

A135303


a(n) = phi(2*n+1)/(2*b(n)), where phi() = A000010 and b() = A003558.


6



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
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OFFSET

1,8


COMMENTS

If n is odd, these are called "coach numbers" ("c"), and studied by J. Pedersen, Byron Walden, Victor QuintanarZilinskas and Linda Velarde of Santa Clara University. Coach Theorem: Let b > 1 be an odd number and let phi(b) be the Euler totient function. Let Sigma(b) be the complete symbol of b, let c be the number of coaches in Sigma(b), and let k = Sum_{i=1..r} k(i). Then phi(b) = 2 * c * k [Hilton & Pedersen, p. 262]. The complete symbols for b = 17 and 43 are shown in the examples.  Gary W. Adamson, Aug 15 2012
Conjecture relating to primes with more than one coach: The combined set of integers in the top rows of all coaches of these primes is composed of a permutation of the first q odd integers, where prime p = (4q1) or (4q+1), (q > 0). Example: As shown for 17, this prime has two coaches with the top rows [1} and [3, 7, 5]. 43 has three coaches with q = 11. The top rows are [1, 21, 11], [3, 5, 19], [7, 9, 17, 13, 15]. The comment of Sep 08 2012 in A216371 applies to primes with one coach, in which case "all coaches" is reduced to one and the set of q odd integers is in the top row of the coach.  Gary W. Adamson, Sep 10 2012
Conjecture [Carl Schick]: If 2*n+1 is prime, then these are the number of distinct cycles of f(k) = (2*n+1)  2*k beginning at an odd number 0 < k < 2*n.  Jonathan Skowera, Aug 03 2013
Comment from Gary W. Adamson, Oct 04 2019: (Start)
Conjecture of Aug 03 2013 proved by Jean Pederson. By way of example, take A003558(5) = 11, such that
2^5 == 1 (mod 11). Then Pederson on p. 98 has:
11  1 = 2^1 * 5 (pick "1", odd, the putative seed number)
11  5 = 2^1 * 3 (then subtract 3 in the next row)
11  3 = 2^3 * 1 (cycle ends). Then Pedersen constructs the "coach" (p. 98) for N= 11: [1, 5, 3]
..........................[1, 1, 3]. The top row represents the angles on the tape used to construct an 11gon at the operative crease lines beginning with Pi/11. (extract the (1,5,3) column). Then extract the exponents of 2: (1,1,3); which are the bottom row terms. The final result is that at successive creases on the tape are at angles of j*Pi/11, j = (1,5,3); alternatively at the top of the tape, then the bottom. The code U(1), D(1), U(3) is understood to be those numbers of bisections at each vertex. The total numbers of bisections = 5 = (1 + 1 + 3), shown to be the entry for N=11 in A003558. (End)


REFERENCES

Peter Hilton & Jean Pedersen, A Mathematical Tapestry: Demonstrating the Beautiful Unity of Mathematics; Cambridge University Press, 2010, pages 260264.
V. I. Levenshtein, Conflictavoiding codes and cyclic triple systems [in Russian], Problemy Peredachi Informatsii, 43 (No. 3, 2007), 3953.


LINKS

T. D. Noe, Table of n, a(n) for n = 1..10000


FORMULA

a(n) = "c", a Coach number; = A000010(n)/(2*A003558(n1)/2)); or phi(b) = 2 * c * k, with c = Coach numbers.


EXAMPLE

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 (n1)/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 (n1)/2) is present without duplication. In this method, k is likewise 7 but is represented by the numbers of terms in the top row. Pederson'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 Pederson's
number of coaches. Another example is the complete system of coaches shown on p. 261 of (Hilton and Pederson):
31: [1, 15}, [3, 7], [5, 13, 9, 11]
[1, 3], [2, 3], [1, 1, 1, 2]
The alternative system, called an rt 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)


MAPLE

A135303 := proc(n)
numtheory[phi](2*n+1)/2/A003558(n) ;
end proc:
seq(A135303(n), n=1..40) ; # R. J. Mathar, Dec 01 2014


CROSSREFS

Cf. A003558, A179480.
Cf. A216371 (odd primes with one coach).
Sequence in context: A292436 A184097 A205399 * A036065 A082907 A146532
Adjacent sequences: A135300 A135301 A135302 * A135304 A135305 A135306


KEYWORD

nonn,changed


AUTHOR

N. J. A. Sloane, Dec 05 2007


STATUS

approved



