Quote:
Originally Posted by sweety439
There are other k's excluded from the Riesel/Sierpinski problems (Riesel is still much more such k's)
* R30 k=1369:
for even n let n=2*q; factors to: (37*30^q  1) * (37*30^q + 1)
odd n: covering set 7, 13, 19
* R88 k=400:
for even n let n=2*q; factors to: (20*88^q  1) * (20*88^q + 1)
odd n: covering set 3, 7, 13
* R95 k=324:
for even n let n=2*q; factors to: (18*95^q  1) * (18*95^q + 1)
odd n: covering set 7, 13, 229
* S55 k=2500:
odd n: factor of 7
n = = 2 mod 4: factor of 17
n = = 0 mod 4: let n=4q and let m=5*55^q; factors to: (2*m^2 + 2m + 1) * (2*m^2  2m + 1)
* S200 k=16:
odd n: factor of 3
n = = 0 mod 4: factor of 17
n = = 2 mod 4: let n = 4*q  2 and let m = 20^q*10^(q1); factors to: (2*m^2 + 2m + 1) * (2*m^2  2m + 1)
* S225 k=114244:
for even n let k=4*q^4 and let m=q*15^(n/2); factors to: (2*m^2 + 2m + 1) * (2*m^2  2m + 1)
odd n: factor of 113
* R10 k=343:
n = = 1 mod 3: factor of 3
n = = 2 mod 3: factor of 37
n = = 0 mod 3: let n=3q and let m=7*10^q; factors to: (m  1) * (m^2 + m + 1)
* R957 k=64:
n = = 1 mod 3: factor of 73
n = = 2 mod 3: factor of 19
n = = 0 mod 3: let n=3q and let m=4*957^q; factors to: (m  1) * (m^2 + m + 1)
* S63 k=3511808:
n = = 1 mod 3: factor of 37
n = = 2 mod 3: factor of 109
n = = 0 mod 3: let n=3q and let m=152*63^q; factors to: (m + 1) * (m^2  m + 1)
* S63 k=27000000:
n = = 1 mod 3: factor of 37
n = = 2 mod 3: factor of 109
n = = 0 mod 3: let n=3q and let m=300*63^q; factors to: (m + 1) * (m^2  m + 1)
* R936 k=64:
n = = 0 mod 2: let n = 2q; factors to: (8*936^q  1) * (8*936^q + 1)
n = = 0 mod 3: let n=3q; factors to: (4*936^q  1) * [16*936^(2q) + 4*936^q + 1]
n = = 1 mod 6: factor of 37
n = = 5 mod 6: factor of 109

Of course,
* In the Riesel case if k and b are both rth powers for an r>1
* In the Sierpinski case if k and b are both rth powers for an odd r>1
* In the Sierpinski case if k is of the form 4*m^4, and b is 4th power
Then this k proven composite by full algebraic factors