The obvious order to tabulate coefficients of a bilateral series: In[744]:= Riffle[1 - #, #] &@Range@5 Out[744]= {0, 1, -1, 2, -2, 3, -3, 4, -4, 5} In[745]:= FindSequenceFunction@% Out[745]= 1/4 (-1)^#1 (-1 + (-1)^#1 + 2 #1) & In[746]:= LeafCount@% Out[746]= 19 (Codegolfers: Do it in 13.) But where in my Fourier table is the, say, -13th harmonic? I.e., what is Out[747]= InverseFunction[1/4 (-1)^#1 (-1 + (-1)^#1 + 2 #1) &] ? Mathematica left it unsolved. Try it at 2: In[748]:= %@2 // N Out[748]= -41.98942986800845 - 1.414587830424065 I And I am the Virgin Mary. The fancy, ordered-pairs version of FindSequenceFunction failed also. It is a nice little puzzle to find a terse expression for this function. But further spazzing with Mathematica turned up the weird identity <terse expression> = In[741]:= 1/2 - I ProductLog[Abs@#, (2 # - 1/2) π]/π & In[742]:= N[% /@ {0, 1, -1, 2, -2, 3, -3, 4, -4}] // Chop // Rationalize Out[742]= {1, 2, 3, 4, 5, 6, 7, 8, 9} This uses non-obvious(?) properties of ProductLog: 1/2 - (I ProductLog[k, -(1/2) (1 - 4 k) π])/π == 2 k (Positive integer k) 1/2 - (I ProductLog[k, -(1/2) (1 + 4 k) π])/π == 1 + 2 k (Integer k) (π - 2 I ProductLog[-k, -(1/2) (1 + 4 (-1 + k)) π])/(2 π) == 2 - 2 k (Positive integer k) --rwg