A130198 Single paradiddle. In percussion, the paradiddle is a four-note drum sticking pattern consisting of two alternating notes followed by two notes on the same hand.
0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1
Offset: 0
Links
- Wikipedia, Paradiddle
- Index entries for sequences related to music
- Index entries for linear recurrences with constant coefficients, signature (1,0,0,-1,1).
Crossrefs
Cf. A165211.
Programs
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Magma
[(1-(-1)^((n+5)*(n+6)*(n^2+11*n+32) div 8))/2 : n in [0..100]]; // Wesley Ivan Hurt, Aug 23 2015
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Maple
A130198:= n -> [0, 1, 0, 0, 1, 0, 1, 1][(n mod 8)+1]: seq(A130198(n), n=0..100); # Wesley Ivan Hurt, Aug 23 2015
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Mathematica
CoefficientList[Series[x*(1 - x + x^3)/((1 - x)*(1 + x^4)), {x, 0, 30}], x] (* Wesley Ivan Hurt, Aug 23 2015 *) -
PARI
a(n)=((n%8>3)+(n%4==1))%2 \\ Jaume Oliver Lafont, Mar 19 2009
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PARI
a(n)=210\2^(n%8)%2; \\ Jaume Oliver Lafont, Mar 24 2009
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PARI
apply( A130198(n)=bittest(210,n%8), [0..99]) \\ M. F. Hasler, May 24 2019
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Python
def A130198(n): return n&1^bool(n+1&4) # Chai Wah Wu, Aug 30 2024
Formula
From R. J. Mathar, Mar 27 2009: (Start)
a(n) = a(n-8) = a(n-1) - a(n-4) + a(n-5).
G.f.: -x*(1+x^3-x)/((x-1)*(1+x^4)). (End)
a(n) = (1-(-1)^((n+5)*(n+6)*(n^2+11*n+32)/8))/2. - Wesley Ivan Hurt, Aug 23 2015
a(n) = A165211(n+5). - Wesley Ivan Hurt, Aug 23 2015
Comments