OFFSET
0,2
COMMENTS
If for every term t > 1 in A002182 there exists a term in A002182 of the form t/p for some prime p|t (Cf. A328523) then (1): a(n+1) is a multiple of a(n) for n >= 1 and (2): this sequence can be used to find terms to A199337 and A330737.
Proof of (1): Let Generation(n) be the terms in A002182 with n prime factors counted with multiplicity. Then a(n) = GCD(Generation(n)). As each term in Generation(n) is of the form a(n) * t for some t and each term in Generation(n + 1) is p * g for some g in Generation(n), a(n + 1) is a multiple of a(n).
Proof of (2): As a(n + 1) is a multiple of a(n) we have that if m | a(n), we also have m | a(n + k), k >= 0 hence the largest number not divisible by m can have at most n - 1 prime factors counted with multiplicity.
EXAMPLE
The terms in A002182 with n = 4 prime divisors counted with multiplicity are 24, 36 and 60. Their GCD is 12 hence a(4) = 12.
MATHEMATICA
(* First, load the function f at A025487, then: *)
Block[{s = Union@ Flatten@ f@ 20, t}, t = DivisorSigma[0, s]; s = Map[s[[FirstPosition[t, #][[1]] ]] &, Union@ FoldList[Max, t]]; t = PrimeOmega[s]; Drop[Array[GCD @@ s[[Position[t, #][[All, 1]] ]] &, Max@ t + 1, 0], -3] ] (* Michael De Vlieger, Jan 12 2020 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
David A. Corneth, Jan 04 2020
STATUS
approved