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A080307
Multiples of the Fermat numbers 2^(2^n)+1.
6
3, 5, 6, 9, 10, 12, 15, 17, 18, 20, 21, 24, 25, 27, 30, 33, 34, 35, 36, 39, 40, 42, 45, 48, 50, 51, 54, 55, 57, 60, 63, 65, 66, 68, 69, 70, 72, 75, 78, 80, 81, 84, 85, 87, 90, 93, 95, 96, 99, 100, 102, 105, 108, 110, 111, 114, 115, 117, 119, 120, 123, 125, 126, 129, 130, 132
OFFSET
1,1
COMMENTS
Since all the Fermat numbers are relatively prime to each other (see link), the probability that a given integer is not a multiple of the first k Fermat numbers is 2^((2^k)-1) / 2^(2^k)-1, the limit of which is 0.5 as k increases infinitely; therefore the probability that an integer is a Fermat multiple, as well as the probability that it is not, is 0.5.
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Matthew Vandermast, Feb 16 2003
STATUS
approved