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Numbers m such that m .. m+11 all have the same number of prime factors, counted with multiplicity.
1

%I #29 Jul 12 2024 02:49:27

%S 3195380868,5208143601,5208143602,5327400945,5604994082,5604994083,

%T 6940533603,6940533604,7109053186,7112231268,19355940562,22180594465,

%U 24073076004,24155988484,29495293764,30997967601,41999754228,42322452483,42322452484,45479198003,46553917683

%N Numbers m such that m .. m+11 all have the same number of prime factors, counted with multiplicity.

%C Since a(3) = a(2) + 1, a(6) = a(5) + 1 and a(8) = a(7) + 1, a(2) = 5208143601, a(5) = 5604994082 and a(7) = 6940533603 are the first three m such that m .. m+12 have the same number of prime factors, counted with multiplicity.

%C For n <= 12, A001222(a(n)) = 4. It must always be at least 4 because at least one of a(n) .. a(n)+11 is divisible by 8.

%H Martin Ehrenstein, <a href="/A374023/b374023.txt">Table of n, a(n) for n = 1..10000</a>

%F A001222(a(n)) = A001222(a(n)+1) = ... = A001222(a(n)+11).

%e 5208143601 is a term because

%e 5208143601 = 3 * 139 * 2153 * 5801

%e 5208143602 = 2 * 47 * 4261 * 13003

%e 5208143603 = 13 * 103 * 419 * 9283

%e 5208143604 = 2^2 * 3 * 434011967

%e 5208143605 = 5 * 7^2 * 21257729

%e 5208143606 = 2 * 37 * 109 * 645691

%e 5208143607 = 3^2 * 647 * 894409

%e 5208143608 = 2^3 * 651017951

%e 5208143609 = 73^2 * 367 * 2663

%e 5208143610 = 2 * 3 * 5 * 173604787

%e 5208143611 = 11 * 29 * 1129 * 14461

%e 5208143612 = 2^2 * 7 * 186005129

%e all have 4 prime factors, counted with multiplicity.

%o (PARI) isok(m) = #Set(apply(bigomega, vector(11, i, m+i-1))) == 1; \\ _Michel Marcus_, Jul 11 2024

%Y Subsequence of A033987.

%Y Cf. A001222.

%Y Numbers m through m+k have the same value of A001222: A045920 (k=1), A045939 (k=2), A045940 (k=3), A045941 (k=4), A045942 (k=5), A123103 (k=6), A123201 (k=7), A358017 (k=8), A358018 (k=9), A358019 (k=10).

%K nonn

%O 1,1

%A _Zak Seidov_ and _Robert Israel_, Jun 25 2024

%E Missing term inserted by, and more terms from _Martin Ehrenstein_, Jul 11 2024