

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


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.


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


AUTHOR

N. J. A. Sloane, Dec 05 2007


STATUS

approved



