This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A319150 #46 Dec 14 2024 07:29:21
%S A319150 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 A319150 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 A319150 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 A319150 a(n) = gcd(A275286(n), A001818(n+1)) / A025549(n+1)^2.
%C A319150 A(n) = a(n)*A025549(n+1)^2 = gcd(B(n), C(n)).
%C A319150 B(n) = A275286(n).
%C A319150 C(n) = A001818(n+1).
%C A319150 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 A319150 Let S(n) = Sum_{k=0..n} ((-1)^k)/(2*k+1)^2 (S(n) is NOT an integer sequence).
%C A319150 Notice that when n approaches +oo, D(n) converges to Catalan's constant (A006752).
%C A319150 A294970(n) is equal to the numerator of S(n) (when reduced).
%C A319150 Therefore B(n)/A(n) = A294970(n)
%C A319150 A294971(n) is equal to the denominator of S(n) (when reduced).
%C A319150 Therefore C(n)/A(n) = A294971(n).
%C A319150 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 A319150 Explicit formula:
%F A319150 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 A319150 A few relations:
%F A319150 gcd(A275286(n), A001818(n+1)) = a(n)*A025549(n+1)^2 = A(n);
%F A319150 A275286(n)/A(n) = A294970(n);
%F A319150 A001818(n+1)/A(n) = A294971(n);
%F A319150 Limit_{n->+oo} A294970(n)/A294971(n) = G (Catalan's Constant, decimal expansion: A006752).
%e A319150 For n = 5:
%e A319150 B(n) = A275286(5) = 98607816;
%e A319150 C(n) = A001818(5+1) = 108056025;
%e A319150 gcd(98607816,108056025) = A(5) = 9;
%e A319150 A025549(5+1)^2 = 3^2 = 9;
%e A319150 So a(5) = A(5)/A025549(5+1)^2 = 9/9 = 1.
%t A319150 a[n_] := GCD[(2n+1)!!^2 * Sum[(-1)^k/(2k+1)^2, {k, 0, n}], (2n+1)!!^2]*
%t A319150 LCM @@ Range[1, 2n+1, 2]^2 / ((2n+1)!!)^2; Array[a, 100, 0] (* _Amiram Eldar_, Nov 16 2018 *)
%o A319150 (PARI) dfo(n) = (2*n)! / n! / 2^n;
%o A319150 a6(n) = dfo(n+1)^2*sum(k=0, n, (-1)^k/(2*k+1)^2);
%o A319150 a8(n) = ((2*n)!/(n!*2^n))^2;
%o A319150 a9(n) = (((2*n)!/n!)/2^n)/lcm(vector(n, i, 2*i-1));
%o A319150 a(n) = gcd(a6(n) , a8(n+1)) / a9(n+1)^2; \\ _Michel Marcus_, Nov 08 2018
%Y A319150 Cf. A275286, A001818, A025549, A294970, A294971, A006752.
%K A319150 nonn
%O A319150 0,28
%A A319150 _Tristan Cam_, Nov 08 2018