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User:M. F. Hasler/Notes on Riesel and Sierpinski numbers
Until today, the sequence A076337 was defined as "Riesel numbers: numbers n such that n*2^k - 1 is composite for all k >= 1".
However, if it is not required that n is odd, then for any term n, 2n (and thus n*2^m for all m >= 0) would also be in the sequence. We do not want this, as is manifest from the sequence A101036 which has A076337(1) = 509203 and terms larger than twice this number, but not twice that number which must be there if even terms were allowed.
If n is odd and such that n*2^k - 1 is composite for all k >= 1, then n*2^k - 1 is also composite for k=0 because this gives n - 1 which is an even number > 2 and therefore composite. (Indeed we know that n > 3 since n = 1 and n = 3 are not in this sequence, i.e., they do not have the Riesel property. (For n = 1 we have all Mersenne primes as counter-examples ; for n = 3 we have 3*2 - 1 = 5, 3*2^2 - 1 = 11, 3*2^3 - 1 = 23, ... are all prime.)
So one can "strengthen" the condition "... for all k >= 1" to "... for all k >= 0" without losing any term (and of course without getting any additional term, since the "stronger" condition implies the "weaker" one; the two are actually equivalent).
However, even though including k=0 is possible because it yields (for odd n) an even term that is composite anyway, it does not remove the necessity of including "odd" in the definition: If n is even, then it may well satisfy the condition (n-1 being odd is compatible with being composite) and actually *does* satisfy the condition if it is equal to a Riesel number times some power of two, as already stated.
Exactly the same considerations hold for Sierpinski numbers, which have the analog definition with +1 instead of -1.
Covering sets
It is conjectured that all Riesel numbers can be proved to be Riesel using a covering set, as explained below. But other authors conjecture not even the opposite but an even stronger version saying that there are infinitely many Riesel numbers that to not possess a covering set.
The idea of a covering set is the following: For any prime p, the residue of n*2^k - 1 (mod p) is always periodic, namely with the multiplicative order of 2 in (Z/pZ)*, o(2;p). If there is a zero residue n*2^k' - 1 == 0 (mod p) for some prime p and exponent k' = k'(p), this proves that the number will be composite for all exponents k that differ from k' by a multiple of the period o(2;p). If one has found a collection of primes with zero residues such that all exponents are covered by the corresponding k'(p) + o(2,p).Z, then the property is established.
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Initial version. — MFH 07:09, 23 August 2020 (EDT)