|
The sequence consists of prime numbers and composite numbers.
Let z = m + p, then
1) for any k > 3, the assertion, k == 1 (mod z)
2) if k is a composite number, then it can be represented as k = (z*x + 1) * (z*y + 1), where x and y are integer
Numbers k for which p = 1, can be reposed in the form (2^z + 1)/3, where z is a prime or a pseudoprime to base 2, this statement is true only for the calculated interval. Is this true statement for all positive numbers, remains unclear.
If m == 0 (mod p), then k = (2^z + 1)/(2^p + 1);
if m == 0 (mod 2) and p = m-2, then k = (2^z + 1)/(2^p + 2^p - 2^(m/2) + 1);
if m+3 = 2*p and z == 0 (mod 3), then k = (2^z + 1)/(2^p + 2^(p-1) + 3);
if p=1, then z=m+p is a prime number, or pseudoprime base 2.
And correspondingly the number k has the form k = (2^z + 1)/3.
Moreover, if the number k is composite, then it appears that some divisor also belongs to our sequence.
Example
29
(2^29+1)/3=178956971=59*3033169
178956971 belongs to our sequence, where m=28 p=1 (z=29)
3033169 belongs to our sequence, where m=22 p=7 (z=29)
37
(2^37+1)/3=45812984491=1777*25781083
45812984491 m=36 p=1 (z=37)
25781083 m=25 p=12 (z=37)
41
(2^41+1)/3=733007751851 = 83*8831418697
733007751851 m=40 p=1 (z=41)
8831418697 m=34 p=7 (z=41)
47
(2^47+1)/3=46912496118443 = 283*165768537521
46912496118443 m=46 p=1 (z=47)
165768537521 m=38 p=9 (z=47)
53
(2^53+1)/3=3002399751580331 = 107*28059810762433
3002399751580331 m=52 p=1 (z=53)
28059810762433 m=45 p=8 (z=53)
59
(2^59+1)/3=192153584101141163 =2833*37171*1824726041
192153584101141163 m=58 p=1 (z=59)
1824726041 m=31 p=28 (z=59)
5169448874153=2833*1824726041 m=43 p=16 (z=59)
etc.
| 