A139767 Let e = 2n+1; consider the cyclotomic cosets C_i of 2 mod e; a(n) = maximal value of minimal number of copies of C_1 needed to add together to get any coset C_i.
1, 1, 2, 2, 1, 1, 3, 2, 1, 3, 2, 2, 2, 1, 4, 3, 3, 1, 2, 2, 2, 4, 2, 3, 3, 1, 2, 3, 1, 1, 5, 3, 1, 3, 2, 3, 4, 3, 2, 2, 1, 4, 2, 3, 3, 5, 2, 2, 3, 1, 2, 4, 1, 2, 2, 2, 3, 3, 3, 2, 3, 2, 6, 4, 1, 3, 4, 2, 1, 3, 2, 3, 3, 1, 3, 4, 5, 2, 2, 3, 1, 4, 2, 2, 3, 1, 3, 3, 1, 1, 2, 3, 2, 6, 2, 2, 4, 1, 2
Offset: 1
Keywords
Examples
Let n=7, e=15: the cyclotomic cosets of 2 mod 15 are C_0 := {0} C_1 := {1 2 4 8} C_3 := {3 6 12 9} C_5 := {5 10} C_7 := {-1 -2 -4 -8} == {14 13 11 7} and to get an element of C_7 we must add three elements of C_1; this is the worst case, so a(7) = 3.
References
- F. J. MacWilliams and N. J. A. Sloane, The Theory of Error-Correcting Codes, North-Holland/Elsevier, 1977; see p. 104 for definition of cyclotomic coset.
Links
- N. J. A. Sloane, Table of n, a(n) for n = 1..1000
- N. J. A. Sloane, Fortran program for this sequence
Comments