OFFSET
1,1
COMMENTS
Since 2^(2^n)+1 is prime for n=0,1,...,4 (Fermat primes), 2^(2^(n-1)+1)-2 is in the sequence for n=1,2,...,6. Conjecture: There are no further terms. - Farideh Firoozbakht, Sep 14 2004
For k of the form 2^m and in the interval [a(n)/a(n-1) - 2, a(n+1)/a(n) - 2], with a(0) = 1, the numbers x such that u^k + (u+1)^k + ... + (u+x-1)^k is prime for some u are the divisors of a(n) (excluding 1 as a divisor for n > 1).
Example: n = 4. The interval [a(4)/a(3) - 2, a(5)/a(4) - 2] = [15, 255]. The numbers of the form 2^m for some m in this interval are 16 = 2^4, 32 = 2^5, 64 = 2^6, and 128 = 2^7. Taking k = 16 (for example), numbers x such that u^16 + (u+1)^16 + ... + (u+x-1)^16 is prime for some u are {2, 3, 5, 6, 10, 15, 17, 30, 34, 51, 85, 102, 170, 255, 510} which are the divisors of a(4). This is also true when the exponent is 32, 64, or 128. - Derek Orr, Jun 13 2014
FORMULA
For n=1, 2, ..., 6, a(n)=2^(2^(n-1)+1)-2. - Farideh Firoozbakht, Sep 14 2004
CROSSREFS
KEYWORD
nonn
AUTHOR
Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Sep 07 2002
EXTENSIONS
8589934590 from Farideh Firoozbakht, Sep 14 2004
STATUS
approved