What are some ("small") Riesel numbers without a covering set?

(Answering my own question.)

Indeed, explicit examples of such Riesel numbers are known. We are talking about Riesel numbers $k$ where algebraic factorizations account for some subset of the exponents $n$, and a partial covering takes care of the remaining cases.

The interesting aspect of such numbers $k$, is that $kcdot 2^n - 1$ is composite for all $ninmathbb{N}$ (proven), yet the least prime factor of $kcdot 2^n - 1$ seems to be unbounded as $n$ tends to infinity. See thread linked in the question above.

The algebraic factorization is maybe a bit simpler in the Riesel case ($-1$, as opposed to the $+1$ in Sierpiński). For every $a>1$, we have the high-school factorization of integer polynomials: $$x^a - 1 = (x-1)(x^{a-1}+x^{a-2}+dots+x^2+x+1)$$ So if we pick a $k$ that is a perfect $a$th power, $k=ell^a$, then for all $n$ divisible by $a$ we have that: $$kcdot 2^n - 1 = ell^acdot 2^{ma} - 1 = (ellcdot 2^m)^a - 1$$ is algebraically factorizable from the above formula (take $ellcdot 2^m = x$). So we only have to cover the $n$ that are not divisible by $a$ with some "partial covering" of fixed primes.

So it seems most natural to start with perfect squares ($a=2$). And in fact this has been done in a published paper, namely Michael Filaseta, Carrie Finch, and Mark Kozek: On powers associated with Sierpiński numbers, Riesel numbers and Polignac's conjecture, Journal of Number Theory, Volume 128, Issue 7, July 2008. In it, they use the Riesel number: $$k=3896845303873881175159314620808887046066972469809^2$$ and so all the even exponents $n$ in the expression $kcdot 2^n-1$ are covered by the factorization, and the odd exponents are covered by the prime set: $${ 7, 31, 127, 17, 151, 257, 41, 337, 241, 113, 65537, 71, 97, 673, 1321, 641, 6700417, 281, 14449, 29191 }$$ for which the modulus (least common multiple of the orders of 2 modulo the primes) is $6720$.

I am not sure how Filaseta et al. came up with this covering set, but it may not be so easy to improve on as you might think at first glance. The problem with square Riesel numbers is that the excellent primes $3$ and $5$ (of orders 2 and 4, respectively) cannot be used to cover any positions (that are not already covered by the algebraic factorization). So without those two primes in the set, which normally cover 75% of all positions, and with other primes "lost" as well (see Sierpiński thread linked in question), the covering set becomes big, and so the least $k=ell^2$ (found with the Chinese remainder theorem) becomes big as well.

While I have not come up with a better set than Filaseta et al., I did find smaller $ell$ values that work with the same covering. The best of them being: $$k=1956890318063807573129004316753152275056499243^2$$

However, I expect smaller square Riesel numbers (with the property that no full covering seems to exist) can be found. Has anyone beaten my example here? New records wanted!

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But when I thought about this, in connection to the linked thread, it occurred to me that smaller Riesel numbers of this type should be possible by going from squares to cubes, i.e. $a=3$. I already knew a partial covering for cubes from the other thread, and its $k$ (a Sierpiński) was smaller. I discussed this with user "Gelly", and he found for me the cube: $$k=1469583304447640330447613742^3$$ To prove it is Riesel, all $n$ (from expression $kcdot 2^n-1$) divisible by 3 are handled by the factorization, and for the remaining, use the covering: $${ 3, 5, 127, 17, 43, 257, 29, 113, 5419, 15790321, 449, 2689, 2017 }$$ of modulus $672$. And note that this, just like the other examples mentioned, seems to be a $k$ that does not admit a full covering set. When $n$ is $39 pmod{672}$ or $375 pmod{672}$, we conjecture that no finite covering suffices. So we think that the least prime factor of $1469583304447640330447613742^3cdot 2^{672m+39}-1$ (and the same with exponent $672m+375$) is unbounded.

Again, I think this cube can be "improved", i.e. a smaller example found. Has this been done by someone already? References or new small/minimal examples are very welcome.

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Conclusion: Perfect powers such as $1956890318063807573129004316753152275056499243^2$ (found by me, with the covering set from the linked paper) and $1469583304447640330447613742^3$ (found by user "Gelly") are proven Riesel numbers for which it seems that no covering sets exist.