OFFSET
1,1
COMMENTS
Does such k exist (so that a(n) is nonzero) for all n? These binary trinomials can also be written as f*2^n+1, where f=2^m+1 for some m, which is reminiscent of the Sierpinski problem (see A076336). Hence if there are no Sierpinski numbers of the form 2^m+1, then a(n) is nonzero for all n.
The PFGW program was used to find a(32), which produces a 138012-digit probable prime. If a(256) is nonzero, it is greater than 10^6.
LINKS
T. D. Noe, Table of n, a(n) for n = 1..255
Henri Lifchitz and Renaud Lifchitz (Editors), Search for 2^n+2^m+1, PRP Top Records.
MATHEMATICA
mx=4000; Table[s=1+2^n; k=1; While[k==n || (k<mx && !PrimeQ[s+2^k]), k++ ]; If[k==mx, 0, k], {n, 100}]
CROSSREFS
Cf. A095056.
KEYWORD
nonn
AUTHOR
T. D. Noe, Sep 26 2007
EXTENSIONS
Edited by Peter Munn, Sep 29 2024
STATUS
approved