, 361, 407, 437, 469, 493, 529, 533, 589, 667, 731, 781, 841, 851, 893, 899, 949, 961, 1147, 1189, 1219, 1333, 1343, 1369, 1517, 1577, 1681, 1711, 1739, 1763, 1849, 1891, 2021, 2047, 2173, 2209, 2479 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 LINKS Table of n, a(n) for n=1..63. Index entries for sequences computed from indices in prime factorization EXAMPLE 9 is a semiprime (9 = 3*3), and as the difference between the indices of the smallest (3) and the largest prime (3) dividing 9 is 0, we have 1+(2^(1+A243055(k))) = 3, which is in A019434, and therefore 9 is included in this sequence, like all squares of primes (A001248). 177 = 3 * 59 = prime(2) * prime(17), therefore A243055(177) = 17-2 = 15, and as 1+(2^16) = 65537 is also in A019434, 177 is included in this sequence. PROG (PARI) A209229(n) = (n && !bitand(n, n-1)); A243055(n) = if(1==n, 0, my(f = factor(n), lpf = f[1, 1], gpf = f[#f~, 1]); (primepi(gpf)-primepi(lpf))); isA350069(n) = if(2!=bigomega(n), 0, my(d=1+A243055(n)); (A209229(d) && isprime(1+(2^d)))); CROSSREFS Positions of ones in A342651. Cf. A019434, A209229, A243055, A334101. Subsequence of A001358. A001248 is a subsequence. Sequence in context: A182150 A108634 A135355 * A245383 A050940 A190434 Adjacent sequences: A350066 A350067 A350068 * A350070 A350071 A350072 KEYWORD nonn AUTHOR Antti Karttunen, Jan 29 2022 STATUS approved

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Last modified December 5 10:56 EST 2023. Contains 367589 sequences. (Running on oeis4.)