Two conjectural series for $\pi$ involving the central trinomial coefficients

Question

For each $n=0,1,2,\ldots$, the central trinomial coefficient $T_n$ is defined as the coefficient of $x^n$ in the expansion of $(x^2+x+1)^n$. It is easy to see that $T_n=\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}\binom{2k}k$.

On December 7, 2019, I conjectured that $$\sum_{k=1}^\infty\frac{(105k-44)T_{k-1}}{k^2\binom{2k}k^23^{k-1}}=\frac{5\pi}{\sqrt3}+6\log3\tag{1}$$ and $$\sum_{k=2}^\infty\frac{(5k-2)T_{k-1}}{k^2\binom{2k}k^2(k-1)3^{k-1}}=\frac{21-2\sqrt3\,\pi-9\log3}{12}.\tag{2}$$ As the two series converge very fast, it is easy to check (1) and (2) numerically. The two identities and related congruences appear in Section 10 of my recent preprint New series for powers of $\pi$ and related congruences. I'm unable to find proofs of $(1)$ and $(2)$. So, here I ask the following question.

Question. How to prove the conjectural identities $(1)$ and $(2)$?

Your comments are welcome!

Answer 1

Not an answer, but a reduction to a definite integral.

First, Lagrange inversion implies that the generating function for $T_k$ is $${\cal T}(z):=(1-2z-3z^2)^{-\frac12}=((1+z)(1-3z))^{-\frac12},$$ and thus $T_k = [z^k]\ {\cal T}(z)$. Notice that $${\cal T}'(z) = \frac{1+3z}{(1+z)(1-3z)}{\cal T}(z).$$

Second, by the property of the beta function, we have $$\frac{1}{k\binom{2k}k} = B(k+1,k) = \int_0^1 x^k(1-x)^{k-1}\,{\rm d}x = \frac12 \int_0^1 x^{k-1}(1-x)^{k-1}\,{\rm d}x.$$

It follows that $$I_1:=\sum_{k\geq 1} \frac{T_{k-1}}{k^2\binom{2k}k^2 3^{k-1}} = \frac14 \int_0^1\int_0^1 {\cal T}\big( \frac{x(1-x)y(1-y)}{3} \big)\,{\rm d}x{\rm d}y$$ and $$I_2 := \sum_{k\geq 1} \frac{(k-1) T_{k-1}}{k^2\binom{2k}k^2 3^{k-1}} = \frac14 \int_0^1\int_0^1\, \frac{x(1-x)y(1-y)}3\, {\cal T}'\big( \frac{x(1-x)y(1-y)}{3} \big)\,{\rm d}x{\rm d}y.$$

Since $105k-44 = 105(k-1)+61$, the first sum in question equals $105 I_2 + 61 I_1$.

The second sum in question can be reduced to a definite integral similarly.