OFFSET
1,2
COMMENTS
It seems that all values are odd. For (b^k+1)/(b+1) to be an integer, it seems k must be odd. 2=(0^0+1)/(0+1) has been excluded since neither b nor k would be positive.
When k is a composite, a(n) is a composite.
These numbers are in the form of 111...1 (k of 1s) base b. - Lei Zhou, Feb 08 2012
LINKS
H. Dubner and T. Granlund, Primes of the Form (b^n+1)/(b+1), J. Integer Sequences, 3 (2000), #P00.2.7.
EXAMPLE
43 is in the sequence since (2^7+1)/(2+1)=129/3=43; indeed also (7^3+1)/(7+1)=344/8=43.
MATHEMATICA
max = 44; maxdata = (1 + max^3)/(1 + max); a = {}; Do[i = 1; While[i = i + 2; cc = (1 + m^i)/(1 + m); cc <= maxdata, a = Append[a, cc]], {m, 2, max}]; Union[a] (* Lei Zhou, Feb 08 2012 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Henry Bottomley, Dec 21 2000
STATUS
approved