[seqfan] Square Riesel numbers

[Tomasz Ordowski]

Dear number lovers!

Is there a Sierpinski number k such that a*P(k) - 2^b = k for some natural
a and b,
where P(k) is the product of all odd primes from the covering set for
k*2^n+1 ?
If so, then k^2 is a Riesel number with the same covering set for
k^2*2^n-1.
Is this a way to construct a square Riesel number smaller than the known
one?

3896845303873881175159314620808887046066972469809^2

In general, the geometric progression a(n) = k*(a*P(k) - 2^b)^n gives
Sierpinski numbers a(2m) and Riesel numbers a(2m+1) for m = 0, 1, 2, ...

Best,

Thomas