

A173406


This sequence starts with any odd, composite number, like 15. There exists a power of two such that every 2^n + s_i is composite, where s_i is a term in the sequence less than 2^n. For example, 128+15=143, 512+15=527, 512+143=655, etc.


0




OFFSET

1,1


COMMENTS

We can easily create a kCNF Boolean IsPrimek() function. Make a list of all composite kbit binary numbers. Assign each bit to a variable and invert the bits to create a kclause. Combining these clauses gives us a CNF IsPrimek() function. For example, the 4clause for 15 (1111 base2) is (~d+~c+~b+~a). The 5clause for 15 (01111) will be (e+~d+~c+~b+~a). We have to add a variable to the clause for each new power of two until we get to 2^7. 128+15 is composite so we can remove the variable for 128.
This sequence is the list of powers of 2 that can be removed from the CNF clause for 15. I still can't prove this is an infinite sequence. Assume this sequence is finite. Then there exists a finite width prime sieve for powers of two. For every large enough power of two we can find a prime by adding one of the numbers in this sequence (assuming the sequence is finite). The sequence can still be used as a prime sieve even if the sequence is infinite, assuming it grows slowly enough.


LINKS

Table of n, a(n) for n=1..8.


CROSSREFS

Sequence in context: A179524 A235864 A177065 * A291619 A071700 A133126
Adjacent sequences: A173403 A173404 A173405 * A173407 A173408 A173409


KEYWORD

nonn,uned


AUTHOR

Russell Easterly, Feb 17 2010


STATUS

approved



