OFFSET
1,1
COMMENTS
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.
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.
LINKS
Martin Ehrenstein, Table of n, a(n) for n = 1..10000
EXAMPLE
5208143601 is a term because
5208143601 = 3 * 139 * 2153 * 5801
5208143602 = 2 * 47 * 4261 * 13003
5208143603 = 13 * 103 * 419 * 9283
5208143604 = 2^2 * 3 * 434011967
5208143605 = 5 * 7^2 * 21257729
5208143606 = 2 * 37 * 109 * 645691
5208143607 = 3^2 * 647 * 894409
5208143608 = 2^3 * 651017951
5208143609 = 73^2 * 367 * 2663
5208143610 = 2 * 3 * 5 * 173604787
5208143611 = 11 * 29 * 1129 * 14461
5208143612 = 2^2 * 7 * 186005129
all have 4 prime factors, counted with multiplicity.
PROG
(PARI) isok(m) = #Set(apply(bigomega, vector(11, i, m+i-1))) == 1; \\ Michel Marcus, Jul 11 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
Zak Seidov and Robert Israel, Jun 25 2024
EXTENSIONS
Missing term inserted by, and more terms from Martin Ehrenstein, Jul 11 2024
STATUS
approved