A364265 The first term in a chain of at least 3 consecutive numbers each with exactly 6 distinct prime factors (i.e., belonging to A074969).

323567034, 431684330, 468780388, 481098980, 577922904, 639336984, 715008644, 720990620, 726167154, 735965384, 769385252, 808810638, 822981560, 831034918, 839075510, 847765554, 879549670, 895723268, 902976710, 903293468, 904796814, 918520420, 940737005, 944087484, 982059364

Offset: 1

Author

R. J. Mathar, Jul 16 2023

Keywords

• nonn

Comments

To distinguish this from A259349: "Numbers n with exactly k distinct prime factors" means numbers with A001221(n) = omega(n) = k, which specifies that in the prime factorization n = Product_{i>=1} p_i^(e_i), e_i >= 1, the exponents are ignored, and only the size of the set of the (distinct) p_i is considered. In A259349, the numbers n are products of k distinct primes, which means in the prime factorization of n, all exponents e_i are equal to 1. (If all exponents e_i = 1, the n are squarefree, i.e., in A005117.) Rephrased: the n which are products of k distinct primes have A001221(n) = omega(n) = A001222(n) = bigomega(n) = k, whereas the n which have exactly k distinct prime factors are the superset of (weaker) requirement A001221(n) = omega(n) = k. - R. J. Mathar, Jul 18 2023

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A259349 (requires squarefree). Subsequence of A273879.

Cf. A364266 (5 distinct factors).

See also A001221, A001222, A005117.

Numbers divisible by d distinct primes: A246655 (d=1), A007774 (d=2), A033992 (d=3), A033993 (d=4), A051270 (d=5), A074969 (d=6), A176655 (d=7), A348072 (d=8), A348073 (d=9).

Programs

Maple

omega := proc(n)
    nops(numtheory[factorset](n)) ;
end proc:
for k from 1 do
    if omega(k) = 6 then
        if omega(k+1) = 6 then
            if omega(k+2) = 6 then
                print(k) ;
            end if;
        end if;
    end if;
end do:

PARI

upto(n) = {my(res = List(), streak = 0); forfactored(i = 2, n, if(#i[2]~ == 6, streak++; if(streak >= 3, listput(res, i[1] - 2)), streak = 0)); res} \\ David A. Corneth, Jul 18 2023

Formula

a(1) = A138206(3).

{k: A001221(k) = A001221(k+1) = A001221(k+2) = 6}.

Extensions

More terms from David A. Corneth, Jul 18 2023