A140461 -id:A140461 - cp's OEIS frontend

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

Offset: 0

Paul Curtz, Sep 12 2012

nonn

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

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
A. Joyal, Les nombres de Bernoulli (in French), 2003.

Cf. A002445, A027762, A165734, A165949.

A010686 := proc(n)
    op((n mod 2)+1,[1,5]) ;
end proc:
A216639 := proc(n)
    A027642(6*n+6)/42/A010686(n) ;
end proc: # R. J. Mathar, Sep 21 2012

a(n)=denominator(bernfrac(6*n+6))/if(n%2,210,42) \\ Charles R Greathouse IV, Sep 12 2012

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

a(n) = A027642(6*n+6)/(42*A010686(n)).

a(20)-a(40) from Charles R Greathouse IV, Sep 12 2012

cp's OEIS Frontend

A216639 A027642(6*n+6)/(sequence of period 2:repeat 42,210).

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