document.write( "Question 962850: 0,3,2,2,4,4,6,6,9,9,_,
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Algebra.Com's Answer #588382 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "The next term is the 11th term, so we only need to look at the\r\n" );
document.write( "odd terms.\r\n" );
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document.write( "The odd numbered terms are the number of primes less than (n+1)²/4\r\n" );
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document.write( "term #1 = 0 = the number of primes less than (1+1)²/4=1. \r\n" );
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document.write( "term #3 = 2 = the number of primes less than (3+1)²/4=4. They are 2,3 \r\n" );
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document.write( "term #5 = 4 = the number of primes less than (5+1)²/4=9. They are 2,3,5,7 \r\n" );
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document.write( "term #7 = 6 = the number of primes less than (7+1)²/4=16. They are 2,3,5,7,11,13\r\n" );
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document.write( "term #9 = 9 = the number of primes less than (9+1)²/4²=25. They are 2,3,5,7,11,13,17,19,23\r\n" );
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document.write( "term #1 = 11 = the number of primes less than [(11+1)/2]²=36. They are 2,3,5,7,11,13,17,19,23,29,31\r\n" );
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document.write( "So the answer is 11.\r\n" );
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document.write( "You don't need the pattern for the even terms.\r\n" );
document.write( "But let's find it anyway: \r\n" );
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document.write( "The even umbered terms go 3,2,4,6,9,6,8,12\r\n" );
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document.write( "3,2, each times 1, gives 3,2\r\n" );
document.write( "2,3, each times 2, gives 4,6\r\n" );
document.write( "3,2, each times 3, gives 9,6\r\n" );
document.write( "2,3, each times 4, gives 8,12\r\n" );
document.write( "3,2, eact times 5, gives 15,10\r\n" );
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document.write( "Edwin
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