A191275 Numbers that are congruent to {0, 1, 3, 5, 7, 9, 11} mod 12.
0, 1, 3, 5, 7, 9, 11, 12, 13, 15, 17, 19, 21, 23, 24, 25, 27, 29, 31, 33, 35, 36, 37, 39, 41, 43, 45, 47, 48, 49, 51, 53, 55, 57, 59, 60, 61, 63, 65, 67, 69, 71, 72, 73, 75, 77, 79, 81, 83, 84, 85, 87, 89, 91, 93, 95, 96, 97, 99, 101, 103, 105, 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, 9, 11]]; // Wesley Ivan Hurt, Jul 21 2016
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Maple
A191275:=n->12*floor(n/7)+[0, 1, 3, 5, 7, 9, 11][(n mod 7)+1]: seq(A191275(n), n=0..100); # Wesley Ivan Hurt, Jul 21 2016
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Mathematica
Select[Range[0,120], MemberQ[{0,1,3,5,7,9,11}, Mod[#,12]]&] (* or *) LinearRecurrence[{1,0,0,0,0,0,1,-1}, {0,1,3,5,7,9,11,12}, 70] (* Harvey P. Dale, Jul 06 2014 *)
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PARI
concat(0,Vec((1+x)^2*(1-x+x^2)*(1+x+x^2)/(1-x)^2/(1+x+x^2+x^3+x^4+x^5+x^6)+O(x^98))) \\ Charles R Greathouse IV, Mar 11 2012
Formula
a(n) = a(n-1) + a(n-7) - a(n-8) for n>8.
G.f.: x^2*(1+x)^2*(1-x+x^2)*(1+x+x^2)/((1-x)^2*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)). - Colin Barker, Mar 11 2012
From Wesley Ivan Hurt, Jul 21 2016: (Start)
a(n) = a(n-7) + 12 for n>7.
a(n) = (84*n - 84 - 2*(n mod 7) - 2*((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-3, 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