Is this function decreasing on $(0,1)$?

Solution 1:

A few more terms for those investigating. From Maple. These coefficients are not listed in the On-line Encyclopedia of Integer Sequences.

$$ \frac{4}{\pi} \sqrt{m} \;K(4 \sqrt{m}) \sinh \biggl(\frac{\pi\; K(\sqrt{1 - 16 m})}{2\;K(4 \sqrt{m})}\biggr) \\ = 1 - m - 6 m^{2} - 54 m^{3} - 575 m^{4} - 6715 m^{5} - 83134 m^{6} - 1071482 m^{7} - \\ \quad{}\quad{} 14221974 m^{8} - 193050435 m^{9} - 2667157340 m^{10} - 37378279402 m^{11} - \\ \quad{}\quad{} 530024062361 m^{12} - 7590192561912 m^{13} - \\ \quad{}\quad{}109610113457650 m^{14} - 1594344146568120 m^{15} - \\ \quad{}\quad{}23336667998911128 m^{16} - 343468859344118109 m^{17} - \\ \quad{}\quad{}5079858166426507168 m^{18} - 75457168334744888190 m^{19} - \\ \quad{}\quad{}1125223725054635766392 m^{20} + \operatorname{O} \bigl(m^{21}\bigr) $$

added

Who knows if this is relevant? See A002849 $$ \frac{2}{\pi}K(4\sqrt{m}) = 1+4m+36m^2+400m^3+4900m^4+\dots =\sum_{n=0}^\infty \binom{2n}{n}^2m^n $$

Solution 2:

See the developments here. It seems all that is left is (reasonable) numerical work.