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A006694
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Number of cyclotomic cosets of 2 mod 2n+1.
(Formerly M0192)
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42
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0, 1, 1, 2, 2, 1, 1, 4, 2, 1, 5, 2, 2, 3, 1, 6, 4, 5, 1, 4, 2, 3, 7, 2, 4, 7, 1, 4, 4, 1, 1, 12, 6, 1, 5, 2, 8, 7, 5, 2, 4, 1, 11, 4, 8, 9, 13, 4, 2, 7, 1, 2, 14, 1, 3, 4, 4, 5, 11, 8, 2, 7, 3, 18, 10, 1, 9, 10, 2, 1, 5, 4, 6, 9, 1, 10, 12, 13, 3, 4, 8, 1, 13, 2, 2, 11, 1, 8, 4, 1, 1, 4, 6, 7, 19, 2, 2, 19, 1, 2
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OFFSET
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0,4
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COMMENTS
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a(0) = 0 by convention.
The number of cycles in permutations constructed from siteswap juggling patterns 1, 123, 12345, 1234567, etc., i.e., the number of ball orbits in such patterns minus one.
Also the number of irreducible polynomial factors of the polynomial (x^(2n+1) - 1) / (x - 1) over GF(2). - V. Raman, Oct 04 2012
Also, a(n) is the number of cycles of the Josephus permutation for n elements and a count of 2. For n >= 1, the Josephus permutation is given by the n-th row of A321298. See Knuth 1997 (exercise 1.3.3-29). - Pontus von Brömssen, Sep 18 2022
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REFERENCES
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Donald E. Knuth, The Art of Computer Programming, Vol. 1, 3rd edition, Addison-Wesley, 1997.
F. J. MacWilliams and N. J. A. Sloane, The Theory of Error-Correcting Codes, Elsevier/North Holland, 1977, pp. 104-105.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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Michael Assis, Folding Pi, arXiv:2403.09277 [math.NT], 2024.
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FORMULA
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Conjecture: a((3^n-1)/2) = n. - Vladimir Shevelev, May 26 2008 [This is correct. 2*((3^n-1)/2) + 1 = 3^n and the polynomial (x^(3^n) - 1) / (x - 1) factors over GF(2) into Product_{k=0..n-1} x^(2*3^k) + x^(3^k) + 1. - Joerg Arndt, Apr 01 2019]
1) a(n)=A037226(n) iff 2n+1 is prime;
2) The only case when a(n) < A037226(n) is n=0;
3) If {C_i}, i=1..a(n), is the set of all cyclotomic cosets of 2 mod (2n+1), then lcm(|C_1|, ..., |C_{a(n)}|) = A002326(n). (End)
a(n) = (Sum_{d|(2n+1)} phi(d)/ord(2,d)) - 1, where phi = A000010 and ord(2,d) is the multiplicative order of 2 modulo d. - Jianing Song, Nov 13 2021
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EXAMPLE
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Mod 15 there are 4 cosets: {1, 2, 4, 8}, {3, 6, 12, 9}, {5, 10}, {7, 14, 13, 11}, so a(7) = 4. Mod 13 there is only one coset: {1, 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7}, so a(6) = 1.
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MAPLE
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with(group); with(numtheory); gen_rss_perm := proc(n) local a, i; a := []; for i from 1 to n do a := [op(a), ((2*i) mod (n+1))]; od; RETURN(a); end; count_of_disjcyc_seq := [seq(nops(convert(gen_rss_perm(2*j), 'disjcyc')), j=0..)];
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MATHEMATICA
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Needs["Combinatorica`"]; f[n_] := Length[ToCycles[Mod[2Range[2n], 2n + 1]]]; Table[f[n], {n, 0, 100}] (* Ray Chandler, Apr 25 2008 *)
f[n_] := Length[FactorList[x^(2n + 1) - 1, Modulus -> 2]] - 2; Table[f[n], {n, 0, 100}] (* Ray Chandler, Apr 25 2008 *)
a[n_] := Sum[ EulerPhi[d] / MultiplicativeOrder[2, d], {d, Divisors[2n + 1]}] - 1; Table[a[n], {n, 0, 99}] (* Jean-François Alcover, Dec 14 2011, after Joerg Arndt *)
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PROG
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(PARI) a(n)=sumdiv(2*n+1, d, eulerphi(d)/znorder(Mod(2, d))) - 1; /* cf. A081844 */
(Python)
from sympy import totient, n_order, divisors
def A006694(n): return sum(totient(d)//n_order(2, d) for d in divisors((n+1<<1)-1, generator=True) if d>1) # Chai Wah Wu, Apr 09 2024
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CROSSREFS
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A001917 gives cycle counts of such permutations constructed only for odd primes.
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KEYWORD
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nonn,nice,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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