Folding a strip of stamps

Abstract This paper answers the problem of determining how many ways a strip of N stamps can be folded along the perforations so that the stamps are piled one on top of the other without destroying the continuity of the strip. Using combinatorial arguments, the problem is first solved by an inductive formula for N odd. Part of the computations involved are seen to yield an answer to the case when N is even. Three cases are considered: (1) both, (2) one of, (3) neither, the left end and top of the strip are labeled. If P(N) is the number of folds for a strip of N stamps of type (1), then 1/2P(N) and 1/4P(N)+1/4S(N) yields formulas for the number of folds for a strip of N stamps of types (2) and (3), where S(N) is the number of symmetric folds of the strip. P(N) also equals the number of ways of joining N dots on a circle by chords of alternating blue and red color without having any chords of the same color intersect. The formulas for P(N) and S(N) are too complicated to give here. However, the first few values are listed by computer as: N P(N) S(N) N P(N) S(N) 3 6 2 10 14060 48 4 16 4 11 46310 178 5 50 6 12 146376 132 6 144 8 13 485914 574 7 462 18 14 1557892 348 8 1392 20 15 5202690 1870 9 4536 56 16 16861984 1008