A260160 a(n) = a(n-2) + a(n-6) - a(n-8) with n>8, the first eight terms are 0 except that for a(5) = a(7) = 1.
0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 2, 0, 2, 0, 2, 0, 3, 0, 3, 0, 3, 0, 4, 0, 4, 0, 4, 0, 5, 0, 5, 0, 5, 0, 6, 0, 6, 0, 6, 0, 7, 0, 7, 0, 7, 0, 8, 0, 8, 0, 8, 0, 9, 0, 9, 0, 9, 0, 10, 0, 10, 0, 10, 0, 11, 0, 11, 0, 11, 0, 12, 0, 12, 0, 12, 0, 13, 0, 13, 0, 13, 0, 14, 0, 14, 0, 14
Offset: 1
Links
- Index entries for linear recurrences with constant coefficients, signature (0,1,0,0,0,1,0,-1).
Programs
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Maple
with(numtheory): P:= proc(q) local n; for n from 0 to q do print((1-(-1)^n)*floor(n/6+1/3)/2); od; end: P(100); # Paolo P. Lava, Nov 12 2015
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Mathematica
LinearRecurrence[{0, 1, 0, 0, 0, 1, 0, -1}, {0, 0, 0, 0, 1, 0, 1, 0}, 100] Table[(1 - (-1)^n) (Floor[n/6 + 1/3]/2), {n, 1, 90}] (* Bruno Berselli, Nov 10 2015 *)
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PARI
concat(vector(4), Vec(x^5/(1-x^2-x^6+x^8) + O(x^100))) \\ Altug Alkan, Nov 10 2015
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Sage
[(1-(-1)^n)*floor(n/6+1/3)/2 for n in (1..90)] # Bruno Berselli, Nov 10 2015
Formula
G.f.: x^5/(1-x^2-x^6+x^8).
a(n) = A264041(n) - n*(n+1)/2, 026).
a(n) = (1-(-1)^n)*floor(n/6+1/3)/2. [Bruno Berselli, Nov 10 2015]
Comments