* Robert Baillie <rjbaillie@frii.com> [Feb 22. 2006 16:39]:
HAKMEM 239 has this interesting question:
"119 (Schroeppel): Can someone square some series for Pi to give the series Pi^2/6 = 1 + 1/2^2 + 1/3^2 + ... ?"
[...]
Somewhat similar vein (but finite sums): sum(k)^2 == sum(k^3) which generalizes to (sum_{d\N}{nu(d)})^2 == sum_{d\N}{nu(d)^3} nu(x) counts divisors of x. N a perfect power gives former. (p.567 of my scribblings) I also like this one (specialized Clausen's product formula): F([1/4,1/4];[1];z)^2==F([1/2,1/2,1/2];[1,1],z) z a sixth power (e.g. z=1) The relation is an identity between the square of a sum of squares and a sum of cubes. (p.502) -- p=2^q-1 prime <== q>2, cosh(2^(q-2)*log(2+sqrt(3)))%p=0 Life is hard and then you die.