Hello math-fun ans seqfan [crossposted to rec.puzzles] This should look like an ordinary clock (without hands) : 12 11 1 10 2 9 3 8 4 7 5 6 Consider the 15 digits above and imagine them suddenly moving clockwise to another place on the clock: --> the "1" digits move 1 step clockwise; --> the "2" digits move 2 steps clockwise; --> the "3" digit moves 3 steps clockwise; etc. A « jump » sees the 14 digits moving simultaneously, accordingly to their nature (0, the 15th digit, never moves, of course). So after jump 1 we would have this configuration (digits on the same place assemble to form the smallest integer): 116 1 1 50 127 . . 4 28 . . 39 Jump 0 to jump 5 configurations are represented here: Jump0 1 2 3 4 5 6 7 8 9 10 11 12 Jump1 1 127 . 28 . 39 . 4 . 50 1 116 Jump2 11 1 159 2 . 26 . . 37 0 . 148 Jump3 1 11 1 147 . 2 . 258 . 0 . 369 Jump4 5 1 113 18 1 6 . 24 9 20 7 . Jump5 . . 1 11 1 13579 . . . 20 . 2468 ... Question: at what jump will the « jump 0 » configuration appear again? Best, É. ---- ObSeqFan : We could associate an integer to every jump : it's "horizontal" sum (81 for jump 0; 366 for jump 1; etc.) Find the sequence linking jump 0 to jump 0' (same config. again)