

A074777


Integers n such that sigma(phi(n))/n = 1/2.


1




OFFSET

1,1


COMMENTS

Since 2^(2^n)+1 is prime for n=0,1,...,4 (Fermat primes), 2^(2^(n1)+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(n1)  2, a(n+1)/a(n)  2], with a(0) = 1, the numbers x such that u^k + (u+1)^k + ... + (u+x1)^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+x1)^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


LINKS



FORMULA



CROSSREFS



KEYWORD

nonn


AUTHOR

Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Sep 07 2002


EXTENSIONS



STATUS

approved



