A218448 and related sequences

[David Corneth]

Hi all,

I was trying to find more terms for A218448 (First of a run of 5
consecutive numbers with same prime signature.).

No term can be squarefree. I initially thought to loop over squarefree
terms. That eases the search and I have a search going using that.
This got me up to 44116900322 and going.

Then I though we can ease the search more by using CRT. I wrote some
code that uses it along with not being squarefree. I take some
squarefree number product of 5 primes p1, p2, p3, p4 and p5. I then
say I take 5 numbers Mod(r1, p1^2), Mod(r2, p2^2),Mod(r3,
p3^2),Mod(r4, p4^2),Mod(r5, p5^2),
where {r1, r2, r3, r4, r5} = {0, -1, -2, -3, -4}. I chose the product
of primes to be at most 10^6. I know, arbitrarily.
It could well be that the signature has a 3 in it in which case the
CRT might differ a bit.
Using this method I found 10000 terms <= 77446019146 so I know
A218448(10000) <= 77446019146.
But it misses terms. If I remove these terms from the ones I found
algorithmically using the first method I listed I see I miss and I
missed 43 terms. The terms I missed all have a 2 and one or more 1's
in the prime signature.
The product of the primes p1, p2, p3, p4, p5 exceeds 10^6, the largest
product being
95843098 which comes with term 18421248937. 18421248939 is divisible by 1487^2.

Using this method I found 554 terms <= 10^14 for A218865 but again it
might miss terms.
If they do then the product of the primes with multiplicity 2 exceeds
10^8 or some multiplicity is more than 2. If anyone cares, A218865(19)
> 2*10^12 from Donovan Johnson, May 11 2013
The next 5 terms over there are at most:
2004863060241, 2048634640921, 2136553370145, 2321356401849, 2322703420473
respectively.
Can this second method become an efficient algorithm? Or are we stuck
searching between squarefree numbers?

Best,
David