A374023 Numbers m such that m .. m+11 all have the same number of prime factors, counted with multiplicity.

3195380868, 5208143601, 5208143602, 5327400945, 5604994082, 5604994083, 6940533603, 6940533604, 7109053186, 7112231268, 19355940562, 22180594465, 24073076004, 24155988484, 29495293764, 30997967601, 41999754228, 42322452483, 42322452484, 45479198003, 46553917683

Offset: 1

Author

Zak Seidov and Robert Israel, Jun 25 2024

Keywords

• nonn

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.

Examples

			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.
		

Links

• Martin Ehrenstein, Table of n, a(n) for n = 1..10000

Crossrefs

Subsequence of A033987.

Cf. A001222.

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).

Programs

PARI

isok(m) = #Set(apply(bigomega, vector(11, i, m+i-1))) == 1; \\ Michel Marcus, Jul 11 2024

Formula

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

Extensions

Missing term inserted by, and more terms from Martin Ehrenstein, Jul 11 2024