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A216639
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A027642(6*n+6)/(sequence of period 2:repeat 42,210).
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1
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1, 13, 19, 13, 341, 9139, 43, 221, 19, 270413, 1541, 667147, 79, 16211, 6479, 21437, 103, 996151, 1, 11086933, 103759, 20033, 6533, 11341499, 51491, 8545667, 3097, 16211, 59, 34408161359, 1, 4137341, 5826521, 1339, 219666403, 72719023, 223, 2977, 1501, 45423164501, 83
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OFFSET
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0,2
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COMMENTS
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Is a(n) always an integer? Is there an a(n) ending with 5?
It appears (tested for n <= 800) that a(n) mod 9 is always one of {1, 2, 4, 5, 7, 8}.
There is a similar sequence of ratios A027642(10n+1)/(66*A010686(n)) which starts 1, 1, 217, 41, 1, 172081, 71, 697, 4123, 101, 23, 7055321, 131, 2059, 32767, 697, 1, 21896102683,...
a(n) is always an integer: 42 = 2*3*7 and 1, 2, and 6 divide 12n+6; 210 = 2*3*5*7 and 1, 2, 4, and 6 divide 12n+12. a(n) never ends in 5 (or 0) since 12n+6 is not divisible by 4 hence the (12n+6)-th Bernoulli denominator is not divisible by 5, and Bernoulli denominators are squarefree and hence the (12n+12)-th Bernoulli denominator, divided by 210, cannot be divisible by 5. - Charles R Greathouse IV, Sep 12 2012
The previous comments argue that 3 or 5 are never prime divisors of a(n). In addition (tested up to n <=900), 7 apparently is also a non-divisor of a(n). In summary, the prime divisors appear all to be in A140461. - Jean-François Alcover, Sep 17 2012
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LINKS
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FORMULA
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MAPLE
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op((n mod 2)+1, [1, 5]) ;
end proc:
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PROG
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(PARI) a(n)=my(t=1); fordiv(3*n+3, d, if(isprime(2*d+1), t*=2*d+1)); t/if(n%2, 105, 21) \\ Charles R Greathouse IV, Sep 12 2012
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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