Note that a uniform distribution isn't possible over an infinite line.
My strategy would be to move to positions 1, -2, +4, -8, 16, etc, then we get there in n + 2*(2^0 + 2^1 + ... 2^r) where 2^r < n.
(We always get there having moved n distance from the origin as our last step, otherwise we travel back to the origin spending distances 2, 4, 8, 16, etc. The sum (1, 2, ... 2^r) = 2^(r+1)-1; We double that again to get ~2^(r+2)
An example of the worst case, we lie on a number of the form -(2^r + 1) we might round-trip to +16 before heading down to -9. In that case we'd travel 2,4,8,16,32 + 9 = 71 steps.
Best case, say we lie on a number of the form 2^2r, e.g. 64.
We travel 1, -2, +4, ... , -32, 64.
We've travelled 127 steps.
Overall this takes method travels approximately between 2n and 8n.
I'm not sure how this compares to 1, -1, 2, -2, 4, -4; That has a better worst case.
> We travel 1, -2, +4, ... , -32, 64.
> We've travelled 127 steps.
I think it's 190 = 64+2*(2^0+2^1+2^2+2^3+2^4+2^5) = 64+2+4+8+16+32+64
> Overall this takes method travels approximately between 2n and 8n.
I think asymptotically the range is [3n,9n]
Program the robot so that it finds the pebble, optimize for the distance traveled as the function of the initial distance between the robot and the pebble. The robot can tell the direction (say east from west) and measure the distance traveled.