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Consider r rectangular cards stacked in a pile with their left and lower edges aligned. Each is of a different color and their widths and heights are independent permutations of the integers 1, 2, ..., r. Then the sequence gives the number of ways in which exactly c colors may be seen, where 0 <= c <= r. The values are entries in a triangular table read from left to right along successive rows from the top, each row giving the value of r and each column giving the value of c. Including a row in the triangle for r = 0 and treating the values as a list a(n) starting with n = 1, n = r(r+1)/2 + c + 1.

For example, r = 2. If the widths of the cards from the top of the stack are 1,2 and the heights are 1,2 then two colors are seen; if the widths are 1,2 and the heights are 2,1 then two colors are seen; if 2,1 and 1,2 then two colors are seen; if 2,1 and 2,1 then only one color is seen. Thus the values for c = 1 and c = 2 are 1 and 3 respectively, i.e., a(5) = 1 and a(6) = 3.

The sum of row r in the table is (r!)^2 and T(r,1) for r > 0 is ((r-1)!)^2.

There are n! circular permutations of the integers from 0 to n. Only some have the property that a complete sequence of these integers can be found by choosing a start value then continuing to move left or right by the number of steps indicated, the direction of each move depending on whether the current value is odd or even. For example, of the six permutations for n = 3, only 0132 and 0213 generate a complete sequence if an odd value gives a leftward move that number of places and an even value gives a rightward move that number of places. If the direction rule is reversed, the two valid permutations are 0231 and 0312, the reverse of the previous two. Thus a(3) = 2.

Putting 0 in position 1 at the left and counting rightwards, the starting position for a complete sequence is 2 + floor(n/2) for the rule odd/left, even/right and floor((n+3)/2) for the opposite rule. As a further example, one of the a(9) = 288 valid permutations using the former rule for n = 9 is 0986423175. Starting at position 2 + floor(9/2), i.e., 6, the sequence 2, 1, 3, 6, 5, 4, 7, 9, 8, 0 is found. Clearly, all such sequences end with zero.

It is conjectured that the sequence continues indefinitely.

From David A. Corneth, Dec 05 2018: (Start)

For even n, if [d1, d2, ..., dn] is a valid permutation then so is [n + 1 - d1, n + 1 - d2, ..., n + 1 - dn] which is a different permutation.

More generally, for any valid permutation where n is even, di and n+1 - di can be interchanged for any value (where n + 1 - di != di) to give another valid permutation. Hence a(n) is divisibly by 2^k for n = 2*k.

a(n) > 0. For n = 1, [0, 1] is valid, for n = 2, [0, 1, 2] is valid; for n = 3, [0, 3, 1, 2] is valid etc. By taking this valid tuple from n by adding n + 1 to the right of the permutation or to the right of the 0, depending on the parity of n, one finds another valid tuple. Hence a(n) > 0. (End)

For the complementary case where stepping-on is always in the same direction, no permutation of 0 to n with n even can generate a complete sequence. For odd n, the number of complete sequences corresponds to A141599((n + 1)/2) for n up to 11, as limited by available computing power - it is conjectured that this correspondence continues indefinitely. - Ian Duff, Dec 25 2018

A090651 gives the 14-year repeating cycle applying between century years not a multiple of 400; this sequence extends that to the complete 400-year cycle.