Comments on A060572 from Gary W. Adamson, July-October 2021

From _Gary W. Adamson_, Jul 24 2021: (Start)

The disc configurations can be found by starting with the binary for n, say 19: 1, 0, 0, 1, 1; and writing 1, 2, 4, 9, 19 underneath (next term is twice current term if 0, otherwise twice plus 1). Take the difference row: 1, 1, 2, 5, 10, and label from the right, indicating the numbers of moves per disc through n = 19.

For odd-labeled discs, take mod 3 for the numbers of moves; and for the evens, take mod 3, then use (1->2, 2->1, 0->0). The results show the peg positions of labeled discs after the 19th move: (1, 2, 2, 1, 1) for (5, 4, 3, 2, 1). The final result: peg 1: 1/2/5; peg 2: 3/4; and no discs on peg 0. (End)

a(n) can be determined from the binary expansion by first getting the disc moved on the n-th move; say n=22, binary 1 0 1 1 0. The 22nd move is by disc: (1 + the numbers of trailing zeros) = 2. Then, for the peg number, follow the instructions in the Jul 24 2021 comment, getting 1, 2, 5, 11, 22; with the difference row 1, 1, 3, 6, 11, which we label 5, 4, 3, 2, 1. The labels are the disc numbers, with 1, 1, 3, 6, 11 representing the numbers of moves those discs moved through n=22. The corresponding term for disc 2 is 6. Then using the substitution rules shown in the Jul 24 2021 comment, 6 == 0 (mod 3), indicating that disc 2 moved to peg 0 on move 22. So a(22) = 0.  The final setup is peg 0: 2/3; peg 1: 5; and peg 2: 1/4.  This agrees with A055662(22) which has 1, 2, 0, 0, 2; meaning discs 1 and 4 are on peg 2; 2 and 3 are on peg 0; and disc 5 is on peg 1. - _Gary W. Adamson_, Aug 04 2021

From Gary W. Adamson, Oct 25 2021: (Start)

To get the disc configurations without peg positions, apply the standard optimal transfer rules as stated in the Kappraff and Adamson link, p.259. Algorithm 2 uses the example 10110, the binary expansion of 22. Record the number of bit runs starting from the right, (1121). Each term denotes the number of discs that that go on a peg, then place the next set on a different peg, etc. The result of using Algorithm 2 for 1121 is (5, 2/3, 1/4).

To get the peg positions for (5, 2/3, 1/4) apply the rules in the previous comment to the largest disc (5), and the next largest not on the same peg, (4). Disc 5 has moved once CW so after 22 moves is on peg 1. Disc 4 has moved once CCW so is on peg 2 along with disc 1. 2/3 are on peg 0. Check: A055662(22) is 12002. (End)