A positive integer N needs floor(log2 N)+1 binary digits; a sign and a convention for zero add only constant space. Thus a magnitude bound 2^B gives O(B+1) bits, not O(2^B) bits. Adding L integers of magnitude at most A gives magnitude at most L*A, so only O(log2 L+log2 A+1) bits are needed, even for signed partial sums. Multiplying L such integers can instead require O(L*log2 A+1) bits. This distinction matters because the algorithm adds exponentially many contributions but multiplies only polynomially many factors within each contribution. An arithmetic-operation count alone does not bound bit-operation cost.