OFFSET
0,2
COMMENTS
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
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
A. Joyal, Les nombres de Bernoulli (in French), 2003.
MAPLE
A010686 := proc(n)
op((n mod 2)+1, [1, 5]) ;
end proc:
A216639 := proc(n)
end proc: # R. J. Mathar, Sep 21 2012
PROG
(PARI) a(n)=denominator(bernfrac(6*n+6))/if(n%2, 210, 42) \\ Charles R Greathouse IV, Sep 12 2012
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul Curtz, Sep 12 2012
EXTENSIONS
a(20)-a(40) from Charles R Greathouse IV, Sep 12 2012
STATUS
approved