Knuth, TAOCP, vol.3, p.644, ex.27 gives the following identity, attributed to Wrench: sum(k>=1, A(k)*x^k / (1-x^k) ) = sum(k>=1, x^(k*k) * (A(k) + sum(j>=1, (A(k) + A(k+j)) * x^(k*j)) ) ) This can be verified by writing the terms in sum(k>=1, sum(j>=1, A(k) * x^(k*j) ) ) into a rectangular table and taking sums starting from the diagonal entries, simultaneously to the right and downwards. Just today I managed to find a generalization: sum(k>=1, A(k)*t^k / (1 - B(k)*x^k) ) = sum(k>=1, x^(k*(k-1)) * t^k * (A(k)*B(k)^(k-1) + sum(j>=1, ( A(k)*B(k)^(k+j-1)*x^j + A(k+j)*B(k+j)^(k-1)*t^j ) * x^(j*(k-1)) ) ) ) Setting t=x and B(k)=1 gives Wrench's identity. This is of a kind of formulas I am extremely fond of. Plus it is (at least computationally) even more general than R. P. Agarwal's identity for sum(n>=0, t^n/(1-x*q^n)) (see http://arxiv.org/abs/1202.6525 ) Should I sent this one to Knuth? Best, jj Pari code to verify: ----------------------- N=33; x='x+O('x^N); A(k)=eval(Str("a" k)); B(k)=eval(Str("b" k)); t2 = sum(k=1,N, A(k)*t^k / (1 - B(k)*x^k) ); { t6 = sum(k=1,N, x^(k*(k-1)) * t^k * (A(k)*B(k)^(k-1) + sum(j=1,N, ( A(k)*B(k)^(k+j-1)*x^j + A(k+j)*B(k+j)^(k-1)*t^j ) * x^(j*(k-1)) ) ) ); } t2-t6 \\ == zero (up to orders x^N and t^N) -----------------------