A190719 Numbers that are congruent to {0, 1, 3, 5, 7, 8, 11} mod 12.
0, 1, 3, 5, 7, 8, 11, 12, 13, 15, 17, 19, 20, 23, 24, 25, 27, 29, 31, 32, 35, 36, 37, 39, 41, 43, 44, 47, 48, 49, 51, 53, 55, 56, 59, 60, 61, 63, 65, 67, 68, 71, 72, 73, 75, 77, 79, 80, 83, 84, 85, 87, 89, 91, 92, 95, 96, 97, 99, 101, 103, 104, 107, 108, 109, 111
Offset: 1
Links
- Wikipedia, Neapolitan scale
- Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,1,-1).
Crossrefs
Cf. A190785.
Programs
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Magma
[n : n in [0..150] | n mod 12 in [0, 1, 3, 5, 7, 8, 11]]; // Wesley Ivan Hurt, Jul 21 2016
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Maple
A190719:=n->12*floor(n/7)+[0, 1, 3, 5, 7, 8, 11][(n mod 7)+1]: seq(A190719(n), n=0..100); # Wesley Ivan Hurt, Jul 21 2016
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Mathematica
Select[Range[0,120], MemberQ[{0,1,3,5,7,8,11}, Mod[#,12]]&] (* Harvey P. Dale, Jun 10 2011 *)
Formula
a(n) = a(n-1) + a(n-7) - a(n-8) for n>8.
G.f.: x^2*(1+2*x+2*x^2+2*x^3+x^4+3*x^5+x^6) / ( (x^6+x^5+x^4+x^3+x^2+x+1)*(x-1)^2 ). - R. J. Mathar, Jun 11 2011
a(n) = floor(12*n/7) - floor((n mod 7)/6) - floor(((n+3) mod 7)/5). - Rolf Pleisch, Jun 12 2011
From Wesley Ivan Hurt, Jul 21 2016: (Start)
a(n) = a(n-7) + 12 for n>7.
a(n) = (84*n - 91 - 9*(n mod 7) + 5*((n+1) mod 7) - 2*((n+2) mod 7) - 2*((n+3) mod 7) - 2*((n+4) mod 7) + 5*((n+5) mod 7) + 5*((n+6) mod 7))/49.
a(7k) = 12k-1, a(7k-1) = 12k-4, a(7k-2) = 12k-5, a(7k-3) = 12k-7, a(7k-4) = 12k-9, a(7k-5) = 12k-11, a(7k-6) = 12k-12. (End)
Comments