A319150 - OEIS

%I #41 Dec 10 2018 10:58:47

%S 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,11,11,11,11,11,

%T 11,11,11,11,11,11,1,1,1,1,1,1,1,1,1,19,19,19,19,19,19,19,19,19,19,19,

%U 19,19,19,19,19,19,19,19,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1

%N a(n) = gcd(A275286(n), A001818(n+1)) / A025549(n+1)^2.

%C A(n) = a(n)*A025549(n+1)^2 = gcd(B(n), C(n)).

%C B(n) = A275286(n).

%C C(n) = A001818(n+1).

%C At first gcd(B(n), C(n)) = A025549(n+1)^2, but from n = 27 to n = 37, gcd(B(n), C(n)) = 11*A025549(n+1)^2, and then comes back to normal, then equals 19*A025549(n+1)^2, comes back to normal again, and so on ...

%C Let S(n) = Sum_{k=0..n} ((-1)^k)/(2*k+1)^2 (S(n) is NOT an integer sequence).

%C Notice that when n approaches +oo, D(n) converges to Catalan's constant (A006752).

%C A294970(n) is equal to the numerator of S(n) (when reduced).

%C Therefore B(n)/A(n) = A294970(n)

%C A294971(n) is equal to the denominator of S(n) (when reduced).

%C Therefore C(n)/A(n) = A294971(n).

%C This sequence was used to study the expression B(n)/C(n) (which equals S(n)) in an attempt to find out if Catalan's constant is irrational.

%F Explicit formula:

%F a(n) = gcd( ((2*n+1)!!)^2 * (Sum_{i=0..n}((-1)^i)/(2*i+1)^2), ((2*n+1)!!)^2 ) / ( (((2*n+1)!!)^2) / ( lcm{1,3,5,...,2*n+1} ) )^2.

%F A few relations:

%F gcd(A275286(n), A001818(n+1)) = a(n)*A025549(n+1)^2 = A(n);

%F A275286(n)/A(n) = A294970(n);

%F A001818(n+1)/A(n) = A294971(n);

%F Lim_{n->+oo) A294970(n)/A294971(n) = G (Catalan's Constant, decimal expansion: A006257).

%e For n = 5:

%e B(n) = A275286(5) = 98607816;

%e C(n) = A001818(5+1) = 108056025;

%e gcd(98607816,108056025) = A(5) = 9;

%e A025549(5+1)^2 = 3^2 = 9;

%e So a(5) = A(5)/A025549(5+1)^2 = 9/9 = 1.

%t a[n_] := GCD[(2n+1)!!^2 * Sum[(-1)^k/(2k+1)^2, {k, 0, n}], (2n+1)!!^2]*

%t LCM @@ Range[1, 2n+1, 2]^2 / ((2n+1)!!)^2; Array[a, 100, 0] (* _Amiram Eldar_, Nov 16 2018 *)

%o (PARI) dfo(n) = (2*n)! / n! / 2^n;

%o a6(n) = dfo(n+1)^2*sum(k=0, n, (-1)^k/(2*k+1)^2);

%o a8(n) = ((2*n)!/(n!*2^n))^2;

%o a9(n) = (((2*n)!/n!)/2^n)/lcm(vector(n, i, 2*i-1));

%o a(n) = gcd(a6(n) , a8(n+1)) / a9(n+1)^2; \\ _Michel Marcus_, Nov 08 2018

%Y Cf. A275286, A001818, A025549, A294970, A294971, A006257.

%K nonn

%O 0,28

%A _Tristan Cam_, Nov 08 2018