A350069 - OEIS

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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.

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

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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