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A328489 Odd numbers k such that the four consecutive odd numbers starting with k have a total of 5 prime factors counting multiplicity. 1
3, 5, 7, 11, 13, 17, 37, 67, 107, 307, 877, 1297, 2267, 2657, 3457, 3847, 3917, 4787, 4967, 5737, 11827, 12037, 14627, 16447, 18127, 18517, 19417, 20477, 20747, 20897, 21377, 21557, 22567, 22637, 23057, 23557, 23627, 25577, 29567, 31387, 32057, 33347, 33767, 34757, 35797, 36467, 36787, 37307 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
Numbers k such that A001222(A190577(k))=5.
There are three cases:
k=3.
k, k+4 and k+6 are primes while k+2 is 3 times a prime.
k, k+2 and k+6 are primes while k+4 is 3 times a prime.
All terms > 13 have final digit 7.
The first n for which a(n+1)-a(n)=10 is 7538. - Robert Israel, Oct 19 2019
LINKS
EXAMPLE
a(3)=7 is in the sequence because 7*9*11*13 is the product of exactly 5 primes: 3*3*7*11*13.
MAPLE
A1:= select(t -> isprime((t+2)/3) and isprime(t) and isprime(t+4) and isprime(t+6), {seq(i, i=7..100000, 30)}):
A2:= select(t -> isprime((t+4)/3) and isprime(t) and isprime(t+2) and isprime(t+6), {seq(i, i=17..100000, 30)});
sort(convert({3, 5, 11, 13} union A1 union A2, list));
CROSSREFS
Sequence in context: A153602 A136187 A333207 * A130122 A288893 A046494
KEYWORD
nonn
AUTHOR
J. M. Bergot and Robert Israel, Oct 16 2019
STATUS
approved

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Last modified April 16 07:57 EDT 2024. Contains 371698 sequences. (Running on oeis4.)