1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 1, 2, 3, 1, 1, 1, 2, 1, 1, 2, 1, 1, 3, 4, 1, 1, 1, 4, 1, 1, 1, 1, 1, 4, 1, 4, 3, 3, 1, 2, 2, 1, 1, 2, 1, 3, 1, 4, 1, 3, 2, 1, 2, 1, 9, 6, 1, 3, 1, 2, 1, 1, 1, 4, 1, 1, 5, 2, 3, 3, 1, 2, 1, 2, 1, 1, 6, 1, 1, 2
Offset: 1
Refer to A003558 for the J. Pedersen definition of a Coach. a(8) for b = 17 = 2 since 17 has two possible Coaches:
17: [1] and [3, 7, 5]
[4] [1, 1, 2];
where sum of the bottom row terms = k = 4 = A003558(8). For b = 43, a(21) = 3 since there are three possible coaches for 43:
43: [1, 21, 11] [3, 5, 19] [7, 9, 17, 13, 15]
[1, 1, 5], [3, 1, 3], [2, 1, 1, 1, 2],
where k = sum of terms in bottom rows of all possible coaches = 7 = A003558(21). For the coach with a "1" in the top row, the numbers of terms in the rows ("j" in A003558), = A179480(22) = 3. Note that the parity of numbers of terms in the bottom coach rows is the same.
From _Gary W. Adamson_, Aug 24 2019: (Start)
An alternative to the coach method of Pedersen and Hilton involves the doubling sequence, mod n; (43 in this case). The top row begins (1, 2, 4, 8, 16, ...) but the next number is 11, not 32. 32 == -11 (mod 43). We pick the least (in absolute value) of the two candidates (11 and 32): 11. The top row ends when the rightmost term is (n-1)/2 = 21. In subsequent rows the leftmost term is the least odd number not previously used, in this case 3. Continue with the doubling sequence and stop when the next row produces a term already used.
"20" ends row 2 since (2 * 20) = 40 == -3 (mod 43). 3 has been used so that row ends and our next row begins with the next unused odd term, a 7. That row ends with 18 since 2 * 18 = 36 == -7 (mod 43).
The entire set is complete when every term (1 through (n-1)/2) is present without duplication. In this method, k is likewise 7 but is represented by the numbers of terms in the top row. Pedersen's [1, 21, 11] appears as the only odd terms of the top row. [3, 5, 19] appears as the odd terms of the middle row, and [7, 9, 17, 13, 15] are the only odd terms of the bottom row. The three completed rows are:
[1, 2, 4, 8, 16, 11, 21;
3, 6, 12, 19, 5, 10, 20;
7, 14, 15, 13, 17, 9, 18]
It appears that the numbers of rows is equal to Pedersen's
number of coaches. Another example is the complete system of coaches shown on p. 261 of (Hilton and Pedersen):
31: [1, 15], [3, 7], [5, 13, 9, 11]
[1, 4], [2, 3], [1, 1, 1, 2]
The alternative system, called an r-t table in A065941, is
[1, 2, 4, 8, 15;
3, 6, 12, 7, 14;
5, 10, 11, 9, 13]
The odd terms of the top row (1, 15) appear in the leftmost coach. The odd terms (3, 7) appear in the middle coach, and (5, 11, 9, 13) are shown in the rightmost coach. (End)
Pedersen's coaches can be derived from the alternative system, doubling (mod N) since her coaches are simply another version: (repeated bisections (mod N)). First write out the doubling terms (mod N). Say N = 23: 1, 2, 4, 8, 7, 9, 5, 10, 3, 6, 11, representing the trajectory of terms 2*cos(j*Pi/23), using (x^2-2); j = 1, 2, 4, .... Begin with ("1"), then jump to the next odd term (11), then to each odd term in succession going left, getting: 23: (1, 11, 3, 5, 9, 7); the top row in Pedersen's coach. ....(1, 2, 2, 1, 1, 4) is the bottom row for 23 as shown on p. 105. Those terms are the numbers of term jumps in the previous operation; for example (1 to 11) = 1, (11 to 3) = 2, (3 to 5) = 2; and so on. Note that the number of terms in the doubling trajectory (11) matches the sum of terms in the bottom row of the coach, satisfying 2^11 == 1 (mod 23). - _Gary W. Adamson_, Oct 23 2019
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